Real Analysis, Fourth Edition
Solutions to Exercises
David Malison
Last updated: September 28, 2020
1
The Real Numbers: Sets, Sequences, and Functions
1.1
The Field, Positivity, and Completeness Axioms
1. For any a 6= 0 and b 6= 0, we have
(ab)(a−1 b−1 ) = (ba)(a−1 b−1 )
= b(aa−1 )b−1
= bb−1
=1
Since multiplicative inverses are unique, we conclude that (ab)−1 = a−1 b−1 .
2.
(i) I first prove some elementary results. Observe that if a + c = b + c, then
a=a+0
= a + (c − c)
= (a + c) − c
= (b + c) − c
= b + (c − c)
=b+0
=b
so that a = b. Since 0 + 0a = 0a = (0 + 0)a = 0a + 0a, the above result implies 0 = 0a = a0 for
all a. But this means
ab + (−a)b = (a − a)b
= 0b
=0
1
and
ab + a(−b) = a(b − b)
= a0
=0
so that (−a)b = −ab = a(−b). Also, observe that
−a + a = a − a = 0
so that a = −(−a). Together, the above results imply
(−a)(−b) = −(a(−b))
= −(−ab)
= ab
In particular (−a)2 = a2 for all a.
Now if a is positive, a2 = a · a > 0. If a is not positive and a 6= 0, then −a is positive. But this
implies (−a)2 = a2 is positive.
(ii) Suppose a is positive but a−1 is not positive. Since aa−1 = 1, a−1 cannot be zero. Therefore
−a−1 must be positive, which implies (−a−1 )a = −1 is positive. But −1 cannot be positive since
1 is positive by part (i).
(iii) If a − b > 0 and c > 0, then (a − b)c > 0. This implies ac − bc > 0, so that ac > bc. If c < 0, then
(−c)(a − b) > 0. But this implies (−c)a + (−b)(−c) = bc − ac > 0, so bc > ac.
3. Suppose E = {x} for some x ∈ R. Since x ≤ x, x is an upper bound of E. If y < x, then y is not
an upper bound of E. Therefore x is the least upper bound. A similar argument can be used to show
that x is also the greatest lower bound.
To prove the converse, first observe that a ≤ b and b ≤ a if and only if a = b. Now suppose inf E = sup E
and let x be a point in E. Then sup E = inf E ≤ x ≤ sup E, which implies inf E = sup E = x.
4.
(i) Suppose a 6= 0 and b 6= 0. Then 1 = a a1 b 1b = ab a1 1b . If ab = 0, this expression would imply 1 = 0.
But this contradicts the non-triviality assumption.
(ii) Observe that
(a − b)(a + b) = a(a + b) − b(a + b)
= a2 + ab − ba − b2
= a 2 − b2
Thus if a2 − b2 = 0, either a − b = 0 or a + b = 0 by part (i).
(iii) Let x =
c
1+c .
Then x < c and 0 < x < 1, so x2 < c. But this means x ∈ E, so E is non-empty.
Now suppose x > c + 1. Then x2 > c2 so x ∈
/ E. Thus c + 1 is an upper bound of E.
2
The completeness axiom therefore implies that there exists a least upper bound x0 > 0 of E.
Suppose x20 < c. Choose h such that 0 < h < 1 and
h<
c − x20
2(x0 + 1)
Then
(x0 + h)2 − x20 = 2x0 h + h2 < 2h(x0 + h) < 2h(x0 + 1) < c − x20
This implies (x0 + h)2 < c so that x0 + h ∈ E. But x0 + h > x0 , contradicting the assumption
that x0 is an upper bound of E.
Now suppose x20 > c. Pick k =
x20 −c
2x0 .
Then 0 < k < x0 . If x ≥ x0 − k, then
x20 − x2 ≤ x20 − (x0 − k)2 = 2x0 k − k 2 < 2x0 k = x20 − c
This implies c < x2 so x ∈
/ E. But this means x0 − k is an upper bound of E smaller than x0 ,
contradicting the assumption that x0 is the least upper bound.
The above results imply x20 = c. Now suppose x2 = c for some x > 0. Then x2 − x20 = 0, so either
x − x0 = 0 or x + x0 = 0 by part (ii). Since x and x0 are positive, we must have x − x0 = 0. This
implies x = x0 .
5.
(i) By using the field axioms and the fact that 1 + 1 = 2, we can complete the square to obtain
2
2
b
1 b
−
+c
2
a 2
2
2
bx bx 1 b
1 b
= ax2 +
+
+
−
+c
2
2
a 2
a 2
2
b
b
1 b
b
+
x+
−
+c
= ax x +
2a
2
2a
a 2
2
b
b
1 b
= ax +
x+
−
+c
2
2a
a 2
2
2
b
1 b
=a x+
−
+c
2a
a 2
1
ax + bx + c = ax + bx +
a
2
2
Setting the right-hand equal to zero and rearranging, we obtain
⇐⇒
⇐⇒
ax2 + bx + c = 0
2 2
b
c
b
=
−
x+
2a
2a
a
2
(2ax + b) = b2 − 4ac
(1)
(2)
If b2 − 4ac > 0, there exists a unique positive square root of the right-hand side by Problem 4(iii).
3
This means
2
(2ax + b) =
p
2
b2 − 4ac
But then either
2ax + b +
p
b2 − 4ac = 0
2ax + b −
p
b2 − 4ac = 0
or
by Problem 4(ii). Solving these two equations, we obtain
x=
−b −
or
x=
−b +
√
b2 − 4ac
2a
√
b2 − 4ac
2a
(ii) If b2 − 4ac < 0, then there are no real solutions to equation (2) because z 2 ≥ 0 for all z ∈ R
(see Problem 2(i)). Since the solutions to (2) are the same as the solutions to (1), the quadratic
equation has no real solutions.
6. Let E denote a non-empty set that is bounded below by b. Define the set −E = {−x | x ∈ E}.
Since x ≥ b for all x ∈ E, −x ≤ −b for all x ∈ E. Therefore −b is an upper bound of −E and the
completeness axiom implies there exists a least upper bound x0 of −E.
Now pick any x ∈ E. Then −x ∈ −E, so that −x ≤ x0
bound of E. Now suppose there exists another lower bound
for any x ∈ E, x ≥
x00
=⇒
−x ≤
−x00 .
=⇒
x00
x ≥ −x0 . Thus −x0 is a lower
of E which satisfies x00 > −x0 . Then
This means −x00 is an upper bound of −E. But
x00 > −x0 =⇒ −x00 < x0 , contradicting the assertion that x0 is the least upper bound of E.
We conclude that −x0 is the greatest lower bound of E, which by construction means
inf E = −x0 = − sup(−E) = − sup{−x | x ∈ E}
7.
(i) First, observe that |0| = 0 and |a| > 0 if a 6= 0.
Now if a ≥ 0 then |a| ≥ 0 and |a|2 = a2 . If −a > 0, then |a|2 = (−a)2 = a2 . Therefore |a|2 = a2
for all a. This implies
|ab|2 = (ab)2 = a2 b2 = |a|2 |b|2 = (|a||b|)2
If either a or b is zero, then |ab| = |0| = 0 = |a||b|. If both a and b are non-zero, then |a||b| and
|ab| are positive. Since positive square roots are unique, the above expression implies |ab| = |a||b|.
4
(ii) If either a or b is zero, the result is trivial. Assume both a and b are nonzero. Observe that
|a + b|2 = (a + b)2
= a2 + 2ab + b2
≤ |a|2 + 2|ab| + |b|2
= |a|2 + 2|a||b| + |b|2
= (|a| + |b|)2
where the inequality follows from the fact that x ≤ |x| for all real x. If |a + b| > |a| + |b|, then
|a + b|2 > (|a| + |b|)2 . But this would contradict the above equation. We therefore must have
|a + b| ≤ |a| + |b|.
(iii) Suppose |x − a| < . Since x − a ≤ |x − a| and −(x − a) ≤ |x − a|, we have
x−a<
=⇒
x<+a
−x + a < =⇒
a−<x
Conversely, suppose a − < x and x < a + . If x − a ≥ 0, then |x − a| = x − a < . If x − a < 0,
then |x − a| = −(x − a) = a − x < .
1.2
The Natural and Rational Numbers
8. S(n): There does not exist a natural number between n and n + 1.
Suppose there exists x ∈ N satisfying 1 < x < 2. Consider the set
E = {1} ∪ {k ∈ N | k ≥ 2}
If k = 1, then k + 1 = 2 ∈ N so that k + 1 ∈ E. If k ∈ N and k ≥ 2, then k + 1 ≥ 2 and k + 1 ∈ N
so that k + 1 ∈ E. This means E is inductive, which implies N ⊆ E. But x ∈
/ E, a contradiction.
Therefore (1, 2) ∩ N = ∅, so S(1) is true.
Now suppose S(k) is true for some k ∈ N and suppose there exists x ∈ N satisfying k + 1 < x < k + 2.
Since S(k) is true and k < x − 1 < k + 1, x − 1 ∈
/ N. Define E = N ∼ {x}. Since x > 1 and 1 ∈ N,
1 ∈ E. If n ∈ E, n ∈ N and therefore n 6= x − 1. But this means n + 1 ∈ N ∼ {x} = E, so that E is
an inductive set. This implies N ⊆ E. But x ∈
/ E, a contradiction. Therefore S(k + 1) is true.
9. S(n): If n > 1 is a natural number, then n − 1 is a natural number.
S(1) is true because 1 > 1 is a false statement.
Suppose S(k) is true for some k ∈ N. Since (k + 1) − 1 = k ∈ N, S(k + 1) is true.
S(n): If m and n are natural numbers with m > n, then m − n is a natural number.
S(1) was proven above.
5
Suppose S(k) is true for some k ∈ N. Let m be a natural number such that m > k + 1. Then m − 1 is
a natural number by S(1). Since m − 1 > k, (m − 1) − k = m − (k + 1) is a natural number by S(k).
Thus S(k + 1) is true.
10. The only integer contained in the interval [0, 1) is 0 because the positive integers bounded below by 1.
Thus the statement is true if n = 0.
Suppose n is a natural number. By Problem 8, we know that (n, n + 1) does not contain any natural
number. Therefore n is the only integer contained in [n, n + 1). Now suppose z is an integer in
[−n, −n + 1). Then −z is a natural number in (n − 1, n], so we must have −z = n by Problem 8. But
this means −n is the only integer in [−n, −n + 1).
11. Let E be a non-empty set of integers that is bounded above. By the completeness axiom, there exists
a least upper bound c of E. Then there must exist m ∈ E such that m + 1 > c. Otherwise m + 1 ≤ c
for all m ∈ E, which would imply c − 1 is an upper bound of E that is smaller than c.
I claim that m ≥ n for all n ∈ E. If not, there exists n ∈ E such that
m<n≤c<m+1
But this would imply n is an integer in (m, m + 1), contradicting the result from Problem 10.
12. Let a and b be two real numbers with a < b. Suppose a ≥ 0. By Theorem 2, there exists a rational
number r such that
a
b
√ <r< √
2
2
The previous expression implies
√
a<r 2<b
√
The number i = r 2 is an irrational number lying between a and b. For if i were rational, i could be
written as
m
n
for some pair of natural numbers m and n. There also exist natural numbers p and q
√
√
such that r = pq . But this would imply 2 = mq
, which would mean that 2 is rational. However it
np
√
was shown in the text that 2 is irrational, a contradiction.
Now suppose a < 0. By the Archimedean property, there exists a natural number n such that n > −a.
We can infer from the previous case that there exists an irrational number i satisfying n+a < i < n+b.
The number i0 = i−n is an irrational number lying between a and b. For if i0 were rational, there would
exist a pair of natural numbers r and s satisfying i0 = rs . But this would mean i =
ns+r
s ,
contradicting
the assumption that i is irrational.
13. Let a be a real number and define the set
E = {x | x is rational and x ≤ a}
By construction, a is an upper bound of E. Now suppose there exists another upper bound a0 of E
satisfying a0 < a. By Theorem 2, there exists a rational number x satisfying a0 < x < a. But then
x ∈ E, contradicting the assumption that a0 is an upper bound of E. We can therefore conclude that
a is the least upper bound of E.
6
Now suppose we define the set
E 0 = {x | x is irrational and x ≤ a}
Using the result from Problem 12, we can apply the same argument to conclude that a is the least
upper bound of E 0 .
14. S(n): If r > 0, (1 + r)n ≥ 1 + n · r
S(1) is true because 1 + r ≥ 1 + r.
Suppose S(k) is true for some k ∈ N and fix r > 0. Then
(1 + r)k+1 = (1 + r)(1 + r)k
≥ (1 + r)(1 + k · r)
= 1 + (k + 1)r + k · r2
≥ 1 + (k + 1)r
where the first inequality follows from the induction hypothesis and the second inequality follows
because k and r2 are positive. Therefore S(k + 1) is true.
15.
(i) S(n):
n
X
j2 =
j=1
n(n + 1)(2n + 1)
6
S(1) is true because
12 =
1(1 + 1)(2(1) + 1)
6
Suppose S(k) is true for some k ∈ N. Then
k+1
X
j2 =
k
X
j 2 + (k + 1)2
j=1
j=1
k(k + 1)(2k + 1)
+ (k + 1)2
6
k(2k + 1) + 6(k + 1)
= (k + 1)
6
2
2k + 7k + 6
= (k + 1)
6
(k + 1)(k + 2)(2k + 3)
=
6
(k + 1)((k + 1) + 1)(2(k + 1) + 1)
=
6
=
Thus S(k + 1) is true.
(ii) As a preliminary step, I first use induction to prove that
7
Pn
j=1
j=
n(n+1)
2
for all n ∈ N.
S(n) :
n
X
j=
j=1
n(n + 1)
2
S(1) is true because
1=
1(1 + 1)
2
Suppose S(k) is true for some k ∈ N. Then
k+1
X
j=
j=1
k
X
j + (k + 1)
j=1
k(k + 1)
+ (k + 1)
2 k
= (k + 1)
+1
2
(k + 1)(k + 2)
=
2
(k + 1)((k + 1) + 1)
=
2
=
Thus S(k + 1) is true.
S(n) :
2
13 + 23 + · · · + n3 = (1 + 2 + · · · + n)
S(1) is true because
13 = 12
Suppose S(k) is true for some k ∈ N. Then
2
13 + 23 + · · · + k 3 + (k + 1)3 = (1 + 2 + · · · + k) + (k + 1)3
2
= (1 + 2 + · · · + k + (k + 1) − (k + 1)) + (k + 1)3
= (1 + 2 + · · · + k + (k + 1))2 − 2(1 + 2 + · · · + k + (k + 1))(k + 1)
+ (k + 1)2 + (k + 1)3
2
= (1 + 2 + · · · + k + (k + 1)) − 2
(k + 1)(k + 2)
2
(k + 1)
+ (k + 1)2 + (k + 1)3
= (1 + 2 + · · · + k + (k + 1))2 − (k + 1)2 ((k + 2) − 1)
+ (k + 1)3
= (1 + 2 + · · · + k + (k + 1))2
where the fourth equality follows from the preliminary result. Thus S(k + 1) is true.
8
(iii) S(n): If r 6= 1, then
1 + r + · · · + rn =
1 − rn+1
1−r
S(1) is true because
1 − r2
1−r
(1 − r)(1 + r)
=
1−r
1+r =
=1+r
Suppose S(k) is true for some k ∈ N. Then
1 − rk+1
+ rk+1
1−r
1 − rk+1 + rk+1 − rk+2
=
1−r
k+2
1−r
=
1−r
1 + r + · · · + rk + rk+1 =
Thus S(k + 1) is true.
1.3
Countable and Uncountable Sets
16. Consider the mapping from N to Z defined by



0




n
f (n) =
2





− n + 1
2
if n = 1
if n is even
if n is odd and n > 1
If n is a natural number, then f (2n) = n and f (2n − 1) = −n. We also have f (1) = 0. Therefore f is
onto.
Now suppose f (n) = f (n0 ). If f (n) equals 0, then n = n0 = 1. If f (n) is positive, then
n
n0
=
2
2
=⇒
n+1
n0 + 1
=−
=⇒ n = n0 . Therefore f is one-to-one.
n = n0 . If f (n) is negative, then −
2
2
17. Let A be a non-empty set. Suppose A is countable. Then there exists a one-to-one correspondence f
between a subset of N and A. The inverse of f defines a one-to-one mapping of A to N.
Conversely, suppose there exists a one-to-one mapping f of A to N. Then f provides a one-to-one
correspondence between A and f (A), so the inverse of f provides a one-to-one mapping of f (A) onto
A. Since f (A) is countable by Theorem 3, we can conclude that A is countable by Theorem 5.
18. As a preliminary result, I first show that every finite set of numbers contains a maximal element.
9
S(n): Let S ⊂ R be a non-empty set. If there exists a one-to-one correspondence between {1, · · · , n}
and S, then S contains a maximal element.
Suppose there exists a one-to-one correspondence f between {1} and S. Then S = {f (1)}, so s ≤ f (1)
for all s ∈ S. Thus S(1) is true.
Now assume S(k) is true and suppose there exists a one-to-one correspondence between {1, · · · , k + 1}
and S. Then S = {f (i) | 1 ≤ i ≤ k} ∪ {f (k + 1)}. By the induction hypothesis, {f (i) | 1 ≤ i ≤ k} has a
maximal element ŝ. If ŝ ≥ f (k + 1), then ŝ is a maximal element of S. If ŝ < f (k + 1), then f (k + 1)
is a maximal element of S. We conclude that S(k + 1) must be true.
n times
z
}|
{
S(n): The Cartesian product N × · · · × N is countably infinite.
The identity function establishes a one-to-one correspondence between N and N, so N is countable.
Now suppose N were finite. Then by the preliminary result, there would exist a maximal element m
of N. But m + 1 would then be a natural number larger than m, a contradiction. We conclude that
N is countably infinite, so S(1) is true.
k times
}|
{
z
Suppose S(k) is true. Then there exists a one-to-one mapping f of N onto N × · · · × N. Consider the
k + 1 times
}|
{
z
mapping from N × · · · × N to N defined by
g(n1 , · · · , nk , nk+1 ) = (f −1 (n1 , · · · , nk ) + nk+1 )2 + nk+1
k + 1 times
z
}|
{
It is straight-forward to check that g is one-to-one using the argument in the text. Thus N × · · · × N
k + 1 times
}|
{
z
is equipotent to g(N × · · · × N), a subset of the countable set N. We infer from Theorem 3 that
k + 1 times
}|
{
z
N × · · · × N is countable.
k + 1 times
z
}|
{
Now suppose N × · · · × N is finite. Then there exists a one-to-one mapping f from {1, · · · , n} onto
k + 1 times
k times
}|
{
}|
{
z
z
N × · · · × N for some n ∈ N. Consider the mapping from N × · · · × N to {1, · · · , n} defined by
g(n1 , · · · , nk ) = f −1 (n1 , · · · , nk , 1)
k times
z
}|
{
This establishes a one-to-one correspondence between N × · · · × N and a subset of {1, · · · , n}, implying
k times
k times
z
}|
{
z
}|
{
that N × · · · × N is finite. This contradicts the assumption that N × · · · × N is countably infinite. We
k + 1 times
z
}|
{
conclude that N × · · · × N is countably infinite, so S(k + 1) is true.
19. Let Λ be a non-empty finite set and for each λ ∈ Λ, let Eλ be a countable set. Then for some m ∈ N,
there exists a one-to-one mapping h of {1, · · · , m} onto Λ. For n ∈ {1, · · · , m}, define λn = h(n) and
10
keep E, N (n) and fn defined as in the text. Define
E 0 = {(n, k) ∈ N × N | 1 ≤ n ≤ m, Eλn is nonempty and 1 ≤ k ≤ N (n) if Eλn is also finite}
and define the mapping f of E 0 onto E by f (n, k) = fn (k). The proof then proceeds as in the text.
20. Suppose g(f (a)) = g(f (a0 )). Since g is one-to-one, we must have f (a) = f (a0 ). Since f is one-to-one,
we must also have a = a0 . But this means g ◦ f is one-to-one. Now fix c ∈ C. Since g is onto, there
exists b ∈ B such that g(b) = c. Since f is onto, there also exists a ∈ A such that f (a) = b. But this
means g(f (a)) = c, so g ◦ f is onto.
Suppose f −1 (b) = f −1 (b0 ). Then b = f (f −1 (b)) = f (f −1 (b0 )) = b0 , so f −1 must be one-to-one. Now
suppose a ∈ A. Then a = f −1 (f (a)), so f −1 is onto.
21. S(n): There does not exist a one-to-one mapping from {1, · · · , m + n} to {1, · · · , n} for any natural
number m.
Fix m ∈ N and let f be a mapping from {1, · · · , m + 1} to {1}. Then f (1) = f (2) = 1, so f cannot be
one-to-one. Thus S(1) is true.
Suppose S(k) is true. Fix m ∈ N and let f be a mapping from {1, · · · , m + (k + 1)} to {1, · · · , k + 1}.
If there does not exist n ∈ {1, · · · , m + (k + 1)} such that f (n) = k + 1, then f defines a mapping from
{1, · · · , (m + 1) + k} to {1, · · · , k}. The induction hypothesis then implies f is not one-to-one. Now
suppose f (n) = k + 1 for some n ∈ {1, · · · , m + (k + 1)}. Define the mapping
g(i) =

f (i)
if 1 ≤ i < n
f (i + 1)
if n ≤ i ≤ m + k
If f is one-to-one, then g defines a one-to-one mapping from {1, · · · , m + k} to {1, · · · , k}. But this
contradicts the induction hypothesis, so f cannot be one-to-one. Therefore S(k + 1) must be true.
22. Suppose 2N is countable. Let {Xn | n ∈ N} denote an enumeration of 2N and define
D = {n ∈ N | n is not in Xn }
Then D ∈ 2N , so D = Xd for some d ∈ N. If d is not in D, then we would have a contradiction
because d would have to be in D by construction. Likewise if d is in D, then we have a contradiction
because d could not be in D by construction. We can conclude that no enumeration can exist, so 2N
is uncountable.
23. Define Λ, m, λn , N (n) and fn as in Problem 19 and define E = Eλ1 × · · · × Eλm . If Eλn is empty for
any n ∈ {1, · · · , m}, then E is empty and therefore countable. So suppose that Eλn is non-empty for
all n ∈ {1, · · · , m}. Define
m times
}|
{
z
E 0 = {(k1 , · · · , km ) ∈ N × · · · × N | 1 ≤ kn ≤ N (n) if Eλn is finite}
Define the mapping f of E 0 onto E by f (k1 , · · · , km ) = (f1 (k1 ), · · · , fm (km )). E 0 is a subset of a
countable set by Corollary 4, so E 0 is countable by Theorem 3. Theorem 5 then implies that E is also
11
countable.
Define the set {1, 2}N as the set of functions from N to {1, 2}. Consider the mapping f from 2N to
{1, 2}N defined by
[f (S)](n) =

1
if n ∈ S
2
otherwise
Now suppose f (S) = f (S 0 ). Since n ∈ S ⇐⇒ [f (S)](n) = 1 ⇐⇒ [f (S)0 ](n) = 1 ⇐⇒ n ∈ S 0 ,
we must have S = S 0 . Thus f is one-to-one. Now suppose g is a mapping from N to {1, 2}. Define
S = g −1 ({1}). Then f (S) = g, so f is onto. Thus |{1, 2}N | = |2N | and therefore {1, 2}N is uncountable
by Problem 22. Since {1, 2}N is a subset of NN , NN must also be uncountable.
24. Let I be a non-degenerate interval of real numbers. In Problem 25, we show that |I| = |R|. But R
is not finite. For if it was, it would contain a maximal element by Problem 18. However, no maximal
element of R exists.
25. I proceed by defining a series of one-to-one correspondences between various intervals. I do not explicitly
show that each mapping is one-to-one and onto, although it is easy to do so.
• |(a, b)| = |(0, 1)|, where a < b: Define f : (a, b) → (0, 1) as
f (x) =
1
(x − a)
b−a
• |(0, 1)| = |(1, ∞)|: Define f : (0, 1) → (1, ∞) as
f (x) =
1
x
• |(1, ∞)| = |(c, ∞)|: Define f : (1, ∞) → (c, ∞) as
f (x) = x + c − 1
• |(−∞, c)| = |(c, ∞)|: Define f : (−∞, c) → (c, ∞) as
f (x) = 2c − x
• |(−∞, ∞)| = |(−1, 1)|: Define f : (−∞, ∞) → (−1, 1) as
 x


1+x
f (x) =

 x
1−x
if x ≥ 0
if x < 0
• The preceding mappings establish the equipotence of all non-degenerate open intervals. We can
12
also use these mappings to prove the following relations:
|[a, b]| = |[0, 1]|
|[a, b)| = |[0, 1)|
|(a, b]| = |(0, 1]|
|(0, 1]| = |[c, ∞)|
|(−∞, c]| = |[c, ∞)|
• |[−1, 1]| → |(−∞, ∞)|: Define f : [−1, 1] → (−∞, ∞) as
f (x) =


1







1





1−x




x



1 − x
if x = 1
x
∈N
1−x
x
if x ≥ 0 and
∈
/N
1−x
if x ≥ 0 and


−1






1



−


1
+
x




x



1 + x
if x = −1
x
∈N
1+x
x
if x < 0 and −
∈
/N
1+x
if x < 0 and −
In essence, this function makes room for the interval’s endpoints by displacing Z ∼ {0}.
• |(−1, 1]| = |(−∞, ∞)|: Define f : (−1, 1] → (−∞, ∞) as


1







1



1 − x
f (x) =
x




1−x




x



1 + x
if x = 1
x
∈N
1−x
x
if x ≥ 0 and
∈
/N
1−x
if x ≥ 0 and
if x < 0
• |[−1, 1)| = |(−∞, ∞)|: Define f : [−1, 1) → (−∞, ∞)
f (x) =
 x




1−x







−1
if x ≥ 0
if x = −1
1


−


1
+
x




 x


1 + x
13
x
∈N
1+x
x
if x < 0 and −
∈
/N
1+x
if x < 0 and −
26. Let G denote the set of irrational numbers in (0, 1) and let {qn | n ∈ N} denote an enumeration of the
rationals in (0, 1). Define
√
in =
2
2n
and construct the mapping f : (0, 1) → G as



i

 2n
f (x) = i2n−1



x
if x = qn
if x = in
otherwise
f defines a one-to-one correspondence between (0, 1) and G, so |(0, 1)| = |G|.
In Problem 25 we showed that |R| = |(0, 1)|, so the above result implies |R| = |G|. This means we can
find a one-to-one mapping g from R onto G. Now consider the mapping h : R × R → G × G defined
by
h(x, y) = (g(x), g(y))
h defines a one-to-one mapping from R × R onto G × G, so |R × R| = |G × G|.
Recall that if x is an irrational number in (0, 1), it can be uniquely written as
1
x=
1
a1 +
a2 +
=: [a1 , a2 , a3 , · · · ]
1
a3 + · · ·
where a1 , a2 , a3 , · · · is an infinite sequence of natural numbers. (This representation is called the
continued fraction expansion of x.) Let x = [a1 , a2 , · · · ] and y = [b1 , b2 , · · · ] denote two elements of G
and consider the mapping m : G × G → G defined by
m(x, y) = [a1 , b1 , a2 , b2 , · · · ]
Then m defines a one-to-one correspondence between G × G and G, so |G × G| = |G|. Combining the
above results, we have |R × R| = |G × G| = |G| = |R|.
1.4
Open Sets, Closed Sets, and Borel Sets of Real Numbers
27. The set of rational numbers is not open because every interval around a rational number contains
an irrational number (see Problem 12). The set of irrational numbers is also not open because every
interval around an irrational number contains a rational number (see Theorem 2). By Proposition 11,
this implies that the set of rational numbers is also not closed.
28. From Propositions 8 and 11, we know that R and ∅ are both open and closed.
Suppose A is a non-empty, proper subset of R that is both open and closed. Then there exists x ∈ A
and y ∈ R ∼ A. Suppose without loss of generality that x < y and define
E = {x ∈ A : x < y}
14
Then E is non-empty (x ∈ E) and bounded above (by y). The completeness axiom implies that there
exists a least upper bound of E. Let x∗ = sup E and suppose x∗ ∈ A. Since y ∈
/ A and y is an upper
bound of E, we must have x∗ < y. Therefore there exists r > 0 such that x∗ + r < y. But since A is
open, we can also find r∗ ∈ (0, r) such that (x∗ − r∗ , x∗ + r∗ ) ⊂ A. But this implies x∗ + 2r ∈ E, so x∗
is not an upper bound for E. This contradicts the definition of x∗ . Now suppose x∗ ∈ R ∼ A. Since
A is closed, R ∼ A is open. Therefore there exists r > 0 such that (x∗ − r, x∗ + r) ⊂ R ∼ A. Thus if
x ∈ A, x ≤ x∗ − r. But this means x∗ − r is an upper bound of E, contradicting the assumption that
x∗ is the least upper bound.
The above argument shows that E cannot have a least upper bound, a contradiction of the completeness
axiom. We conclude that no non-empty, proper subset of R that is both open and closed can exist.
29. Let A = (0, 1) and B = {1}. Then Ā = [0, 1] and B̄ = B, so Ā ∩ B̄ = {1} =
6 ∅.
30.
(i) Let x ∈ Ē 0 . Suppose there exists an open interval Ix containing x such that Ix ∩ E 0 = {x}. Then
x ∈ E 0 . Now suppose for any open interval Ix containing x, Ix ∩ E 0 6= {x}. Fix an interval Ix .
Since x is a limit point of E 0 , we know Ix ∩ E 0 6= ∅. Therefore there must exist x0 ∈ Ix ∩ E 0
satisfying x0 6= x. Since Ix is open, there exists r ∈ (0, |x − x0 |) such that (x0 − r, x0 + r) ⊆ Ix .
Now (x0 − r, x0 + r) is an open interval containing x0 and x0 ∈ E 0 , so there must exist x00 ∈
(x0 − r, x0 + r) ∩ (E ∼ {x0 }). But since
0 < |x − x0 | − |x00 − x0 | ≤ |x00 − x|
we know that x00 6= x. Therefore x00 ∈ Ix ∩ (E ∼ {x}). Since Ix was arbitrary, we can conclude
that x is a point of closure of E ∼ {x}. This means x ∈ E 0 . Since every point of closure of E 0 is
in E 0 , E 0 is closed.
(ii) Suppose x ∈ Ē. If x ∈ E, then x ∈ E ∪ E 0 . Suppose x ∈ R ∼ E and let Ix be an open interval
containing x. Since x is a point of closure of E, there exists x0 ∈ E ∩ Ix . But since x ∈ R ∼ E,
x0 6= x. Therefore x0 ∈ Ix ∩ (E ∼ {x}). Since Ix was arbitrary, x is a point of closure of E ∼ {x}.
Thus x ∈ E 0 and therefore x ∈ E ∪ E 0 . We conclude that Ē ⊆ E ∪ E 0 .
Now suppose x ∈ E ∪ E 0 . If x ∈ E, then x ∈ Ē because E ⊆ Ē. If x ∈ E 0 and Ix is an interval
containing x, then there exists x0 ∈ Ix ∩ (E ∼ {x}). But then x0 ∈ Ix ∩ E. Since Ix was arbitrary,
x is a point of closure of E. Therefore x ∈ Ē and we conclude E ∪ E 0 ⊆ Ē.
31. Suppose E is a set containing only isolated points. For each x ∈ E, define f (x) = (p, q) where p and q
are rational numbers such that p < x < q and (p, q) ∩ E = {x}. f defines a one-to-one mapping from
E to Q × Q. By Corollary 4 and Problem 23, Q × Q is a countable set. This means there exists a
one-to-one mapping g from Q × Q onto N. The composition g ◦ f defines a one-to-one mapping from
E to N (see Problem 20), which implies E is countable (see Problem 17).
32.
(i) Suppose E is open and x ∈ E. Then there exists an r > 0 such that the interval (x − r, x + r) is
contained in E. But this means x ∈ int E, so E ⊆ int E. Since int E ⊆ E by definition, E = int E.
Conversely, suppose E = int E. If x is a point in E, then x ∈ int E. But this means there exists
an r > 0 such that the interval (x − r, x + r) is contained in E, so E is open.
(ii) Let E be dense in R and suppose x ∈ int(R ∼ E). Then there exists r > 0 such that (x−r, x+r) ⊆
15
R ∼ E. But this means there does not exist an element of E between any two numbers in
(x − r, x + r), contradicting the assumption that E is a dense set. We conclude that no such x
can be found, so int(R ∼ E) = ∅.
Conversely, suppose int(R ∼ E) = ∅. Let x and y be two real numbers satisfying x < y
and suppose (x, y) ⊂ R ∼ E. Let z ∈ (x, y) and choose r ∈ (0, min(z − x, y − z)). Then
(z − r, z + r) ⊂ (x, y), so (z − r, z + r) ⊂ R ∼ E. But this means z ∈ int(R ∼ E), contradicting
the assumption that int(R ∼ E) = ∅. Therefore (x, y) 6⊂ R ∼ E, which means there must be an
element of E between x and y. But since x and y were arbitrary, this means E is dense in R.
33. Let Fn = [n, ∞). Then {Fn }∞
n=1 is a descending, countable collection of non-empty closed sets. The
T∞
intersection n=1 Fn is empty because of the Archimedean property.
34. The text establishes the following relationships:
Completeness Axiom
=⇒
Heine-Borel Theorem + Archimedean Property
Heine-Borel Theorem
=⇒
Nested Set Theorem
I show below that
Nested Set Theorem + Archimedean Property
=⇒
Completeness Axiom
This implies that the following equivalences hold:
Completeness Axiom
⇐⇒
Heine-Borel Theorem + Archimedean Property
⇐⇒
Nested Set Theorem + Archimedean Property
Suppose the assertions of the Nested Set Theorem and the Archimedean Property are true. Let E be
a non-empty set bounded above by b and let a be an element of E. Define the following sequence of
sets:
[a1 , b1 ] = [a, b]

an + bn
an + bn


if
is an upper bound of E
 an ,
2
2
[an+1 , bn+1 ] = 
a + bn

 n
, bn
otherwise
2
By construction, each bn is an upper bound of E and there exists at least one element of E in [an , bn ]
for all n ∈ N. [a1 , b1 ] is bounded and {[an , bn ]}∞
n=1 is a descending, countable collection of nonempty
T∞
closed sets of real numbers, so the Nested Set Theorem implies that the intersection n=1 [an , bn ] is
non-empty. Let x∗ denote a point in the intersection.
It is easy to establish via induction that
|bn − an | =
16
|b1 − a1 |
2n−1
for all n ∈ N. We can also establish via induction that 2n−1 ≥ n for all n ∈ N. Now fix any x > x∗ .
By the Archimedean property, there exists n ∈ N such that
|x − x∗ |
n>1
|b1 − a1 |
=⇒
|x − x∗ | >
|b1 − a1 |
|b1 − a1 |
= |bn − an |
≥
n
2n−1
=⇒
x > bn
Since bn is an upper bound of E, we must have x ∈
/ E. But since every number greater than x∗ is not
in E, x∗ is an upper bound of E.
Now suppose x < x∗ . Arguing as before, we can find an n ∈ N such that x < an . But since there is
an element of E greater than an , x cannot be an upper bound of E. Thus if x is an upper bound of
E, we must have x ≥ x∗ . We conclude that x∗ is the least upper bound of E.
35. For any collection of sets F, let σ(F) denote the smallest σ-algebra containing F.
Let F denote the collection of closed subsets of R and fix F ∈ F. Since R ∼ F is an open set,
R ∼ F ∈ B. We conclude that F = R ∼ (R ∼ F ) ∈ B, which means F ⊂ B. This implies
σ(F) ⊆ B. Now let G denote the collection of open subsets of R and pick G ∈ G. Then R ∼ G ∈ F,
so G = R ∼ (R ∼ G) ∈ σ(F). But this implies G ⊂ σ(F), so B = σ(G) ⊆ σ(F).
36. Let F denote the collection of intervals of the form [a, b), where a < b. Fix [a, b) ∈ F. Then
[a, b) =
∞ \
a−
n=1
1
,b
n
1
a − , b ∈ B for all n ∈ N and B is closed under countable intersections, we have [a, b) ∈ B.
n
This implies F ⊆ B, so σ(F) ⊆ B.
Since
Now suppose (a, b) is an open interval, where a < b. Pick m ∈ N such that
(a, b) =
1
m
< b − a. Then
∞ [
1
a + ,b
n
n=m
1
Since a + , b ∈ F for all n ≥ m and σ(F) is closed under countable unions, we have (a, b) ∈ σ(F).
n
We can likewise show (−∞, b) ∈ σ(F) and (a, ∞) ∈ σ(F) by using the expressions
∞ [
(−∞, b) =
b − n, b
(a, ∞) =
n=1
∞ [
a+
n=1
1
,a + n + 1
n
Thus all open intervals are in σ(F). By Proposition 9, every non-empty open set can be written as
the countable union of open intervals. Since every open interval is in σ(F) and σ(F) is closed under
countable unions, every non-empty open set is in σ(F). σ(F) also contains ∅, so all open sets are in
σ(F). This mean B ⊆ σ(F).
37. Let (a, b) be an open interval, where a < b. There exists m ∈ N such that
17
1
m
<
b−a
2 .
We can then
write
(a, b) =
∞ [
1
1
a + ,b −
n
n
n=m
so (a, b) is a countable union of closed sets. We can also write
∞ [
1
(−∞, b) =
b − n, b −
n
n=1
∞
[
1
(a, ∞) =
a + ,a + n
n
n=1
Therefore any open interval can be written as a countable union of closed sets. By Proposition 9,
every non-empty open set can be written as a countable union of open intervals. Since the empty set
is closed and a countable collection of countable sets is countable, we can conclude that any open set
is an Fσ set.
1.5
Sequences of Real Numbers
As a preliminary result, I prove that the conclusion of Theorem 18 holds for sequences that converge
to extended real numbers.
Theorem 18 (extended reals): Let {an } and {bn } be two sequences of real numbers. Suppose {an } → a
and {bn } → b, where a and b are extended real numbers. Then for each pair of non-zero real numbers
α and β,
{α · an + β · bn } → α · a + β · b
(1)
provided that α · a + β · b is not of the form ∞ − ∞. Moreover,
if an ≤ bn for all n, then a ≤ b
(2)
Proof: All that remains to be shown is that (1) and (2) hold when a or b is infinite. Suppose a = ∞,
b ∈ R, and α > 0. For any real number c, choose N such that
an ≥
c−β·b+1
,
α
β·b−1
β
bn ≥
for all n ≥ N . Then
α · an + β · bn ≥ α · an + β · b − 1 ≥ c
for all n ≥ N . This implies {α · an + β · bn } → ∞. The cases where a = −∞ or α < 0 can be handled
similarly. If a = b = ∞ and α, β > 0, choose N such that
an ≥
c
,
2·α
bn ≥
c
2·β
for all n ≥ N . Then
α · an + β · bn ≥ c
18
for all n ≥ N , so {α · an + β · bn } → ∞. Other cases in which a and b are both infinite can be dealt
with analogously.
Now suppose an ≤ bn and a = ∞. For any c, there exists N such that an ≥ c for all n ≥ N . But
this means bn ≥ c for all n ≥ N , so b = ∞. The case when b = −∞ can be handled with a similar
argument.
38. Define bn = sup{ak | k ≥ n} and let b = lim sup {an } = limn→∞ bn .
Suppose b = ∞. Then by Proposition 19(ii), {an } is not bounded above. This means it is possible to
define the sequence
n1 = min {n ≥ 1 : an > 1} ,
nk+1 = min {n ≥ nk + 1 : an > k + 1}
For any c, we have ank ≥ c for all k ≥ c. This implies limk→∞ ank = ∞.
Now suppose that {ak } is bounded above. Consider the sequence
n0 = 0,
nk+1 = min n ≥ nk + 1 : bnk +1 < an +
n1 = min {n ≥ 1 : b1 < an + 1} ,
1
k+1
Suppose b > −∞ and fix > 0. Because bk converges to b and nk−1 + 1 ≥ k, by Theorem 17 there
exists an N1 such that |bnk−1 +1 − bk | < /3 for all k ≥ N1 . We can also find N2 such that |b − bk | < /3
for k ≥ N2 and N3 such that 1/k < /3 for k ≥ N3 . Define N = max{N1 , N2 , N3 }. Then
|ank − b| ≤ |ank − bnk−1 +1 | + |bnk−1 +1 − bk | + |b − bk | < for all k ≥ N . But since was arbitrary, we have limk→∞ ank = b.
Now suppose b = −∞. Fix c and pick N such that bnk−1 +1 ≤ c − 1 for all k ≥ N . By construction,
ank − bnk−1 +1 <
1
≤1
k
which implies
ank ≤ |ank − bnk−1 +1 | + bnk−1 +1 < c
for k ≥ N . Thus limk→∞ ank = −∞.
We conclude that lim sup{an } is a cluster point. To see that lim sup{an } is the largest cluster point, let
a denote any cluster point of {an }. Then there exists a subsequence {ank } such that limk→∞ ank = a.
If a = ∞, then {an } is not bounded above so lim sup{an } = ∞ by Proposition 19(ii). If a < ∞, then
for any > 0 there exists N such that ank > a − for all k ≥ N . But this implies bk ≥ a − for
all k, so lim sup{an } = limk→∞ bk ≥ a − by Proposition 14. Since was arbitrary, we conclude that
lim sup{an } ≥ a.
To see that lim inf{ak } is a cluster point, observe that lim inf{an } = − lim sup{−an } by Proposition
19(iii). We know there exists a subsequence {−ank } which satisfies limk→∞ −ank = lim sup{−an }. This
implies limk→∞ ank = limk→∞ −(−ank ) = − lim sup{−an } = lim inf{an } by the extension to Theorem
18 proven above. To see that lim inf{ak } is the smallest cluster point, suppose a is a cluster point of
19
{an }. Then there exists a subsequence {ank } such that limk→∞ ank = a. This means limk→∞ −ank =
−a ≤ lim sup{−an }, so we must have a ≥ − lim sup{−an } = lim inf{an }.
39.
(i) Suppose lim sup{an } = ` ∈ R and fix > 0. Then there exists N such that sup{ak | k ≥ n} > `−
for all n ≥ N . But this means for any k, there exists n ≥ k such that an > ` − . Thus there
must be infinitely many indices n for which an > ` − . Likewise, there exists N such that
sup{ak | k ≥ n} < ` + for all n ≥ N . But this means an < ` + for all n ≥ N . Thus an > ` + for at most N indices.
Conversely, suppose that for any > 0, there are infinitely many indices n for which an > ` − .
Then sup{ak | k ≥ n} ≥ ` − for all n, so lim sup{an } ≥ ` − . Since was arbitrary, we must
have lim sup{an } ≥ `. Now suppose there exists an N such that an ≤ ` + for all n ≥ N . Then
sup{ak | k ≥ n} ≤ ` + for all n ≥ N , so lim sup{an } ≤ ` + . Since was arbitrary, we must
have lim sup{an } ≤ `.
(ii) Suppose {an } is unbounded above. Fix n ∈ N. For any c, there must be k ≥ n such that ak ≥ c.
Otherwise, max{a1 , · · · , an , c} would be an upper bound of {an }. Thus {ak | k ≥ n} is unbounded
for all n, so sup{ak | k ≥ n} = ∞ for all n. This means lim sup{an } = ∞.
Conversely, observe that if an ≤ c for all n, then sup{ak | k ≥ n} ≤ c for all n. But this means
lim sup{an } ≤ c. Thus if lim supn {an } = ∞, {an } cannot be bounded.
(iii) From Problem 6, we know that sup{ak | k ≥ n} = − inf{−ak | k ≥ n}. Therefore lim sup{an } =
− lim inf{−an } by the extension to Theorem 18 proven above.
(iv) Combining parts (i) and (iii), we see that lim inf{an } = ` ∈ R if and only if for each > 0,
an < ` + for infinitely many indices and an < ` − for finitely many indices.
If lim inf{an } = lim sup{an } = a ∈ R, then for any > 0 there exists N such that an ≥ a − and an ≤ a + for all n ≥ N . But this means {an } → a. If lim inf{an } = lim sup{an } = ∞, then
{an } → ∞ by part (ii). The case when lim inf{an } = lim sup{an } = −∞ follows analogously.
To check the converse, suppose that {an } → a ∈ R. Then for any > 0 there are only finitely
many indices for which an ≥ a + and an ≤ a − . But this means there are infinitely many
indices for which an < a + and an > a − , so lim inf{an } = lim sup{an } = a by part (i). Now
suppose that {an } → ∞. Then for any c there exists N such that an ≥ c for all n ≥ N . But
this implies inf{ak | k ≥ n} ≥ c for n ≥ N . Thus lim inf{an } = lim sup{an } = ∞. The case when
{an } → −∞ can be handled similarly.
(v) If an ≤ bn for all n, then sup{ak | k ≥ n} ≤ sup{bk | k ≥ n} for all n. But this implies
lim sup{an } ≤ lim sup{bn } by the extension to Theorem 18 proven above.
40. By Proposition 19 (iv), we know that {an } → a if and only if lim inf{an } = lim sup{an } = a. But by
Problem 38, lim inf{an } = lim sup{an } = a if and only if all the cluster points are the same.
41. This follows immediately from the result in Problem 38.
42. I first prove the following preliminary result:
Claim: If {xn } → x and {yn } → y, where x · y is not of the form 0 · ∞, then {xn · yn } → x · y.
20
Proof: Suppose x and y are both in R. Observe that
xn · yn − x · y = (xn − x) · (yn − y) + y · (xn − x) + x · (yn − y)
For any > 0, we can pick N such that |xn − x| and |yn − y| are smaller than
√
for all n ≥ N . This
implies
|xn − x| · |yn − y| < for all n ≥ N , so {(xn − x) · (yn − y)} → 0. Since {y · (xn − x)} → 0 and {x · (yn − x)} → 0, we must
have {xn · yn − x · y} → 0.
Now suppose x = ∞ and 0 < y < ∞. Fix c and choose N large enough that |yn −y| <
for all n ≥ N . Then
xn · yn >
y
2·c
and xn ≥
2
y
xn · y
≥c
2
for all n ≥ N , so {xn · yn } → ∞. The case when x = ∞ and 0 > y > −∞ can be handled similarly.
√
√
Suppose x = ∞ and y = ∞. Fix c and choose N such that xn ≥ c and yn ≥ c for all n ≥ N . Then
xn · yn > c
for all n ≥ N , so {xn · yn } → ∞. The case when x = ∞ and y = −∞ can be handled similarly.
Observe that for any n, ak ≤ sup{ak | k ≥ n} and bk ≤ sup{bk | k ≥ n} for all k ≥ n. Since ak and bk
are positive, we have
ak · bk ≤ sup{ak | k ≥ n} · sup{bk | k ≥ n}
for all k ≥ n. But this implies
sup{ak · bk | k ≥ n} ≤ sup{ak | k ≥ n} · sup{bk | k ≥ n}
The preliminary result then implies
lim sup{an } ≤ lim sup{an } · lim sup{bn }
provided the product on the right is not of the form 0 · ∞.
43. Let {an } be a sequence of real numbers. Define the set
E = {n ∈ N : an > ak for all k > n}
Suppose E is infinite. Since E is a subset of N, E is countable. Let {nk | k ∈ N} denote an enumeration
of E such that n1 < n2 < · · · . By construction an1 > an2 > · · · , so {ank } is a monotonically decreasing
subsequence. Now suppose E is finite. Then there exists N such that for all n ≥ N , we can find kn > n
such that an ≤ akn . Define the sequence
n1 = N,
21
nj+1 = knj
By construction an1 ≤ an2 ≤ · · · , so {ank } is a monotonically increasing subsequence.
Let {an } be a bounded sequence. By the above result, we can find a monotone subsequence of {an }.
Since this subsequence is bounded, it must converge by Theorem 15.
44. Define the floor function b·c : R → Z as
bxc = max{m ∈ Z : m ≤ x}
This function is well-defined by Problem 11. It is easy to check that 0 ≤ x − bxc < 1 for all x ∈ R.
Now fix x ∈ (0, 1) and consider the sequence
$
a1 = bxpc,
an+1 =
x−
n
X
ai
i=1
pi
!
%
n+1
p
Since 0 ≤ xp < p, we must have 0 ≤ a1 < p. The expression
$
an+1
!
%
ai
an
n+1
=
x−
− n p
pi
p
i=1
!
$
! %
n−1
X ai
n
p − an p
=
x−
pi
i=1
n−1
X
Pn−1
implies 0 ≤ an+1 < p for n ≥ 1, since by construction 0 ≤ x − i=1
ai
pi
pn − an < 1. Thus all the
terms in the sequence {an } are integers satisfying 0 ≤ an < p.
Next, observe that
x−
n
X
ai
i=1
(
This implies
n
X
ai
i=1
pi
pi
= x−
=
1
pn
<
1
pn
n−1
X
ai
an
− n
i
p
p
i=1
!
n−1
X ai
x−
pn − an
i
p
i=1
)
→ x by the Archimedean property, so we can write x =
Now suppose there exists a sequence of integers {bn } with 0 ≤ bn < p such that
E = {n ∈ N | an 6= bn }
22
∞
X
an
.
pn
n=1
P∞
n=1
bn
. Define
pn
If E is non-empty, we can define k = min E by Theorem 1. Suppose without loss that ak < bk . Then
n
X
ai − bi
0 = lim
n→∞
pi
i=1
k
X
ai − bi
=
pi
i=1
+ lim
n→∞
n
X
ai − bi
pi
i=k+1
n
X
ai − bi
a k − bk
+
lim
=
n→∞
pk
pi
i=k+1
Since ak − bk ≥ 1 and bn − an ≤ p − 1, this expression implies
∞
X
ak − bk
1
bn − an
≤
=
pk
pk
pn
n=k+1
≤
∞
p−1 X 1
pk+1 n=0 pn
1
p−1
pk+1 1 − p−1
1
= k
p
=
We can therefore conclude
ak = bk + 1 and
∞
X
bn+k − an+k
=1
pn
n=1
But the latter equality holds if and only if bn+k = p − 1 and an+k = 0 for all n. This means we can
write x as
k
X
ai
x=
i=1
pi
Pk
=
ai pk−i
pk
i=1
and the sequence {bn } must be defined as


an


bn = ak − 1



p − 1
if n < k
if n = k
if n ≥ k
To check the converse, suppose {an } is a sequence of integers with 0 ≤ an < p. Then
0≤
n+m
X
k=n
Fix > 0 and pick N such that
n+m
m
X p−1
ak
p−1 X 1
pm+1 − 1
≤
=
=
k
k
n
k
p
p
p
p
pm+n
k=n
1
pn−1
<
n+m
X
k=n
k=0
for all n ≥ N . Then
2
ak
1
1
≤ n−1 + m+n < k
p
p
p
23
for all n, m ≥ N . But this means the series converges by Proposition 20. We also have
0≤
n
X
ak
pk
k=1
which implies 0 ≤
∞
X
ak
k=1
45.
pk
≤
n
X
p−1
pk
k=1
=1−
1
<1
pn
≤ 1 by Theorem 18.
(i) Let {sn } denote the sequence of partial sums. Suppose {sn } converges and fix > 0. By Theorem
17, there exists an N such that
if n, n0 ≥ N , then |sn0 − sn | < Define N 0 = N + 1. Then for any n ≥ N 0 , we have
n+m
X
ak = |sn+m − sn−1 | < k=n
for any natural number m.
To prove the converse, first fix > 0. Let N be an index such that
n+m
X
k=n
ak <
for n ≥ N and any natural number m
2
Choose natural numbers n and n0 such that n, n0 ≥ N . Without loss of generality, assume n0 > n.
If n0 > n + 1, define m = n0 − n − 1. Then
|sn0 − sn | =
n+1+m
X
ak <
k=n+1
<
2
If n0 = n + 1, then
|sn0 − sn | = |an+1 |
= |an+1 + an+2 + an+3 − (an+2 + an+3 )|
≤ |an+1 + an+2 + an+3 | + |an+2 + an+3 |
< +
2 2
<
Therefore the sequence {sn } is Cauchy, which implies that {sn } converges by Theorem 17.
P∞
(ii) Suppose k=1 |ak | is summable and fix > 0. By part (i), there exists N such that
n+m
X
k=n
ak ≤
n+m
X
k=n
|ak | =
n+m
X
for n ≥ N and any natural number m. But this implies
24
|ak | < i=n
P∞
k=1
ak is summable by part (i).
(iii) If each term an is non-negative, the sequence of partial sums forms an increasing sequence. Thus
by Theorem 15, the sequence of partial sums converges if and only if it is bounded.
46. The text establishes the following relationships:
Completeness Axiom
=⇒
Monotone Convergence Theorem + Archimedean Property
We proved in Problem 43 that
Monotone Convergence Theorem
=⇒
Bolzano-Weierstrass Theorem
I show below that
Bolzano-Weierstrass Theorem
=⇒
Nested Set Theorem
From Problem 34, we know
Nested Set Theorem + Archimedean Property
=⇒
Completeness Axiom
Thus we will have shown the following equivalences:
Completeness Axiom
⇐⇒
Monotone Convergence Theorem + Archimedean Property
Bolzano-Weierstrass Theorem + Archimedean Property
⇐⇒
⇐⇒
Nested Set Theorem + Archimedean Property
Suppose the assertion of the Bolzano-Weierstrass Theorem holds. Let {Fn }∞
n=1 denote a descending,
countable collection of nonempty closed sets of real numbers for which F1 is bounded. Let an = sup Fn
for all n ∈ N. Since Fn is closed, an ∈ Fn . Since {Fn }∞
n=1 is descending, an ∈ F1 . Since F1 is bounded,
the sequence {an } is bounded. By the Bolzano-Weierstrass Theorem, there exists a subsequence {ank }
that converges. Let x = limk→∞ ank . Suppose x ∈
/ Fn for some n. Since Fn is closed, R ∼ Fn is open.
Therefore there exists r > 0 such that (x − r, x + r) ⊆ R ∼ Fn . But since {ank } → x, we can find an
index N such that |ank − x| < r for all k ≥ N . This means there exists nk ≥ n such that ank ∈
/ Fn .
T∞
But ank ∈ Fnk ⊆ Fn , a contradiction. Therefore x ∈ Fn for all n, so n=1 Fn 6= ∅.
1.6
Continuous Real-Valued Functions of a Real Variable
47. Since E is closed, R ∼ E is open. By Proposition 9, we can write R ∼ E =
S∞
n=1 (an , bn )
where
{(an , bn )} is a collection of disjoint open intervals. If x ∈ E, define g(x) = f (x). If x ∈ (an , bn ), define
25
g(x) as
g(x) =

f (bn ) − f (an )


(x − an ) if an , bn ∈ R
f (an ) +


bn − an






f (an )
if an ∈ R and bn = ∞



f (bn )







0
if bn ∈ R and an = −∞
if an = −∞ and bn = ∞
48. I first show that f is continuous at 0. By construction, f (0) = 0. Fix any > 0. Since limn→∞ n sin n1 =
1, there exists an index N such that
n sin
for all n ≥ N . Define δ = min
n
1
N , 1+
1
<1+
n
o
. For any non-zero rational number x =
p
q,
where
p
q
is in
lowest terms and q is chosen to be positive for convenience, we have
|x| =
p
<δ
q
=⇒
1
|p|
≤
<q
δ
δ
=⇒
q ≥ N,
=⇒
=⇒
|p| ≤ δ · q
1
< 1 + ,
|p| ≤ δ · q
q
1
1
≤ δ · q · sin
< δ · (1 + ) ≤ |f (x)| = p · sin
q
q
q sin
If x is irrational, then
|x| < δ
|f (x)| = |x| < =⇒
since δ < . Thus |x − 0| < δ implies |f (x) − 0| < , so f (x) is continuous at zero.
Next, I show that f is discontinuous at all non-zero rational numbers. Let x =
p
q
be a non-zero rational
number. Define z = |p · q| + 1 and
pn = p · (z n + q) ,
qn = q · z n
Since p, q and z are pairwise coprime, pn and qn are relatively prime for all n. Define xn =
pn
qn
and
observe that
pn
1
· qn · sin
=
n→∞ qn
qn
lim f (xn ) = lim
n→∞
pn
n→∞ qn
However
f (x) = p sin
lim
1
p
· lim qn · sin
=
n→∞
qn
q
1
p
6=
q
q
because the left-hand side is an irrational number. Since limn→∞ xn = x but limn→∞ f (xn ) 6= f (x), f
is not continuous at x.
Lastly, we can also show that f is continuous at all irrational numbers. Let x be an irrational number
and fix > 0. Let x = [a0 ; a1 , a2 , · · · ] denote the continued fraction expansion of x and define the
26
sequences
hn = an hn−1 + hn−2
h−1 = 1
h−2 = 0
kn = an kn−1 + kn−2
k−1 = 0
k−2 = 1
Since an ≥ 1, kn ≥ n for all n. It is possible to show
x−
for any rational number
p
q
hn
kn
→ x and
p
hn
< x−
kn
q
satisfying 0 < q < kn . This means we can find an index N such that
x−
hn
· |x|
<
kn
+ 2 · |x|
and
1 − n · sin
for all n ≥ N . Let δ = x −
hN
kN
1
<
n
2 · |x|
. Then for any rational
p
q
satisfying x −
p
q
< δ, we must have
q ≥ kN ≥ N . But this implies
x − p · sin
p
1
p
1
≤ x−
+
− p · sin
q
q
q
q
= x−
p
p
1
+
1 − q · sin
q
q
q
≤ x−
p
1
p
1
+ x−
1 − q · sin + |x| 1 − q · sin
q
q
q
q
p
= x−
q
<δ 1+
1
1 + 1 − q · sin
q
+
2 · |x|
2
+ |x| 1 − q · sin
1
q
<
For any irrational x0 satisfying |x0 − x| < δ, we have |x0 − x| < because δ < . Therefore |x0 − x| < δ
implies |f (x0 ) − f (x)| < , so f is continuous at x.
49.
(i)
• f + g: Fix > 0 and pick x ∈ E. There exists δ1 , δ2 > 0 such that
|x0 − x| < δ1 =⇒ |f (x0 ) − f (x)| <
2
|x0 − x| < δ2 =⇒ |g(x0 ) − g(x)| <
2
for all x0 ∈ E. Define δ = min{δ1 , δ2 }. Then
|x0 − x| < δ =⇒ |f (x0 ) + g(x0 ) − (f (x) + g(x))| ≤ |f (x0 ) − f (x)| + |g(x0 ) − g(x)| < 27
for all x0 ∈ E. Thus f + g is continuous at x.
• f g: Fix > 0 and pick x ∈ E. There exists δ1 , δ2 > 0 such that
0
0
0
0
r
,
3 3 max{2|g(x)|, 1}
r
,
3 3 max{2|f (x)|, 1}
|x − x| < δ1 =⇒ |f (x ) − f (x)| < min
|x − x| < δ2 =⇒ |g(x ) − g(x)| < min
for all x0 ∈ E. Define δ = min{δ1 , δ2 }. Then |x0 − x| < δ implies
3
|g(x)| · |f (x0 ) − f (x)| ≤
3
|f (x)| · |g(x0 ) − g(x)| ≤
3
|f (x0 ) − f (x)| · |g(x0 ) − g(x)| <
|g(x)|
<
max{2|g(x)|, 1}
3
|f (x)|
·
<
max{2|f (x)|, 1}
3
·
for all x0 ∈ E. But this implies
|f (x0 ) · g(x0 ) − f (x) · g(x)| =
≤
|(f (x0 ) − f (x)) · (g(x0 ) − g(x))
+ f (x) · (g(x0 ) − g(x)) + g(x) · (f (x0 ) − f (x))|
|f (x0 ) − f (x)| · |g(x0 ) − g(x))|
+ |f (x)| · |g(x0 ) − g(x)| + |g(x)| · |f (x0 ) − f (x)|
<
for all x0 ∈ E satisfying |x0 − x| < δ. Thus f g is continuous at x.
(ii) Let E 0 denote the domain of h. Fix > 0 and choose x ∈ E 0 . Define y = h(x). There exists
δ > 0 such that for any y 0 ∈ E,
|y 0 − y| < δ =⇒ |f (y 0 ) − f (y)| < There also exists η > 0 such that for any x0 ∈ E 0 ,
|x0 − x| < η =⇒ |h(x0 ) − h(x)| < δ
But this means
|x0 − x| < η =⇒ |f (h(x0 )) − f (h(x))| < for any x0 ∈ E 0 , so f ◦ h is continuous at x.
(iii) Observe that
max{f (x), g(x)} =
1
(f (x) + g(x) + |f (x) − g(x)|)
2
The result then follows from parts (i) and (iv).
28
(iv) For any x, x0 ∈ R we have
|x| − |x0 | ≤ |x − x0 | and |x0 | − |x| ≤ |x − x0 |
which means
|x| − |x0 | ≤ |x − x0 |
But then for any > 0, we have
|x − x0 | < =⇒ |x| − |x0 | < so | · | is continuous.
50. Suppose f is Lipschitz on E. Then there exists a c ≥ 0 such that
|f (x0 ) − f (x)| ≤ c|x0 − x| for all x0 , x ∈ E
Fix > 0 and define δ =
. Then for all x, x0 ∈ E, we have
max{2c, 1}
|x0 − x| < δ =⇒ |f (x0 ) − f (x)| ≤
c
<
max{2c, 1}
Thus f is uniformly continuous on E.
√
Consider the function f (x) = x on the set E = [0, 1]. Fix > 0 and x ∈ E. If x = 0, define δ = 2 .
Then
√
|x0 | < δ =⇒
x0 =
p
|x0 | < √
If x > 0, define δ = x. Then
√
x0
−
√
√
x =
x0
−
so that
|x0 − x| < δ =⇒
√
√
√
x0 + x
|x0 − x|
x ·√
=√
√
√
x0 + x
x0 + x
√
x0 −
√
x ≤
(1)
|x0 − x|
√
<
x
Thus f is continuous on E. Since E is closed and bounded, f is uniformly continuous on E by Theorem
23. To see that f is not Lipschitz, fix any c ≥ 0. From equation (1), we have
√
√
x0 − x
1
=√
√
0
x0 − x
x + x
as long as x and x0 do not both equal 0. If c = 0, then |f (x0 ) − f (x)| > c|x0 − x| for all x0 , x ∈ E
√
√
satisfying x 6= x0 . If c > 0, choose x, x0 ∈ 0, max 4c12 , 1 with x 6= x0 . Then x0 + x < 1c , so
√
√
x0 − x
>c
x0 − x
Thus for any c ≥ 0, we can find x0 , x ∈ E violating the Lipschitz condition.
29
51. Let Ak = {ak,1 , ak,2 , · · · , ak,2k +1 } denote the sequence of sets defined recursively by
a1,1 = a,
a1,2 =
a1,3 = b



ak, i+1
if i ∈ {1, · · · , 2k+1 + 1} is odd
a i − ak, 2i


 k, 2 +1
2
if i is even
2
ak+1,i =
a+b
,
2
The elements of Ak partition the interval [a, b] into subintervals of length
b−a
.
2k
Define the function
ϕk : [a, b] → R as
ϕk (x) = f (ak,i ) +
f (ak,i+1 ) − f (ak,i )
(x − ak,i )
ak,i+1 − ak,i
if ak,i ≤ x ≤ ak,i+1
By construction, ϕk is piecewise linear for each k. If ak,i ≤ x ≤ ak,i+1 , then
|f (x) − ϕk (x)| =
ak,i+1 − x
x − ak,i
(f (x) − f (ak,i )) +
(f (x) − f (ak,i+1 ))
ak,i+1 − ak,i
ak,i+1 − ak,i
≤ |f (x) − f (ak,i )| + |f (x) − f (ak,i+1 )|
By Theorem 23, there exists a positive number δ such that for all x, x0 ∈ [a, b],
|x − x0 | < δ =⇒ |f (x) − f (x0 )| <
Choose K such that
b−a
2K
2
< δ. Then
aK,i ≤ x ≤ aK,i+1 =⇒ |x − aK,i | < δ and |x − aK,i+1 | < δ
=⇒ |f (x) − f (aK,i )| < and |f (x) − f (aK,i+1 )| <
=⇒ |f (x) − ϕK (x)| < 2
2
Since any x ∈ [a, b] must belong to [aK,i , aK,i+1 ] for some i ∈ {1, · · · , 2K }, |f (x) − ϕK (x)| < for all
x ∈ [a, b].
52. Suppose E is not bounded above. Then for any c, there exists x ∈ E satisfying x > c. But this means
the function f (x) = x does not obtain its maximum on E. Likewise, if E is not bounded below the
function f (x) = −x does not obtain its maximum on E. Now suppose E is open. Then for any x ∈ E,
there exists an x0 ∈ E such that x0 > x. But this means the function f (x) = x does not obtain its
maximum on E. We conclude that if every continuous, real-valued function on E takes its maximum
on E, then E must be closed and bounded.
The converse is given by the Extreme Value Theorem.
53. Suppose E is unbounded and consider the open cover {(x−1, x+1)}x∈E . Let {(xi −1, xi +1)}ni=1 denote
a finite collection of sets from this open cover. There exists x ∈ E satisfying x > max{x1 , · · · , xn } + 1,
so {(xi − 1, xi + 1)}ni=1 cannot cover E. Now suppose E is a set which is not closed. Then there must
30
exist a point of closure x∗ of E which is not in E. For each x ∈ E, define Ix as

x + x∗


−∞,

2
Ix = 
x + x∗


,∞
2
Then
S
x∈E Ix
is an open cover of E. Let
Sn
i=1 Ixi
if x < x∗
if x > x∗
denote a finite collection of sets from this open
cover. Since x∗ is a point of closure of E, there exists a point x ∈ E satisfying |x − x∗ | < 12 min{|x1 −
Sn
Sn
x∗ |, · · · , |xn − x∗ |}. But by construction, x0 ∈
/ i=1 Ixi . Therefore i=1 Ixi cannot cover E.
The converse is given by the Heine-Borel Theorem.
54. Let E be an interval and suppose y and y 0 are two points in f (E). Then there exists x, x0 ∈ E such
that f (x) = y and f (x0 ) = y 0 . Without loss of generality, assume y < y 0 and pick any z ∈ (y, y 0 ). Since
E is an interval, [min{x, x0 }, max{x, x0 }] ⊆ E. By the Intermediate Value Theorem, there is a point
x0 ∈ (min{x, x0 }, max{x, x0 }) at which f (x0 ) = z. But this implies z ∈ f (E).
Conversely, suppose every continuous real-valued function on E has an interval as its image. Take f
to be the identity function. Then f (E) = E, so E is an interval.
55. Let E be an open interval and let f be a monotone function. If f is continuous, then f (E) is an interval
by Problem 54.
To prove the converse, fix x0 ∈ E and suppose {xn } is a sequence in E converging upward to x0 .
By the Monotone Convergence Theorem for Real Sequences, f (xn ) must converge to a limit f (x+
0 ).
Because f is monotone, we have
x < x0 =⇒ f (x) ≤ f (x+
0 ),
x0 ≤ x =⇒ f (x0 ) ≤ f (x)
for any x ∈ E. Therefore if f (x+
0 ) < f (x0 ), there does not exist any x ∈ E such that f (x) ∈
(f (x+
0 ), f (x0 )). An analogous argument applies to sequences converging downward to x0 . We can
conclude that if f (E) is an interval, any monotone sequence {xn } in E converging to x0 must satisfy
{f (xn )} → f (x0 ).
Now suppose f is not continuous at x0 . Then we can find a positive such that for any n there exists
an xn ∈ E that satisfies |xn − x0 | <
1
n
and |f (xn ) − f (x0 )| > . But this means we can find a monotone
subsequence {xnk } such that {xnk } converges to x but {f (xnk )} does not converge to f (x). Therefore
if any monotone sequence {xn } in E converging to x0 satisfies {f (xn )} → f (x0 ), f must be continuous
at x0 . Thus if f (E) is an interval, f is continuous on E.
56. For any x ∈ R and n ∈ N, say x satisfies property Cn if there exists a δx > 0 such that for all
x0 , x00 ∈ R,
|x0 − x| < δx and |x00 − x| < δx =⇒ |f (x0 ) − f (x00 )| <
1
n
(2)
Let An denote the collection of points in E which satisfy property Cn . Suppose x ∈ An and pick y
such that |x − y| <
|x0 − y| <
δx
2 .
Then for all x0 , x00 ∈ R,
δx
δx
1
and |x00 − y| <
=⇒ |x0 − x| < δx and |x00 − x| < δx =⇒ |f (x0 ) − f (x00 )| <
2
2
n
31
δx
2 ,
so y ∈ An . But this means x −
δx
2 ,x
δx
2
⊆ An , so x
T∞
is an interior point of An . Since x was an arbitrary point of An , An is an open set. Thus n=1 An is
Therefore y satisfies property Cn for δy =
+
a Gδ set. I now show that this set is equal to the set of all points at which f is continuous.
T∞
Now suppose x ∈ n=1 An . Fix > 0 and pick n such that > n1 . Since x ∈ An , there exists a δx > 0
such that (2) is satisfied. By choosing x00 = x in (2), we see that
|x0 − x| < δ =⇒ |f (x0 ) − f (x)| <
1
<
n
for any x0 ∈ R. Therefore f is continuous at x.
Now suppose f is continuous at x. Then for any n ∈ N , there exists a δ > 0 such that
|x0 − x| < δ =⇒ |f (x0 ) − f (x)| <
1
2n
for all x0 ∈ R. But this implies
|x0 − x| < δ and |x00 − x| < δ =⇒ |f (x0 ) − f (x00 )| ≤ |f (x0 ) − f (x)| + |f (x00 ) − f (x)| <
so x satisfies property Cn for all n. But this means x ∈
T∞
n=1
1
n
An .
57. For any k, n, N ∈ N, define the set
Ak,n (N ) =
x ∈ R : |fk (x) − fn (x)| ≤
1
N
Since the function |fk (x) − fn (x)| is continuous and ∞, N1 is closed, Ak,n (N ) is closed for all k, n, N .
This implies that the set
Bn (N ) =
∞
\
Ak,n (N ) =
x ∈ R : sup |fk (x) − fn (x)| ≤
k≥n
k=n
is closed for all n, N . Define the set
B=
∞ [
∞
\
1
N
Bn (N )
N =1 n=1
This set is the intersection of a countable collection of Fσ sets. I now show that B is the set of points
for which {fn (x)} converges.
Suppose x ∈ B. Fix > 0 and pick N such that >
2
N.
There exists n such that x ∈ Bn (N ). For any
n0 , m0 ≥ n, we have
|fn0 (x) − fm0 (x)| ≤ |fn0 (x) − fn (x)| + |fm0 (x) − fn (x)| ≤ 2 sup |fk (x) − fn (x)| ≤
k≥n
2
<
N
But this means {fn (x)} is Cauchy, so {fn (x)} converges to a real number by Theorem 17.
Conversely, suppose {fn (x)} converges to a real number. By Theorem 17, {fn (x)} is Cauchy. Now fix
32
any N . There exists n such that for all n0 , m0 ≥ n
|fn0 (x) − fm0 (x)| <
1
N
But this means
sup |fn0 (x) − fn (x)| ≤
n0 ≥n
1
N
so x ∈ Bn (N ). Since N was arbitrary, x ∈ B.
58. Suppose A ⊆ R is an open set. Then f −1 (A) is open by Proposition 22.
Suppose A ⊆ R is a closed set. Then R ∼ A is open, so f −1 (R ∼ A) is open. Since f −1 (R ∼ A) =
{x ∈ R : f (x) ∈
/ A} = R ∼ f −1 (A), f −1 (A) must be closed.
Define C as
C := A ∈ B : f −1 (A) ∈ B
C is a σ-algebra:
• f −1 (∅) = ∅ ∈ B
• Suppose A ∈ C. Then f −1 (R ∼ A) = R ∼ f −1 (A) ∈ B because f −1 (A) ∈ B and B is closed under
complements.
S
• Suppose Aλ ∈ C for λ ∈ Λ, where Λ is a countable set. Then f −1
λ∈Λ Aλ = {x ∈ R : f (x) ∈
S
S
S
−1
(Aλ ) ∈ B because f −1 (Aλ ) ∈ B for all λ ∈ Λ
λ∈Λ Aλ } =
λ∈Λ {x ∈ R : f (x) ∈ Aλ } =
λ∈Λ f
and B is closed under countable intersections.
We already showed that all open subsets of R are in C, so B ⊆ C. But C ⊆ B by construction, so B = C.
59. Fix > 0 and x∗ ∈ E. Choose N such that for all x ∈ E and n ≥ N , |fn (x) − f (x)| < 3 . Since fN is
continuous, it is also possible to find a δ > 0 such that for all x ∈ E,
|x − x∗ | < δ =⇒ |fN (x) − fN (x∗ )| <
3
Thus for all x ∈ E, we have
|x − x∗ | < δ =⇒ |f (x) − f (x∗ )| ≤ |f (x) − fN (x)| + |fN (x) − fN (x∗ )| + |f (x∗ ) − fN (x∗ )| < Therefore f is continuous at x∗ . Since x∗ was an arbitrary point in E, f is continuous on E.
60. Suppose f is a continuous, real-valued function defined on a set E. Fix x∗ ∈ E and let {xn } be a
sequence in E converging to x∗ . Pick > 0 and choose δ > 0 such that for all x ∈ E,
|x∗ − x| < δ =⇒ |f (x∗ ) − f (x)| < Since {xn } → x∗ , we can find an N such that |x − xn | < δ for all n ≥ N . But this means |f (xn ) −
f (x∗ )| < for all n ≥ N . Therefore {f (xn )} converges to f (x∗ ).
Now fix any x∗ ∈ E and suppose f (x∗ ) is not continuous at x∗ . Then there exists > 0 such that for
33
any n ∈ N, we can find xn ∈ E satisfying
|xn − x∗ | <
1
and |f (xn ) − f (x∗ )| ≥ n
The sequence {xn } is in E and converges to x∗ but its image sequence {f (xn )} fails to converge to
f (x∗ ).
Let f be a continuous real-valued function on the non-empty, closed and bounded set E. The argument
in the text can be used to establish that f is bounded. Define y = sup{f (x) : x ∈ E}. For each n,
we can find xn ∈ E such that |y − f (xn )| <
1
n.
Thus the sequence {f (xn )} converges to y. By the
Bolzano-Weierstrass Theorem, there exists a subsequence of {xn }, call it {xnk }, that converges to some
x∗ . Since {xnk } ∈ E for all k and E is closed, x∗ must also be in E. But Proposition 21 then implies
{f (xnk )} converges to f (x∗ ). We therefore have
f (x∗ ) = lim f (xnk ) = lim f (xn ) = y
n→∞
k→∞
2
Lebesgue Measure
2.1
Introduction
1. Since m is countably additive and B = A ∪ (B ∼ A), we have
m(B) = m(A) + m(B ∼ A) ≥ m(A)
where the inequality follows because m(B ∼ A) ≥ 0.
2. Since A = A ∪ ∅, we have
m(A) = m(A) + m(∅)
by the countable additivity property of m. If m(A) < ∞, then this expression implies m(∅) = 0.
3. Define the sequence of sets
E10 = E1 ,
0
En+1
= En+1 ∼
n
[
Ei
i=1
Then En0 ⊆ En , the collection of sets {En0 }∞
n=1 is pairwise disjoint, and
∞
[
En0 =
n=1
∞
[
En
n=1
By the countable additivity and monotonicity of m, we have
m
∞
[
n=1
!
En
=m
∞
[
!
En0
n=1
=
∞
X
n=1
m(En0 ) ≤
∞
X
m(En )
n=1
4. Let E be a subset of R and fix y ∈ R. The sets E and E + y have the same number of elements, so
34
c(E) = c(E + y).
Let {En }∞
n=1 denote a countable, disjoint collection of sets. If EN is infinite for some N , then
∞≥
∞
X
N
X
c(En ) ≥
n=1
But
P∞
S∞
n=1
n=1
En ⊇ EN , so
S∞
n=1
c(En ) = ∞
n=1
En is infinite and thus c (
S∞
n=1
S∞
En ) = ∞. Therefore c ( n=1 En ) =
c(En ) = ∞.
Now suppose En is finite for all n. Define
I = {n : c(En ) > 0}
If I is infinite, then
∞≥
∞
X
c(En ) =
n=1
X
c(En ) ≥
n∈I
X
1=∞
n∈I
P∞
S∞
so that n=1 c(En ) = ∞. n=1 En is infinite because the sets in the collection are disjoint. Therefore
S∞
P∞
c ( n=1 En ) = n=1 c(En ) = ∞. If I is finite, there must exist an index N such that En = ∅ for all
n ≥ N . But this implies
c
∞
[
!
En
=c
n=1
N
[
!
En
=
n=1
N
X
c (En ) =
n=1
∞
X
c (En )
n=1
where the second equality follows because the number of elements in a finite union of finite disjoint
sets equals the sum of the number of elements in each set.
2.2
Lebesgue Outer Measure
5. By Proposition 1, the interval [0, 1] has outer measure equal to 1. The example on page 31 shows that
any countable set has outer measure 0. Therefore [0, 1] cannot be countable.
6. Because the outer measure is monotone, m∗ (A) ≤ m∗ ([0, 1]) = 1. Since the outer measure of a
countable set is 0 and the set of rational numbers in [0, 1] is countable, m∗ ([0, 1] ∼ A) = 0. By the
subadditivity of m∗ , we have 1 = m∗ ([0, 1]) ≤ m∗ ([0, 1] ∼ A) + m∗ (A) = m∗ (A). We can therefore
conclude m∗ (A) = 1.
7. Since E is bounded m∗ (E) is finite. Because m∗ (E) is finite, for any k there exists a countable collection
S∞
of nonempty, open, bounded intervals {Ik,n }∞
n=1 such that E ⊆
n=1 Ik,n and
m∗ (E) +
Define Gk =
S∞
n=1 Ik,n .
∞
X
1
>
`(Ik,n )
k n=1
Gk is open for all k, so G :=
35
T∞
n=1
Gk is a Gδ set. By the definition of m∗ , we
must have m∗ (Gk ) ≤
P∞
n=1
`(Ik,n ). Since E ⊆ G ⊆ Gk , we can conclude
m∗ (E) ≤ m∗ (G) ≤ m∗ (Gk ) ≤
∞
X
`(Ik,n ) < m∗ (E) +
n=1
1
k
by the monotonicity of m∗ . Since this expression holds for all k, m∗ (E) = m∗ (G).
8. I first prove a preliminary result:
Sn
Let Ai , i = 1, · · · , n, be a finite collection of sets of real numbers and define B = i=1 Ai . Then
Sn
B̄ = i=1 Āi .
Sn
Proof: Suppose x ∈ i=1 Āi and let I be an open interval containing x. Because x is in Āi for some i,
there must exist a point y in I that is also in Ai . But y is also in B, so we can conclude x ∈ B̄. Thus
Sn
i=1 Āi ⊆ B̄.
Tn
To see the reverse inclusion, suppose that x ∈ i=1 ĀC
i . Then for each i, there exists i > 0 such
C
that (x − i , x + i ) ⊆ AC
i . Define = mini i . Then (x − , x + ) ⊆ Ai for all i, so (x − , x + ) ⊆
Tn
Tn
Sn
C
C
C
C
C
i=1 Ai = B . But this means x ∈ B̄ . Thus
i=1 Āi ⊆ B̄ , which implies B̄ ⊆
i=1 Āi .
Now since the rationals are dense in R, B̄ = [0, 1]. We therefore have
n
[
[0, 1] = B̄ ⊆
Ik =
n
[
I¯k
k=1
k=1
where the last equality follows from the preliminary result. But Proposition 1 and the monotonicity
and subadditivity of outer measure then imply
∗
∗
1 = µ ([0, 1]) ≤ µ
n
[
!
I¯k
k=1
≤
n
X
µ∗ (I¯k ) =
k=1
n
X
µ∗ (Ik )
k=1
9. By the monotonicity of m∗ , m∗ (B) ≤ m∗ (A ∪ B). Since m∗ is countably subadditive, we have m∗ (A ∪
B) ≤ m∗ (A) + m∗ (B) = m∗ (B). Therefore m∗ (A ∪ B) = m∗ (B).
10. By the subadditivity property of m∗ , we know that m∗ (A ∪ B) ≤ m∗ (A) + m∗ (B). Therefore we only
need to show that the reverse inequality holds. Fix > 0. Since A and B are bounded, A ∪ B is
bounded and m∗ (A ∪ B) is finite. We can therefore find a countable collection of non-empty, open,
bounded intervals {Ik }∞
k=1 which covers A ∪ B and satisfies
m∗ (A ∪ B) >
∞
X
`(Ik ) − k=1
Without loss, assume the length of each interval in the collection is less than
α
2
(the intervals can be
subdivided until this condition holds). Then by construction, each interval only intersects either A or
B. Define
A = {k : Ik ∩ A 6= ∅},
36
B = {k : Ik ∩ B 6= ∅}
Since {Ik }k∈A and {Ik }k∈B form open covers of A and B, respectively, we can conclude
m∗ (A ∪ B) >
X
`(Ik ) +
k∈A
X
`(Ik ) − ≥ m∗ (A) + m∗ (B) − k∈B
This expression holds for all > 0, so we must have m∗ (A ∪ B) ≥ m∗ (A) + m∗ (B). Therefore
m∗ (A ∪ B) = m∗ (A) + m∗ (B).
2.3
The σ-Algebra of Lebesgue Measurable Sets
11. Since σ-algebras are closed under countable unions, intersections and complements, the result follows
from the following identities:
[a, ∞) =
∞ \
a−
k=1
1
,∞
k
(−∞, a] = (a, ∞)C
(−∞, a) = [a, ∞)C
(a, b) = (−∞, b) ∩ (a, ∞)
[a, b) = (−∞, b) ∩ [a, ∞)
(a, b] = (−∞, b] ∩ (a, ∞)
[a, b] = (−∞, b] ∩ [a, ∞)
12. Since all intervals of the form (a, ∞) are open sets and thus in the Borel σ-algebra, the result follows
from Problem 11.
13.
(i) and (ii) I first prove two preliminary results:
Claim 1: The translate of an open set is open.
Proof: Let A be an open set and suppose x ∈ A + y. Then x − y ∈ A. Since A is open, there
exists r > 0 such that (x − y − r, x − y + r) ⊆ A. Now pick z ∈ (x − r, x + r). Then z − y ∈ A, so
z = (z − y) + y is in A + y. Therefore (x − r, x + r) ⊆ A + y, so A + y is open.
Claim 2: For any set of real numbers A, AC + y = (A + y)C for all y ∈ R.
Proof: Since (AC + y) ∩ (A + y) = ∅, we must have AC + y ⊆ (A + y)C . To see the reverse
inclusion, suppose x ∈ (A + y)C . Then for all z ∈ A, we must have x 6= z + y. But this implies
x − y ∈ AC , so x = (x − y) + y is in AC + y.
Now suppose G is a Gδ set. Then there exists a countable collection of open sets {Gk }∞
k=1 such
T∞
that G = k=1 Gk . Fix y ∈ R and suppose x ∈ G + y. Then x = z + y for some z ∈ G. But
T∞
T∞
since z ∈ Gk for all k, x ∈ k=1 (Gk + y). Therefore G + y ⊆ k=1 (Gk + y). Now suppose
T∞
x ∈ k=1 (Gk + y). Then for every k, there exists xk ∈ Gk such that x = xk + y. But this implies
T∞
x − y ∈ Gk for all k, so x − y ∈ G. Thus x = (x − y) + y is in G + y, so k=1 (Gk + y) ⊆ G + y.
T∞
Therefore G + y = k=1 (Gk + y). By Claim 1, Gk + y is open for all k. Thus G + y is a Gδ set.
37
Now suppose F is an Fσ set. Then there exists a countable collection of closed sets {Fk }∞
k=1 such
T∞
S∞
C
C
that F = k=1 Fk . Since F = k=1 Fk is a Gδ set, we have
(F + y)C = F C + y =
∞
\
(FkC + y) =
k=1
∞
\
(Fk + y)C
k=1
where the first and last equality follow from Claim 2 and the middle equality follows from the
S∞
result for Gδ sets. Taking complements, we have F + y = k=1 (Fk + y). Since FkC + y is open
and Fk + y = (FkC + y)C by Claim 2, Fk + y is closed for all k. Thus F + y is an Fσ set.
(iii) This follows directly from Propositions 2 and 4.
14. Suppose every bounded subset of E has outer measure equal to zero. Then for all n, we have
n
X
m∗ ((−n, n) ∩ E) = 0
i=1
so that
P∞
n=1
m∗ ((−n, n) ∩ E) = 0. By the countable subadditivity of m∗ , we must have
∗
∗
m (E) = m
∞
[
!
(−n, n) ∩ E
≤
∞
X
m∗ ((−n, n) ∩ E) = 0
n=1
n=1
Thus if m∗ (E) > 0, there must exist an n such that the m∗ ((−n, n) ∩ E) > 0.
15. Define In as
h
h
∪ (n − 1) · , n ·
In = −n · , (−n + 1) ·
2
2
2
2
Then {In }∞
n=1 is a collection of disjoint and measurable sets that cover R, so E =
∗
S∞
n=1 (E
∩ In ). We
∗
also have m (E ∩ In ) ≤ m (In ) = for all n. To obtain a finite collection of sets, observe that
∗
∗
m (E) = m
∞
[
!
E ∩ In
=
n=1
∞
X
m∗ (E ∩ In )
n=1
by countable additivity. Since m∗ (E) < ∞, we can find an index m such that
m
∗
∞
[
!
(E ∩ In )
n=m+1
=
∞
X
m∗ (E ∩ In ) < n=m+1
Thus we can write E as the union of the finite collection of sets
(
)
∞
[
m
{E ∩ In }n=1 ∪ E ∩
In
n=m+1
where the sets are disjoint and measurable and each set has measure at most .
2.4
16.
Outer and Inner Approximation of Lebesgue Measurable Sets
• Measurability + (i) =⇒ (iii): Let E be a measurable set and fix > 0. Since E is measurable, E C
is measurable. By part (i), there exists an open set O containing E C for which m∗ (O ∼ E C ) < .
38
Define F = OC . Then F is closed, contained in E and
m∗ (E ∼ F ) = m∗ (E ∩ F C ) = m∗ (O ∩ E) = m∗ (O ∼ E C ) < • (iii) =⇒ (iv): By part (iii), for any k there exists a closed set Fk contained in E that satisfies
S∞
m∗ (E ∼ Fk ) < k1 . Define F = k=1 Fk . F is an Fσ set contained in E, and by the monotonicity
of m∗ we have
∗
∗
m (E ∼ F ) = m
∞
\
!
(E ∼ Fk )
≤ m∗ (E ∼ Fk ) <
k=1
1
k
for all k. Thus m∗ (E ∼ F ) = 0.
• (iv) =⇒ Measurability: Since a set of measure zero is measurable, as is an Fσ set, and the
measurable sets are an algebra, the set
E = (E ∼ F ) ∪ F
is measurable.
17. Suppose E is a measurable set and fix > 0. By Theorem 11, there is an open set O containing E and
a closed set F contained in E for which m∗ (O ∼ E) <
2
and m∗ (E ∼ F ) < 2 . By the measurability
of E, we have
m∗ (O ∼ F ) = m∗ (O ∩ F C ∩ E) + m∗ (O ∩ F C ∩ E C )
= m∗ (E ∼ F ) + m∗ (O ∼ E)
<
Now fix > 0 and suppose that there is a closed set F and an open set O for which F ⊆ E ⊆ O and
m∗ (O ∼ F ) < . Then
m∗ (O ∼ E) ≤ m∗ (O ∼ F ) < If such sets can be found for all , E is measurable by Theorem 11(i).
18. Since E has finite outer measure, we can use the argument in Problem 7 to construct a Gδ set G
satisfying E ⊆ G and m∗ (E) = m∗ (G).
According to Theorem 11(iv) and the excise property, E is measurable if and only if we can find an
Fσ set F that satisfies F ⊆ E and m∗ (F ) = m∗ (E).
19. By Theorem 11(i), we know that if E is not measurable there exists an 0 > 0 such that m∗ (O ∼ E) ≥ 0
for any open set O containing E. Since E has finite outer measure, we can find a collection of open
intervals {Ik }∞
k=1 that contains E and satisfies
m∗ (E) >
∞
X
k=1
39
`(Ik ) − 0
Let O =
S∞
k=1 Ik .
Then
m∗ (O) − m∗ (E) ≤
∞
X
`(Ik ) − m∗ (E) < 0 ≤ m∗ (O ∼ E)
k=1
20. Suppose E is measurable. Then
b − a = m∗ ((a, b)) = m∗ ((a, b) ∩ E) + m∗ ((a, b) ∼ E)
(1)
where the first equality follows from Proposition 1 and the second equality from the definition of
measurability.
Conversely, suppose (1) holds for all open, bounded intervals. Fix a set A with finite outer measure.
For any > 0, we can choose a countable collection of nonempty, open, bounded intervals {(ak , bk )}∞
k=1
that covers A and satisfies
∗
m (A) >
∞
X
(bk − ak ) − (2)
k=1
We then have
m∗ (A) >
∞
X
(m∗ ((ak , bk ) ∩ E) + m∗ ((ak , bk ) ∼ E)) − k=1
≥ m∗
∞
[
!
(ak , bk ) ∩ E
+ m∗
k=1
∞
[
!
(ak , bk ) ∼ E
−
k=1
≥ m∗ (A ∩ E) + m∗ (A ∼ E) − where the first inequality follows from applying (1) to (2), the second inequality follows from the
countable subadditivity of m∗ , and the third inequality follows the monotonicity of m∗ . Since this
expression must hold for all , we conclude that m∗ (A) ≥ m∗ (A ∩ E) + m∗ (A ∼ E). Thus E is
measurable.
21. Let E1 and E2 be two sets of real numbers. Suppose we can find Gδ sets G1 and G2 that satisfy
E1 ⊆ G1 and m∗ (G1 ∼ E1 ) = 0
E2 ⊆ G2 and m∗ (G2 ∼ E2 ) = 0
Write G1 =
T∞
k=1
G1,k and G2 =
T∞
k=1
∞
G2,k , where {G1,k }∞
k=1 and{G2,k }k=1 are countable collections
∞
of open sets. Without loss, assume that {G1,k }∞
k=1 and {G2,k }k=1 are decreasing sequences (otherwise,
T∞
Tk
Tk
use the sequences defined by G̃1,k = n=1 G1,n and G̃2,k = n=1 G2,n ). Define G = k=1 (G1,k ∪G2,k ).
Then G1 ∪ G2 ⊆ G, so G is a Gδ set containing E1 ∪ E2 . We also have
G ⊆ lim sup G1,n ∪ lim sup G2,n = G1 ∪ G2
40
where the equality follows because the sequences are decreasing. Therefore G = G1 ∪ G2 , so
m∗ (G ∼ (E1 ∪ E2 )) = m∗ ((G1 ∪ G2 ) ∼ (E1 ∪ E2 ))
≤ m∗ (G1 ∼ (E1 ∪ E2 )) + m∗ (G2 ∼ (E1 ∪ E2 ))
≤ m∗ (G1 ∼ E1 ) + m∗ (G2 ∼ E2 )
=0
Now suppose we can find Fσ sets F1 and F2 that satisfy
F1 ⊆ E1 and m∗ (E1 ∼ F1 ) = 0
F2 ⊆ E2 and m∗ (E2 ∼ F2 ) = 0
S∞
∞
F1,k and F2 = k=1 F2,k , where {F1,k }∞
k=1 and{F2,k }k=1 are countable collections of
S∞
closed sets. Define F = k=1 (F1,k ∪ F2,k ) = F1 ∪ F2 . Then F is an Fσ set contained in E1 ∪ E2 and
Write F1 =
S∞
k=1
m∗ (E1 ∪ E2 ∼ F ) = m∗ (E1 ∪ E2 ∼ (F1 ∪ F2 ))
≤ m∗ (E1 ∼ (F1 ∪ F2 )) + m∗ (E2 ∼ (F1 ∪ F2 ))
≤ m∗ (E1 ∼ F1 ) + m∗ (E2 ∼ F2 )
=0
22. Let A denote a set of real numbers. Suppose {Ik }∞
k=1 is a countable collection of bounded, open
intervals that contains A. Then
!
∞
∞
[
X
Ik ≥ m∗∗ (A)
`(Ik ) ≥ m∗
k=1
k=1
so m∗ (A) ≥ m∗∗ (A).
If m∗∗ (A) = ∞, then m∗∗ (A) ≥ m∗ (A) trivially. Suppose m∗∗ (A) < ∞. Then for any , we can find
an open set O containing A such that
m∗∗ (A) > m∗ (O) − ≥ m∗ (A) − Since this inequality holds for all , m∗∗ (A) ≥ m∗ (A). We therefore conclude that the two set functions
are equal.
23. Claim 1: Suppose A is a bounded set of real numbers. Then m∗∗∗ (A) = m∗ (A) if and only if A is
measurable.
Proof: Suppose m∗∗∗ (A) = m∗ (A). For any > 0, we can find a closed set F ⊆ A such that
m∗∗∗ (A) < m∗ (F ) + . But the excision property then implies
m∗ (A ∼ F ) = m∗ (A) − m∗ (F ) = m∗∗∗ (A) − m∗ (F ) < Since was arbitrary, we can apply Theorem 11(iii) to conclude that A is measurable.
41
Conversely, suppose A is measurable. Fix > 0. By Theorem 11(iii), there exists a closed set F
contained in A for which m∗ (A ∼ F ) < . Applying the excision property, we have
m∗ (A) < m∗ (F ) + ≤ m∗∗∗ (A) + Since this expression holds for all , m∗ (A) ≤ m∗∗∗ (A). Since m∗ (F ) ≤ m∗ (A) for all F ⊆ A, we also
have m∗∗∗ (A) ≤ m∗ (A). We therefore conclude that m∗ (A) = m∗∗∗ (A).
Claim 2: Suppose A is an unbounded set of real numbers. Then m∗∗∗ (A ∩ I) = m∗ (A ∩ I) for every
bounded interval I if and only if A is measurable.
Proof: Suppose m∗∗∗ (A ∩ I) = m∗ (A ∩ I) for every bounded interval I. Let In = (−n, n). By Claim
S∞
1, A ∩ In is measurable for all n. Therefore A = n=1 (A ∩ In ) is measurable by Proposition 7.
Conversely, suppose A is measurable. Pick any bounded interval I. By Proposition 8, I is measurable.
Because the set of measurable sets is closed under finite intersections, A ∩ I is measurable. Claim 1
then implies m∗∗∗ (A ∩ I) = m∗ (A ∩ I).
2.5
Countable Additivity, Continuity, and the Borel-Cantelli Lemma
24. Using the identities
E1 = (E1 ∼ E2 ) ∪ (E1 ∩ E2 )
E2 = (E2 ∼ E1 ) ∪ (E1 ∩ E2 )
E1 ∪ E2 = (E1 ∼ E2 ) ∪ (E2 ∼ E1 ) ∪ (E1 ∩ E2 )
and the finite additivity of Lebesgue measure, we obtain
m(E1 ∪ E2 ) + m(E1 ∩ E2 ) = m(E1 ∼ E2 ) + m(E1 ∩ E2 ) + m(E2 ∼ E1 ) + m(E1 ∩ E2 )
= m(E1 ) + m(E2 )
25. Consider the sequence of sets {Ek }∞
k=1 defined by Ek = (k, ∞). Then m(Ek ) = ∞ for all k and
T∞
k=1 Ek = ∅, so
!
∞
\
0=m
Ek 6= lim m(Ek ) = ∞
k→∞
k=1
26. By countable subadditivity of m∗ , we know that
m
∗
A∩
∞
[
k=1
If m∗ (A ∩
S∞
k=1
!
Ek
∗
=m
∞
[
!
(A ∩ Ek )
k=1
≤
∞
X
m∗ (A ∩ Ek )
k=1
Ek ) = ∞, the reserve inequality holds trivially. If m∗ (A ∩
42
S∞
k=1
Ek ) < ∞, we can use
the result from Problem 7 to find a Gδ set G that contains A ∩
∗
∗
∞
[
A∩
m (G) = m
S∞
k=1
Ek and satisfies
!
Ek
k=1
Since G is measurable, {G ∩ Ek }∞
k=1 is a countable collection of measurable, disjoint sets. We therefore
have
∗
m
∞
[
A∩
!
= m∗ (G)
Ek
k=1
≥m
∗
∞
[
G∩
!
Ek
k=1
∞
[
= m∗
!
(G ∩ Ek )
k=1
=
≥
∞
X
k=1
∞
X
m∗ (G ∩ Ek )
m∗ (A ∩ Ek )
k=1
where the inequalities follow from the monotonicity of m∗ .
27.
(i) Let {Mk }nk=1 be a finite collection of disjoint sets in M0 . Define Mk = ∅ for k = n + 1, n + 2, · · ·
S∞
0
and consider the countable collection {Mk }∞
k=1 . Then
k=1 Mk ∈ M and
m
0
n
[
!
0
Mk
=m
k=1
∞
[
!
Mk
=
∞
X
m0 (Mk ) =
∞
X
m0 (Mk ) +
m0 (Mk ) =
n
X
m0 (Mk )
k=1
k=n+1
k=1
k=1
k=1
n
X
where the second equality follows by countable additivity and the final equality follows because
m0 (∅) = 0. Thus m0 is finitely additive.
Let A and B be sets in M0 satisfying A ⊆ B. Then B ∼ A ∈ M0 and
m0 (B) = m0 (A ∪ (B ∼ A)) = m0 (A) + m0 (B ∼ A) ≥ m0 (A)
where the second equality follows from finite additivity and the inequality follows because m0 (B ∼
A) ≥ 0. Thus m0 is monotone.
0
Fix M ∈ M0 and let {Mk }∞
k=1 be a countable collection of sets in M that covers M . Define
S
S
S∞
k−1
∞
M̃0 = ∅ and M̃k = Mk ∼ n=0 Mn . Then M̃k ⊆ Mk for all k, k=1 Mk = k=1 M̃k , and the
collection of sets {M̃k }∞
k=1 is pairwise disjoint. We therefore have
0
0
m (M ) ≤ m
∞
[
!
Mk
0
=m
k=1
∞
[
!
M̃k
=
k=1
∞
X
k=1
∞
X
m0 M̃k ≤
m0 (Mk )
k=1
where the first inequality follows from monotonicity, the second equality follows from countable
additivity, and the final inequality follows monotonicity.
43
Let A and B be sets in M0 satisfying A ⊆ B. Suppose m0 (B) < ∞. By finite additivity,
m0 (B) = m0 (A) + m0 (B ∼ A)
so m0 (B ∼ A) = m0 (B) − m0 (A).
(ii) We can follow the proof of Theorem 15 in the text with m0 replacing m.
28. Let M0 be a σ-algebra of subsets of R and let m0 be a set function on M0 which takes values in [0, ∞].
0
Assume m0 is finitely additive and that if {Mk }∞
k=1 is an ascending collection of sets in M , then
m
∞
[
0
!
Mk
k=1
= lim m0 (Mk )
k→∞
Sk
0
∞
Let {Mk }∞
k=1 be a collection of disjoint sets in M . Define M̃k =
n=1 Mn . Then {M̃k }k=1 is an
S
S
∞
∞
ascending collection of sets in M0 and k=1 Mk = k=1 M̃k . We therefore have
∞
[
0
m
!
Mk
=m
0
k=1
∞
[
k=1
!
M̃k
= lim m0 (M̃k ) = lim
k→∞
k→∞
k
X
m0 (Mn ) =
n=1
∞
X
m0 (Mk )
k=1
where the second equality follows from the continuity property of m0 and the third equality follows
from the finite additivity of m0 .
2.6
Nonmeasurable Sets
29. Let ∼ denote the rational equivalence relation.
(i) Let x be a real number. Then x − x = 0 ∈ Q, so x ∼ x. Therefore ∼ is reflexive.
Suppose x ∼ y. Then x−y is a rational number, so y−x = (−1)(x−y) is a rational number because
−1 is a rational number and the rational numbers are closed under multiplication. Therefore ∼
is symmetric.
Suppose x ∼ y and y ∼ z. Then x − y and y − z are rational numbers, so x − z = x − y + (y − z)
is a rational number because the rational numbers are closed under addition. Therefore ∼ is
transitive.
(ii) Since the difference between any two rational numbers is rational, all the elements of Q belong
to a single equivalence class. Thus the singleton set {q} for any q ∈ Q is a valid choice set for ∼
on Q.
(iii) Since x is not irrationally equivalent to itself, irrational equivalence is not reflexive. Therefore
irrational equivalence is not an equivalence relation on R or Q.
30. Let E be a nonempty set of real numbers and let CE denote a choice set for the rational equivalence
relation on E. For any x ∈ CE , let Ex ⊆ E denote the equivalence class to which x belongs. Then
Ex ⊆ {x + q : q ∈ Q}
44
for all x ∈ CE . Since Q is countable, Ex is countable by Theorems 3 and 5 of Chapter 1. If CE is
countable, then
[
E=
Ex
x∈CE
is countable by Corollary 6 of Chapter 1. A countable set has outer measure equal to zero (see page
31). Therefore if E has positive outer measure, CE must be uncountably infinite.
31. It suffices to only consider bounded sets because if a set has positive outer measure, it contains a
bounded subset with positive outer measure (Problem 14).
32. Define E = (0, 1) and Λ = {1, 2}. Then E is a bounded, measurable set of real numbers, Λ is a finite
set of real numbers, and {λ + E}λ∈Λ is disjoint. If we take Λ = N, then Λ is a countably infinite,
unbounded set of real numbers, and {λ + E}λ∈Λ is disjoint. However, in both cases m(E) = 1 6= 0.
Therefore the lemma does not remain true if Λ is allowed to be finite or unbounded.
If Λ is uncountably infinite and bounded, we can find a countable subset of Λ that satisfies the conditions
of the lemma. Thus the conclusion of the lemma still holds.
33. By Problem 7, we can find a Gδ set G that contains E and satisfies m∗ (G) = m∗ (E). If m∗ (G ∼ E) = 0,
then G ∼ E is measurable. But if G ∼ E is measurable, then E = G ∩ (G ∼ E)C is measurable.
Therefore if E is not measurable, m∗ (G ∼ E) > 0.
2.7
The Cantor Set and the Cantor-Lebesgue Function
34. Consider the function f (x) =
ψ(x)
2 ,
where ψ : [0, 1] → [0, 2] is defined in Proposition 21. f is continuous
and strictly increasing because it is the composition of two continuous, strictly increasing functions.
The range of f is [0, 1] by the Intermediate Value Theorem. Define O = [0, 1] ∼ C. Through a minor
adaptation of the argument in the proof of Proposition 21, we see that f (O) and f (C) are disjoint and
m(f (O)) = 21 . But since [0, 1] = f ([0, 1]) = f (O) ∪ f (C), we must have m(f (C)) =
1
2
by the additivity
of m. Therefore the inverse of f is a continuous, strictly increasing function defined on [0, 1] that maps
f (C) (a set of positive measure) onto the Cantor set (a set of measure zero).
35. Suppose f is continuous at x0 ∈ I. Since I is open, there exists r > 0 such that (x0 − r, x0 + r) ⊆ I.
Pick sequences bn ∈ (x0 , x0 + r) and an ∈ (x0 − r, x0 ) that converge to x0 . Then an < x0 < bn , the
sequences {an } and {bn } are in I, and limn→∞ [f (bn ) − f (an )] = 0 by Proposition 21 and Theorem 18
of Chapter 1.
Conversely, suppose there are sequences {an } and {bn } in I such that for each n, an < x0 < bn
and limn→∞ [f (bn ) − f (an )] = 0. Fix > 0 and choose n such that |f (bn ) − f (an )| < . Define
δ = min{x0 − an , bn − x0 }. If |x − x0 | < δ, then
an < x0 − δ < x < x0 + δ < bn
which implies
f (an ) ≤ f (x) ≤ f (bn )
45
because f is increasing. But this means
f (an ) − f (bn ) ≤ f (x) − f (x0 ) ≤ f (bn ) − f (an )
which implies
|f (x) − f (x0 )| ≤ |f (bn ) − f (an )| < 36. Define O = [0, 1] ∼ C and fix any point x ∈ (0, 1) ∩ C. Then ϕ(t) = f (t) ≤ f (x) for any t ∈ O ∩ [0, x).
Therefore ϕ(x) ≤ f (x). To prove the reverse inequality, fix > 0. Since ϕ is continuous, there exists
a δ > 0 such that ϕ(t) < ϕ(x) + for all t ∈ [0, 1] satisfying |x − t| < δ. Since m(C) = 0, C cannot
contain the open interval (x, x + δ). Therefore there exists t ∈ O ∩ (x, x + δ) such that
f (x) ≤ f (t) = ϕ(t) < ϕ(x) + But since f (x) ≤ ϕ(x) + for arbitrary positive , f (x) ≤ ϕ(x).
Note that under the given assumptions, f may not agree with ϕ at 0 or 1.
37. The strictly increasing, continuous function function ψ defined in Proposition 21 maps a measurable
set onto a non-measurable set. Since ψ is one-to-one, the inverse of ψ is a continuous function that
maps a non-measurable set onto a measurable set. Thus the pre-image of a measurable set need not
be measurable.
38. Let E ⊆ [a, b] be a non-empty set of measure zero. Fix > 0. By Theorem 11(a), there exists an open
set O containing E for which m(O) = m(O ∼ E) < c . By Proposition 9 of Chapter 1, we can write O
as the union of a countable collection of pairwise disjoint open intervals {Ik }∞
k=1 . For any interval Ik ,
define Ik0 = Ik ∩ [a, b], ak = inf u∈Ik0 f (u) and bk = supu∈Ik0 f (u). We then have
m(f (Ik0 )) ≤ bk − akk = sup |f (u) − f (v)| ≤ c sup |u − v| ≤ c `(Ik )
u,v∈Ik0
u,v∈Ik0
where the first inequality follows because f (Ik0 ) ⊆ (ak , bk ) and the second inequality follows from the
Lipschitz condition. We therefore have
m(f (E)) ≤ m(f (O ∩ [a, b]))
!!
∞
[
=m f
Ik0
k=1
=m
∞
[
!
f
(Ik0 )
k=1
≤
∞
X
m(f (Ik0 ))
k=1
∞
X
≤c
`(Ik )
k=1
= c m(O)
<
46
Since this expression holds for any positive , we must have m(f (E)) = 0.
Before proceeding to the next step, I prove a preliminary lemma.
Claim: A continuous image of a closed and bounded set of real numbers is also closed and bounded.
Proof: Let A be a closed and bounded set of real numbers and let f be a continuous function on A.
−1
Let {Ek }∞
(Ek ) = A ∩ Uk where Uk
k=1 be an open cover of f (A). By Proposition 22 of Chapter 1, f
is an open set. Then
∞
[
f
−1
(Ek ) = f
∞
[
−1
k=1
!
Ek
⊇ f −1 (f (A)) ⊇ A
k=1
∞
Therefore {Uk }∞
k=1 is an open cover of A. By the Heine-Borel Theorem, {Uk }k=1 contains a finite
subcover {Uk }N
k=1 . But then
f (A) = f
N
[
!
(A ∩ Uk )
=f
f
−1
k=1
N
[
!!
Ek
⊆
k=1
N
[
Ek
k=1
Therefore {Ek }N
k=1 is a finite cover of f (A). Since every open cover of f (A) contains a finite subcover,
f (A) is closed and bounded by Problem 53 of Chapter 1.
Now suppose E ⊆ [a, b] is an Fσ set. Then there exist a countable collection of closed sets {Fk }∞
k=1 such
S∞
that E = k=1 Fk . By the preliminary lemma and the fact that Lipschitz functions are continuous
S∞
(see page 25), f (Fk ) is closed and bounded for all k. Thus f (E) = k=1 f (Fk ) is an Fσ set.
Now suppose E ⊆ [a, b] is measurable. By Theorem 11(iv), we can find an Fσ set F contained in E
for which m(E ∼ F ) = 0. But since f (E) = f (F ) ∪ f (E ∼ F ) and both f (F ) and f (E ∼ F ) are
measurable, f (E) is measurable.
k
39. Let Fk denote the set of points that remain
operations. Fk is the union of 2 disjoint
after k removal
k
. Since a finite union of closed sets is closed,
closed intervals, each of length lk := 2−k 1 − α + α 23
T∞
each Fk is closed. Since an intersection of closed sets is closed and F = k=1 Fk , F is closed. Define
O = [0, 1] ∼ F and pick x, y ∈ [0, 1]. Since O is open, if either x or y belongs to O we can find a point
between x and y which also belongs to O. So suppose both x and y belong to F . Choose k ∈ N such
S2k
that lk < |x − y|. Write Fk = n=1 In , where each In is a closed interval of length lk . Since x and
k
y both belong to Fk , they must belong to one of the intervals in {In }2n=1 . However x and y cannot
belong to the same interval since lk < |x − y|. Since the intervals are disjoint, there must exist a point
between x and y which is not in Fk and therefore not in F . Thus O is dense in [0, 1]. Finally, observe
that O is the countable union of the disjoint collection of open intervals which are removed during the
construction of F . At the k-th deletion stage, 2k−1 intervals of length α3−k are removed. Therefore
the length of O is given by
∞
αX
m (O) =
2
k=1
k
2
=α
3
which implies m(F ) = 1 − α by the excision property.
40. Let F denote a generalized Cantor set (see Problem 39) of measure 1 − α. Define O = [0, 1] ∼ F . Since
O is open, it does not contain any of its boundary points. But since O is dense in [0, 1], every point in
F is a boundary point of O. Thus the boundary of O equals F and therefore has measure 1 − α.
47
41. Suppose x ∈ C and fix > 0. Choose k such that 3−k < and let Ck denote the points that remain
after k removal operations. Ck is the union of a pairwise disjoint collection of 2k closed intervals, each of
which has length 3−k . Since x ∈ Ck , there exists an interval I = [a, b] in this collection that contains x.
The endpoints of the intervals in the collection are never removed in the construction of the Cantor set,
so both a and b belong to C. Furthermore, both a and b belong to (x − , x + ) since I ⊆ (x − , x + ).
But x cannot equal both a and b, so one of the endpoints is in the set C ∩ (x − , x + ) ∼ {x}. This
process can be repeated for all k 0 > k to generate an infinite collection of points in C ∩ (x − , x + ).
Since was arbitrary and C is closed, C is perfect.
42. Suppose X is countable. Let {xk }∞
k=1 denote an enumeration of X. Define n1 = 1 and U1 = (xn1 −
1, xn1 + 1). Since X is a perfect set, U1 ∩ X is infinite. Therefore we can define n2 = min{k : xk ∈
U1 ∼ {xn1 }}. Let U2 be an open interval satisfying
xn2 ∈ U2
xn1 6∈ Ū2
Ū2 ⊂ U1
By continuing in this fashion, we obtain a sequence of sets {Un }∞
n=1 that satisfies
xk ∈
/ Ūn+1 for k = 1, · · · , n
(1)
Un ∩ X 6= ∅ for n ∈ N
Ūn+1 ⊂ Un for n ∈ N
Let En = Ūn ∩X. Then {En }∞
n=1 is a decreasing sequence of non-empty closed sets, and E1 is bounded.
T∞
T∞
T∞
Therefore by the Nested Set Theorem, n=1 En is non-empty. Since n=1 En ⊆ X, xk ∈ n=1 En for
some k. But then xk ∈ Ūk+1 , a contradiction with the condition in (1).
43. This conclusion follows immediately from Problems 41 and 42.
44. Observe that C cannot contain an open interval. For if we could find an open interval I ⊆ C, then
m(C) ≥ m(I) > 0 by the monotonicity of measure. However this expression contradicts the result
from Proposition 19 that m(C) = 0.
Now fix any interval I. By the previous argument, there must exist x ∈ I ∼ C. Since C is closed,
I ∼ C is open. Therefore, we can find an interval Ix that contains x and is contained in I ∼ C. But
then Ix is an open interval contained in I that is disjoint from C. Since any open set contains an open
interval, the proof is complete.
45. Suppose A is an interval and f is a strictly increasing function defined on A. Define B = f (A). Since
f is strictly monotone, it defines a one-to-one correspondence between the sets A and B. Therefore
the inverse function f −1 : B → A is well-defined.
Now fix y ∈ B and define the function
fy (α) = f f −1 (y) + α
for α ∈ A − f −1 (y). Note that fy (0) = y, fy is strictly increasing in α, and fy−1 (y 0 ) = f −1 (y 0 ) − f −1 (y)
48
for y 0 ∈ B.
Fix > 0. If f −1 (y) is on the interior of A, we can find 0 ∈ (0, ) such −0 , 0 ∈ A − f −1 (y). Define
δ = min{y − fy (−0 ), fy (0 ) − y}
If f −1 (y) is a lower (upper) bound of A, choose 0 ∈ (0, ) such that 0 ∈ A − f −1 (y) (−0 ∈ A − f −1 (y))
and define δ = fy (0 ) − y (δ = y − fy (−0 )). Now pick y 0 ∈ B satisfying |y − y 0 | < δ. If y 0 ≥ y, then
fy (fy−1 (y 0 )) − y = |y − y 0 | < δ ≤ fy (0 ) − y
=⇒
fy−1 (y 0 ) < y − fy (fy−1 (y 0 )) = |y − y 0 | < δ ≤ y − fy (−0 )
=⇒
fy−1 (y 0 ) > −
If y 0 < y, then
Therefore |fy−1 (y 0 )| = |f −1 (y 0 ) − f −1 (y)| < .
46. Define
F = {E ⊆ R : f −1 (E) ∈ B}
Suppose O is an open set. Then f −1 (O) is open (and thus Borel) by Proposition 22 of Chapter 1.
Therefore O ∈ F.
Next observe that F is a σ-algebra:
• f −1 (∅) = ∅ ∈ B, so ∅ ∈ F.
• Suppose E ∈ F. Then f −1 (E C ) = f −1 (E)C ∈ B, since f −1 (E) is in B and B is closed under
complements. Therefore E C ∈ F.
S∞
S∞
• Suppose Ek ∈ F for k ∈ N. Then f −1 ( k=1 Ek ) = k=1 f −1 (Ek ) ∈ B, since f −1 (Ek ) is in B for
S∞
all k and B is closed under countable unions. Therefore k=1 Ek ∈ B.
Since B is the smallest σ-algebra containing the open sets, we must have B ⊆ F. But this implies
f −1 (B) ∈ B for all B ∈ B.
47. Let f be a continuous, strictly increasing function defined on an interval A. Let g denote the inverse
of f . By Problem 45, g is a continuous function. Therefore if B is a Borel set contained in the range
of g, g −1 (B) ∈ B by Problem 46. But g −1 (B) = {y : g(y) ∈ B} = {f (x) : x ∈ B} = f (B), so f (B) is a
Borel set.
3
Lebesgue Measurable Functions
3.1
Sums, Products, and Compositions
1. Suppose f (x) 6= g(x) for some x ∈ [a, b]. Then the set
E := {x ∈ [a, b] : f (x) 6= g(x)} = (f − g)−1 ((−∞, 0) ∪ (0, ∞))
49
is non-empty. Since f and g are continuous functions, f − g is continuous. Therefore E must be open
by Proposition 22 of Chapter 1. But then E must contain an open interval and therefore have non-zero
measure. Since {x ∈ [a, b] : f (x) 6= g(x)} cannot be contained in a set of measure zero, f cannot equal
g a.e. on [a, b].
To see that the statement does not hold for an arbitrary measurable set, suppose f is a continuous
function defined over a set E of measure zero. Let g(x) = f (x) + 1. Then f (x) 6= g(x) for all x ∈ E,
but f = g a.e. on E.
2. If f is continuous on D ∪ E, then f is continuous on its restrictions to D and E. However, the converse
is false. For example, suppose D = [0, 1), E = [1, 2], define f as
f (x) =

1
if 0 ≤ x < 1
2
if 1 ≤ x ≤ 2
3. Let E denote the measurable domain of f and define
E0 = {x ∈ E : f is not continuous at x}
Since E0 is finite, m(E0 ) = 0 and f is measurable on E0 . By Proposition 3, f is measurable on E ∼ E0 .
We conclude from Proposition 5(ii) that f is measurable on E.
4. Let E denote a non-measurable subset of (0, 1). We know such a set exists from Theorem 17 of Chapter
2. Consider the function f defined as
f (x) = ex · (2χE − 1)
where χE is the characteristic function of the set E. Then {x ∈ R : f (x) > 0} = E is not a measurable
set. However f is one-to-one, so f −1 (c) is either empty or a singleton set and therefore measurable.
5. Fix c ∈ R. Then
∞
[
{x ∈ E : f (x) > c} =
{x ∈ E : f (x) > q}
q∈Q : q>c
Since {q ∈ Q : q > c} is a countable set and each {x ∈ E : f (x) > q} is measurable, the above
expression implies {x ∈ E : f (x) > c} is a measurable set.
6. From Proposition 5(ii), we know that g is measurable on R if and only if the restrictions of g to D and
R ∼ D are measurable. But since the restriction of g to D is f and the restriction of g to R ∼ D is
always measurable, the result follows immediately.
7. By Proposition 2, f is measurable if and only if for each open set O, f −1 (O) is measurable. If f −1 (A) is
measurable for each Borel set A, then f is measurable because all open sets are Borel sets. Conversely,
suppose f is measurable. Define F as
F = {A ⊆ R : f −1 (A) is measurable}
Then F is a σ-algebra:
50
• Since f −1 (R) = E is measurable, R ∈ F.
• Suppose A ∈ F. Then f −1 (AC ) = (f −1 (A))C is measurable because f −1 (A) is measurable and
the collection of measurable sets is closed under complementation.
S∞
−1
• Suppose {An }∞
( n=1 An ) =
n=1 is a countable collection of sets satisfying An ∈ F for all n. Then f
S∞
−1
(An ) is measurable because f −1 (An ) is measurable for all n and the collection of mean=1 f
surable sets is closed under countable unions.
Since all open sets are in F and B is the smallest σ-algebra containing the Borel sets, we must have
B ⊆ F. But this implies f −1 (A) is measurable for each Borel set A.
8. The proof of Proposition 1 only relies on the fact that the collection of Lebesgue measurable sets is
a σ-algebra. Since the collection of Borel sets is also a σ-algebra, the statement remains valid. The
proof of Theorem 6 also remains valid if we define a Borel measurable function f on a Borel set E to
be finite almost everywhere if there exists a Borel set B0 ⊆ E of measure zero that satisfies
{x ∈ E : f is finite} ⊇ E ∼ B0
(This slight modification is necessary because not all sets of measure zero are Borel sets.)
(i) This statement follows immediately from the fact that all Borel sets are Lebesgue measurable.
(ii) This statement follows from the fact that the collection of sets {A ⊆ R : f −1 (A) is a Borel set}
is a σ-algebra containing the open sets.
(iii) Observe that for any real number c, we have
(f ◦ g)−1 ((c, ∞)) = g −1 (f −1 ((c, ∞)))
Since (c, ∞) is a Borel set, g −1 (f −1 ((c, ∞))) is a Borel set by part (ii). Therefore f ◦ g is a Borel
measurable function.
(iv) Since f is Borel measurable, f −1 ((c, ∞)) is a Borel set for any real number c. By Problem
7, (f ◦ g)−1 (−∞, c) = g −1 (f −1 ((−∞, c))) = g −1 (f −1 ((c, ∞))) is a Lebesgue measurable set.
Therefore f ◦ g is a Lebesgue measurable function.
9. Since |fn (x) − fm (x)| is a measurable function by Theorem 6 and Proposition 7, each set of the form
x ∈ E : |fn (x) − fm (x)| < k1 is measurable. Because a sequence of points converges if and only if it
is Cauchy (Theorem 17, Chapter 1), we have
E0 =
∞ [
∞
\
\ x ∈ E : |fn (x) − fm (x)| <
k=1 N =1 n,m≥N
1
k
Since the collection of measurable sets is a σ-algebra, E0 is measurable.
10. See the counter-example on page 57.
11. Let h denote the inverse of g. For any open set O, we have
(f ◦ g)−1 (O) = g −1 f −1 (O) = {x : g(x) ∈ f −1 (O)} = {h(y) : y ∈ f −1 (O)} = h f −1 (O)
51
Since f is measurable and h is Lipschitz, f −1 (O) and h(f −1 (O)) are measurable sets by Proposition 3
and Problem 38 of Chapter 2. Thus (f ◦ g)−1 (O) is measurable, so f ◦ g is a measurable function.
3.2
Sequential Pointwise Limits and Simple Approximation
12. For any natural number n, we can use the Simple Approximation Lemma to construct simple functions
ϕn and ψn such that
1
on E
n
ϕn ≤ f ≤ ψn and 0 ≤ ψn − ϕn <
Fix > 0 and pick N > 1 . Then
|f − ϕn | = f − ϕn ≤ ψn − ϕn <
1
1
≤
< on E for all n ≥ N
n
N
|f − ψn | = ψn − f ≤ ψn − ϕn <
1
1
≤
< on E for all n ≥ N
n
N
and
Thus the sequences {ϕn } and {ψn } converge to f uniformly on E. By replacing ϕn with max{ϕ1 , · · · , ϕn }
and ψn with min{ψ1 , · · · , ψn }, we have {ϕn } increasing and {ψn } decreasing.
13. For any natural number n and integer k, define
En,k =
x ∈ E : f (x) ∈
k k+1
,
n
n
Since the function f is bounded on En,k , we can use the Simple Approximation Lemma to find a simple
function ϕn,k such that
|f − ϕn,k | <
1
on En,k
n
Extend each ϕn,k to E by setting ϕn,k (x) = 0 if x ∈
/ En,k . Let {ki } denote an enumeration of the
integers and define
ϕn =
∞
X
ϕn,ki
i=1
The range of ϕn is a countable union of finite sets of real numbers and is therefore countable (see
PN
i=1 ϕn,ki (x), ϕn is the pointwise limit of a
Corollary 6 of Chapter 1). Since ϕn (x) = limN →∞
sequence of measurable functions and therefore measurable by Proposition 9. We can conclude that
ϕn is a semisimple function on E.
Now fix > 0. Choose x ∈ E and N > 1 . For any n ≥ N , there exists k such that
k
n
< f (x) ≤
k+1
n .
But this means x ∈ Ek,n and x ∈
/ Ek0 ,n for all k 0 6= k, so
|f (x) − ϕn (x)| = |f (x) − ϕn,k (x)| <
1
1
≤
<
n
N
Therefore ϕn converges uniformly to f on E.
14. Define
En = {x ∈ E : |f (x)| < n}
f is bounded on each En and each En is a measurable subset of E. Since f is finite a.e. on E, we know
52
that
m E∼
∞
[
!
= m ({x ∈ E : f (x) is not finite}) = 0
En
n=1
Since m(E ∼ E1 ) ≤ m(E) < ∞ and E ∼ E1 ⊇ E ∼ E2 ⊇ · · · , Theorem 15(ii) of Chapter 2 implies
0=m E∼
∞
[
n=1
!
En
=m
∞
\
!
(E ∼ En )
= lim m(E ∼ En )
n=1
n→∞
Thus for any > 0, we can find an n such that m(E ∼ En ) < .
15. By Problem 14, for any > 0 we can find a measurable set F contained in E such that f is bounded
on F and m(E ∼ F ) < . By Problem 12, we can find a sequence of measurable functions {ϕn } on F
that converges uniformly to f on F . Each simple function can be extended to E by defining ϕn (x) = 0
if x ∈ E ∼ F .
16. By Theorem 12 of Chapter 2, there exists a finite disjoint collection of open intervals {Ik } such that
Sn
Sn
Pn
m (E∆ k=1 Ik ) < . Define O = k=1 Ik , h = χO = k=1 χIk and
F = I ∼ (E∆O) = I ∩ ((E ∩ O) ∪ (E c ∩ Oc ))
If x ∈ I ∩ O ∩ E, then h(x) = 1 = χE (x); if x ∈ I ∩ Oc ∩ E c , then h(x) = 0 = χE (x). Thus h is a step
function on I for which h = χE on F and m(I ∼ F ) = m(E∆O) < .
Pn
17. Let k=1 ck χEk denote the canonical representation of ψ. By Problem 16, we know that for each Ek
we can find a step function hk defined on I and a measurable subset Fk of I for which hk = χEk on
Pn
Tn
Fk and m(I ∼ Fk ) < n . Let h = k=1 ck hk and F = k=1 Fk . If x ∈ F , then hk (x) = χEk (x)
Pn
Pn
for k = 1, · · · , n. Thus h(x) = k=1 ck hk (x) = k=1 ck χEk (x) = ψ(x), so h = ψ on F . We also
Sn
Pn
know that m(I ∼ F ) = m ( k=1 (I ∼ Fk )) ≤ k=1 m(I ∼ Fk ) < by the subadditivity of Lebesgue
measure.
18. By the Simple Approximation Lemma, we can find a simple function ψ defined on E such that |ψ−f | < on E. By Problem 17, we can find a step function h on I and a measurable subset F of I for which
h = ψ on F and m(I ∼ F ) < .
19. Let ϕ and ψ be two simple functions and let n and m denote the finite number of distinct values each
function takes, respectively. Then nm is the maximum number of distinct values ϕ + ψ, ϕψ, min{ϕ, ψ}
or max{ϕ, ψ} can take. Since these compositions are measurable by Theorem 6 and Proposition 8,
they are simple functions.
20. If x ∈ A ∩ B, then χA∩B (x) = χA (x) = χB (x) = 1. If x ∈
/ A ∩ B, then χA∩B (x) = 0 and either
χA (x) = 0 or χB (x) = 0. In either case, χA∩B (x) = χA (x)χB (x).
Suppose A and B are disjoint. If x ∈ A, then χA∪B (x) = 1 = χA (x) and χB (x) = 0. If x ∈ B, then
χA∪B (x) = 1 = χB (x) and χA (x) = 0. If x ∈
/ A ∪ B, then χA∪B (x) = 0 = χA (x) = χB (x). In each
case, χA∪B (x) = χA (x) + χB (x). If A and B are not disjoint, decompose A ∪ B into the disjoint sets
53
A ∩ B C , AC ∩ B, and A ∩ B. Then using our result for disjoint sets, we have
χA∪B = χA∩B C + χB∩AC + χA∩B =
(χA∩B C + χA∩B ) + (χAC ∩B + χA∩B ) − χA∩B = χA + χB − χA∩B
where the final equality follows because A = (A ∩ B) ∪ (A ∩ B C ) and B = (AC ∩ B) ∪ (A ∩ B).
Since A and AC are disjoint and χA∪AC = 1, we have χA + χAC = χA∪AC = 1. Thus χAC = 1 − χA .
21. Fix c ∈ R and suppose inf n fn (x) < c. Then there exists n such that fn (x) < c, for otherwise c
S∞
would be a lower bound of {fn (x)} that exceeds inf n fn (x). Thus x ∈ n=1 {x ∈ E : fn (x) < c}. Now
suppose fn (x) < c for some n. Then inf n fn (x) < c because inf n fn (x) is a lower bound of {fn (x)}.
Therefore x ∈ {x ∈ E : inf n fn (x) < c}. We conclude that
∞
n
o
[
x ∈ E : inf fn (x) < c =
{x ∈ E : fn (x) < c}
n
n=1
Each set {x ∈ E : fn (x) < c} is measurable because each function fn is measurable. Since the collection
of measurable sets is closed under countable unions, the above expression implies {x ∈ E : inf n fn (x) < c}
is measurable.
Now supn fn = − inf n (−fn ), so supn fn is measurable by Theorem 6. Since lim inf n fn = limn inf k≥n fk
and lim supn fn = limn supk≥n fk , lim inf n fn and lim supn fn are measurable by Proposition 9.
22. Fix > 0. For each natural number n, define
En = (−∞, a) ∪ {x ∈ [a, b] | f (x) − fn (x) < } ∪ (b, ∞)
Suppose x ∈ En . If x ∈ (−∞, a) ∪ (b, ∞), then we can find a δ > 0 such that (x − δ, x + δ) ⊆
(−∞, a) ∪ (b, ∞). If x ∈ {x ∈ [a, b] | f (x) − fn (x) < }, pick ˜ such that f (x) − fn (x) < ˜ < . Since f
and fn are continuous on [a, b], there exist δ̃, δ̃n > 0 such that
− ˜
2
− ˜
0
0
0
If x ∈ [a, b] and |x − x | < δ̃n , then |fn (x) − fn (x )| <
2
o
n
Define δ = min δ̃, δ̃n and suppose |x − x0 | < δ. If x0 ∈
/ [a, b], then x0 ∈ (−∞, a) ∪ (b, ∞) ⊆ En . If
If x0 ∈ [a, b] and |x − x0 | < δ̃, then |f (x) − f (x0 )| <
x0 ∈ [a, b], then
f (x0 ) − fn (x0 ) = f (x0 ) − f (x) + f (x) − fn (x) + fn (x) − fn (x0 ) ≤
|f (x0 ) − f (x)| + f (x) − fn (x) + |fn (x) − fn (x0 )| < so that x0 ∈ {x ∈ [a, b] | f (x) − fn (x) < } ⊆ En . Thus (x − δ, x + δ) ⊆ En . We conclude that each
En is open. Since limn fn (x) = f (x), for any x ∈ [a, b] there exists an n such that f (x) − fn (x) < .
Thus {En } is an open cover of [a, b]. By the Heine-Borel Theorem, we can find a finite subcover
K
K
{Enk }K
k=1 . Define N = max{nk }k=1 . Since {Enk }k=1 covers [a, b], we know that for any x ∈ [a, b]
there exists n0 ≤ N such that f (x) − fn0 (x) < . But since the sequence {fn } is increasing, we must
54
have f (x) − fn (x) ≤ f (x) − fn0 (x) < for all n ≥ N . Thus {fn } converges to f uniformly on [a, b].
23. Let f be an extended real-value function on a measurable set E. Observe that f = f + − f − , where
f + = max{f, 0} and f − (x) = max{−f, 0}. Since f + ≥ 0 and f − ≥ 0, the proof for the non-negative
case implies the existence of sequences of non-negative simple functions {ϕn }, {ψn } on E that satisfy
{ϕn } ↑ f + pointwise on E
{ψn } ↑ f − pointwise on E
Let φn = ϕn − ψn . By Problem 19, φn is a simple function on E for all n. If f (x) ≥ 0, then
f − (x) = 0. Thus ψn (x) = 0 for all n and φn (x) = ϕn (x) → f + (x) = f (x). Likewise, if f (x) < 0 then
φn (x) = −ψn (x) → −f − (x) = f (x). Therefore φn converges pointwise to f on E. We also have
|φn | ≤ ϕn + ψn ≤ f + + f − = |f |
on E for all n.
24. For each natural number n, define gn (x) = f (x) + x/n. Fix a real number c and define
A = {x ∈ I : gn (x) > c}
Suppose x ∈ A and x0 ∈ A and without loss of generality assume that x < x0 . Pick any x00 lying
between x and x0 . Since gn is strictly increasing, we must have c < gn (x) < gn (x00 ) < gn (x0 ). But this
implies x00 ∈ A. Thus the set A is an interval and therefore measurable by Proposition 8 of Chapter 2.
Since gn is measurable for all n and f is the pointwise limit of {gn }, f is measurable by Proposition 9.
3.3 Littlewood’s Three Principles, Egoroff ’s Theorem, and Lusin’s Theorem
25. See Problem 47 of Chapter 1.
26. Pick x ∈ F and fix > 0. Since g is continuous, there exists a δ > 0 such that
if |x0 − x| < δ, then |g(x0 ) − g(x)| < Since f = g on F , we have
if x0 ∈ F and |x0 − x| < δ, then |f (x0 ) − f (x)| < Therefore f is continuous on F .
Let E = R, A = {x ∈ R : x is rational} and f = χA . Then f is real-valued and measurable but not
continuous on E at any point.
27. Let E = R and define fn = χ(n,∞) and f = 0. Let = 1 and choose any closed set F ⊆ E. If F is
bounded, then m(F ) is finite and m(E ∼ F ) = m(E) − m(F ) = ∞ > 1 = . Therefore any closed set
satisfying m(E ∼ F ) < cannot be bounded. So assume F is unbounded and pick any natural number
55
N . Since F is unbounded, there must exist x ∈ F satisfying x > N . But for this x, we have
|fN (x) − f (x)| = 1 ≥ Thus there is no index N such that
|fn − f | < on F for all n ≥ N
Therefore even though {fn } converges pointwise to f on E, {fn } cannot converge uniformly to f on
any closed set F ⊆ E satisfying m(E ∼ F ) < .
28. Assume E has finite measure and suppose {fn } is a sequence of measurable functions on E that
converges pointwise a.e. to a function f that is finite a.e. Define E0 = {x ∈ E : limn→∞ fn (x) = f (x)}
and E1 = {x ∈ E : f (x) is finite}. By assumption, m(E ∼ E0 ) = m(E ∼ E1 ) = 0. Let Ẽ = E0 ∩ E1
and observe that
0 ≤ m E ∼ Ẽ ≤ m(E ∼ E1 ) + m(E ∼ E2 ) = 0
Therefore m(E ∼ Ẽ) = 0.
Fix > 0. By Egoroff’s Theorem, we can find a closed set F contained in Ẽ for which
{fn } → f uniformly on F and m Ẽ ∼ F < But then F is also a closed set contained in E that satisfies
m(E ∼ F ) ≤ m Ẽ ∼ F + m E ∼ Ẽ ∼ F < + m E ∼ Ẽ = 29. Let f be a real-valued measurable function on a set of real numbers E and fix > 0. Let {kn : n ∈ N}
denote an enumeration of the integers and define
En = E ∩ [kn , kn + 1)
Since m(En ) ≤ 1 for all n, we know from the finite case of Lusin’s Theorem that we can find a collection
of continuous functions {gn } on R and a collection of closed sets {Fn } such that
f = gn on Fn , Fn ⊆ En , and m(En ∼ Fn ) < /2n
for all n. Let F =
S∞
n=1
Fn and let
g=
∞
X
χEn gn
n=1
Pick x ∈ F and > 0. Since the sets in {Fn } are closed and disjoint, there must exist an index n and
S
a δ̃ ∈ (0, 1/2) such that the interval I = (x − δ̃, x + δ̃) satisfies I ∩ Fn 6= ∅ and I ∩ n0 6=n Fn0 = ∅.
Otherwise, x would be a limit point (and thus an element of) two sets in {Fn }. Since gn is continuous,
there exists a δ 0 > 0 such that
if |x0 − x| < δ 0 , then |gn (x) − gn (x0 )| < 56
Let δ = min{δ̃, δ 0 }. Notice that if x0 ∈ F and |x − x0 | < δ, then x0 ∈ Fn and g(x0 ) = gn (x0 ). We can
therefore conclude
if x0 ∈ F and |x0 − x| < δ then |g(x) − g(x0 )| < Thus g is continuous on F .
Now suppose {xk } is a sequence in F that converges to x. Then there exists an index n such that
xk ∈ Fn for all k large enough. But since Fn is closed, we must have x ∈ Fn ⊆ F . Thus F contains all
its limit points and is therefore a closed set.
Since g is continuous on F , we can find a continuous extension of g to all of R by Problem 25. To
check that m(E ∼ F ) < , observe that
m(E ∼ F ) = m
∞
[
!
(En ∼ Fn )
≤
n=1
∞
X
m (En ∼ Fn ) = n=1
30. Suppose f is an extended real-valued measurable function on E that is finite a.e. Fix > 0 and let
E0 = {x ∈ E : f is finite}. We know from Lusin’s Theorem that there exists a continuous function g
on R and a closed set F contained in E0 for which
f = g on F and m(E0 ∼ F ) < But since E = (E ∼ E0 ) ∪ E0 and m(E ∼ E0 ) = 0, we also have
m(E ∼ F ) ≤ m(E ∼ E0 ∼ F ) + m(E0 ∼ F ) = m(E0 ∼ F ) < 31. Let {kn : n ∈ N} denote an enumeration of the integers and define
Ẽn = E ∩ [kn , kn + 1)
Since m(Ẽn ) ≤ 1, we know from Egoroff’s Theorem that for any natural number m we can find a
closed set Fn,m contained in Ẽn for which
{fn } → f uniformly on Fn,m and m(Ẽn ∼ Fn,m ) <
Let E0 = E ∼
S∞
m=1
S∞
m E∼
n=1
∞
[
n=1
1
m2n
Fn,m . Observe that
!
Fn,m
=m
∞ [
Ẽn ∼ Fn,m
n=1
!
≤
∞
X
n=1
57
m(Ẽn ∼ Fn,m ) <
1
m
By the continuity of measure, we have
m(E0 ) = m E ∼
N
\
Fn,m
E∼
m=1
≤ lim m E ∼
N →∞
N →∞
!!
n=1
= lim m
≤ lim
Fn,m
E∼
m=1
N →∞
!
m=1 n=1
∞
[
∞
\
=m
∞ [
∞
[
∞
[
!!
Fn,m
n=1
∞
[
!
Fn,N
n=1
1
N
=0
Let g(n) = (g1 (n), g2 (n)) denote a 1-to-1 correspondence from N onto N × N and define En =
S∞
n=0 En , each En is measurable, {fn } → f uniformly on En if n ≥ 1, and
Fg1 (n),g2 (n) . Then E =
m(E0 ) = 0.
4
Lebesgue Integration
4.1
The Riemann Integral
1. Let = 1 and let N denote a natural number. Then
|f (qn ) − fn (qn )| = |1 − 0| = 1 ≥ for all n ≥ N . Thus there is no index N such that |f − fn | < on [0, 1] for all n ≥ N .
2. Let P = {x0 , x1 , · · · , xn } denote a partition of [a, b] and let P 0 denote a refinement of P . Let ni denote
the number of points in P 0 that lie between xi and xi+1 for i = 0, · · · , n − 1. Label the points of P 0 as
P 0 = {x0,0 , x0,1 , · · · , x0,n0 , x1,0 , x1,1 , · · · , x1,n1 , · · · , xn−1,0 , xn−1,1 , · · · , xn−1,nn−1 , xn,0 }
where
xi = xi,0 < xi,1 < · · · < xi,ni < xi+1,0 = xi+1 ,
i = 0, · · · , n − 1
Define
mi = inf{f (x) | xi < x < xi+1 }
Mi = sup{f (x) | xi < x < xi+1 }
mi,ni = inf{f (x) | xi,ni < x < xi+1,0 }
Mi,ni = sup{f (x) | xi,ni < x < xi+1,0 }
58
and
mi,j = inf{f (x) | xi,j < x < xi,j+1 },
j = 0, · · · , ni − 1
Mi,j = sup{f (x) | xi,j < x < xi,j+1 },
j = 0, · · · , ni − 1
if ni ≥ 1. Since (xi,ni , xi+1,0 ) ⊆ (xi , xi+1 ), we know that
mi,ni ≥ mi and Mi,ni ≤ Mi
Likewise,
mi,j ≥ mi and Mi,j ≤ Mi for j = 0, · · · , ni − 1
if ni ≥ 1. Also observe that xi+1 − xi = xi+1,0 − xi,ni if ni = 0 and
xi+1 − xi = xi+1,0 − xi,ni +
nX
i −1
(xi,j+1 − xi,j )
(1)
j=0
if ni ≥ 1. We therefore have
L(f, P ) =
n−1
X
mi (xi+1 − xi )
i=0
=
n−1
X
mi (xi+1,0 − xi,ni ) +
n−1
X
mi (xi,j+1 − xi,j )
i: ni ≥1 j=0
i=0
≤
i −1
X nX
mi,ni (xi+1,0 − xi,ni ) +
i −1
X nX
mi,j (xi,j+1 − xi,j )
i: ni ≥1 j=0
i=0
= L(f, P 0 )
and
U (f, P ) =
n−1
X
Mi (xi+1 − xi )
i=0
=
n−1
X
Mi (xi+1,0 − xi,ni ) +
n−1
X
Mi (xi,j+1 − xi,j )
i: ni ≥1 j=0
i=0
≥
i −1
X nX
Mi,ni (xi+1,0 − xi,ni ) +
i −1
X nX
Mi,j (xi,j+1 − xi,j )
i: ni ≥1 j=0
i=0
= U (f, P 0 )
3. Let P and P 0 be two partitions of [a, b]. Define P 00 to be a refinement of both P and P 0 . It is apparent
from the definitions of Darboux sums that L(f, P 00 ) ≤ U (f, P 00 ). Combining this observation with the
result from Problem 2, we can conclude
L(f, P ) ≤ L(f, P 00 ) ≤ U (f, P 00 ) ≤ U (f, P 0 )
59
Thus U (f, P 0 ) is an upper bound of the set {L(f, P ) | P a partition of [a, b]}, so U (f, P 0 ) must exRb
Rb
ceed (R) a f = sup{L(f, P ) | P a partition of [a, b]}. But then (R) a f is a lower bound of the
Rb
Rb
set {U (f, P ) | P a partition of [a, b]}, which means (R) a f is no greater than (R) a f {U (f, P ) |
P a partition of [a, b]}.
4. Let {Pn0 } and {Pn00 } denote sequences of partitions of [a, b] that satisfy
lim
n→∞
L(f, Pn0 )
b
Z
= (R)
f
and
a
lim
n→∞
U (f, Pn00 )
Z
= (R)
b
f
a
For each n, define Pn to be a refinement of both Pn0 and Pn00 . Then
0 ≤ U (f, Pn ) − L(f, Pn ) ≤ U (f, Pn00 ) − L(f, Pn0 )
where the latter equality follows from Problem 2. Taking limits of this expression, we conclude that
lim [U (f, Pn ) − L(f, Pn )] = 0
n→∞
5. By definition of upper and lower Riemann integrals, we know that
b
Z
Z
f ≤ U (f, Pn )
(R)
and
a
b
f ≥ L(f, Pn )
(R)
a
for all n. Therefore
Z
0 ≤ (R)
b
Z
f − (R)
b
f ≤ U (f, Pn ) − L(f, Pn )
a
a
where the first inequality follows from Problem 3. Taking limits, we can conclude that
Z
b
Z
f − (R)
0 = (R)
a
b
Z
f
=⇒
a
(R)
b
Z
f = (R)
a
b
f
a
Therefore f is Riemann integrable.
6. Since f is a continuous function on the closed, bounded interval [a, b], we know by the Extreme Value
Theorem that f is bounded. For each natural number n, define Pn = {x0,n , x1,n , · · · , xn,n } to be a
partition of [a, b] into n subintervals of length (b − a)/n. Fix > 0. By Theorem 23 of Chapter 1, there
exists a δ > 0 such that for all x, x0 ∈ [a, b],
if |x − x0 | < δ, then |f (x) − f (x0 )| <
2(b − a)
Choose an index N that satisfies N > (b − a)/δ and let n denote a natural number greater than N .
Let
Mi,n = sup{f (x) | xi−1,n < x < xi,n }
and
60
mi,n = inf{f (x) | xi−1,n < x < xi,n }
Then
xi,n − xi−1,n
xi−1,n < x < xi,n +
= sup f (x) − f
2
xi,n − xi−1,n
sup f
− f (x) xi−1,n < x < xi,n
2
+
≤
2(b − a) 2(b − a)
=
b−a
Mi,n − mi,n
where the second line follows because |x − (xi,n − xi−1,n )/2| < δ for all x satisfying xi−1,n < x < xi,n .
We therefore have
U (f, Pn ) − L(f, Pn ) =
n
X
b−a
(Mi,n − mi,n )
≤
n
i=1
for all n ≥ N . We can conclude that limn→∞ [U (f, Pn ) − L(f, Pn )] = 0, so f is Riemann integrable
over [a, b] by Problem 5.
7. Since f is a real-valued increasing function on the interval [a, b], we know that f is bounded below by
f (a) and bounded above by f (b). We also know that
sup f (x)
Therefore
i
i−1
<x<
n
n
i
≤f
n
and
inf f (x)
i
i−1
<x<
n
n
≥f
i−1
n
n X
i
i−1
1
f (1) − f (0)
=
0 ≤ U (f, Pn ) − L(f, Pn ) ≤
f
−f
n
n
n
n
i=1
Taking limits of the above expression, we conclude that limn→∞ [U (f, Pn ) − L(f, Pn )] = 0. Thus f is
Riemann integrable over [a, b] by Problem 5.
8. We first check that f is bounded on [a, b]. Since {fn } converges uniformly to f on [a, b], there exists
an index N such that
|f − fn | < 1 on [a, b] for all n ≥ N
(2)
Pick a natural number n that exceeds N . Since fn is bounded, there exists a real number M such that
|fn | ≤ M on [a, b]. If f is unbounded on [a, b], there exists x ∈ [a, b] such that |f (x)| ≥ M + 1. But
then
|f (x) − fn (x)| ≥ |f (x)| − |fn (x)| ≥ M − (M + 1) = 1
which is a contradiction to (2).
Since fn is Riemann integrable, we know from Problem 4 that we can find a partition Pn = {x0,n , x1,n , · · · , xNn ,n }
of [a, b] such that U (fn , Pn ) − L(fn , Pn ) < 1/n. Define
Mi,n = sup{fn (x) | xi−1,n < x < xi,n }
mi,n = inf{fn (x) | xi−1,n < x < xi,n }
0
Mi,n
= sup{f (x) | xi−1,n < x < xi,n }
m0i,n = inf{f (x) | xi−1,n < x < xi,n }
61
for i ∈ {1, · · · , Nn }. Because {fn } converges to f uniformly on [a, b], for any > 0 there exists an
index N such that
and fn (x) < f (x) +
on [a, b]
b−a
b−a
f (x) < fn (x) +
for all n ≥ N . Taking sups and infs of these inequalities over the interval (xi−1,n , xi,n ), we see that
and |m0i,n − mi,n | ≤
b−a
b−a
0
|Mi,n
− Mi,n | ≤
if n ≥ N . We therefore have
|U (f, Pn ) − U (fn , Pn )| ≤
Nn
X
0
Mi,n − Mi,n
· (xi,n − xi−1,n ) ≤ i=1
Nn
X
|L(f, Pn ) − L(fn , Pn )| ≤
mi,n − m0i,n · (xi,n − xi−1,n ) ≤ i=1
for all n ≥ N . We conclude that limn→∞ |U (f, Pn ) − U (fn , Pn )| = limn→∞ |L(f, Pn ) − L(fn , Pn )| = 0.
We also know that
U (f, Pn ) − L(f, Pn ) ≤ |U (f, Pn ) − U (fn , Pn )| + U (fn , Pn ) − L(fn , Pn ) + |L(fn , Pn ) − L(f, Pn )|
1
≤ |U (f, Pn ) − U (fn , Pn )| + + |L(fn , Pn ) − L(f, Pn )|
n
Taking limits of this expression, we see that limn→∞ [U (f, Pn )−L(f, Pn )] = 0. Therefore, f is Riemann
integrable by Problem 5.
Rb
Rb
Since U (f, Pn ) − (R) a f ≤ |U (f, Pn ) − L(f, Pn )| → 0 and U (fn , Pn ) − (R) a fn ≤ |U (fn , Pn ) − L(fn , Pn )| ≤
1/n → 0, we also have
Z
b
Z
f − (R)
(R)
a
b
Z
Z
f − U (f, Pn ) + |U (f, Pn ) − U (fn , Pn )| + U (fn , Pn ) − (R)
fn ≤ (R)
a
b
b
fn
a
a
→0
Therefore,
Z
(R)
b
Z
f = lim (R)
n→∞
a
b
fn
a
4.2 The Lebesgue Integral of a Bounded Measurable Function over a Set
of Finite Measure
9. Since any set contained in a set of measure zero is measurable, {x ∈ E : f (x) ≤ c} ⊆ E is measurable
for any real number c. Therefore f is measurable.
Since f is bounded, there exists a real number M such that |f | ≤ M on E. We then have
Z
Z
f ≤
E
Z
|f | ≤
E
M · χE = M · m(E) = 0
E
62
where the first inequality follows from Corollary 7 and the second inequality from Theorem 5. We can
R
therefore conclude E f = 0.
10. Let {ϕn } and {ψn } be two sequences of simple functions that satisfy
Z
Z
ϕn ≤ f · χA ≤ ψn on E and lim
Z
f · χA = lim
ϕn =
n→∞
E
n→∞
E
ψn
E
Then ϕn ≤ f ≤ ψn on A, so we must have
Z
Z
Z
ϕn ≤
f≤
A
A
ψn
A
R
for all n. Taking limits of this expression, we see that
A
f=
R
E
f · χA .
11. See Dirichlet’s Function from Section 4.1 for a counter-example.
12. There exists a set E0 ⊆ E such that m(E0 ) = 0 and f = g on E ∼ E0 . We therefore have
Z
Z
Z
f=
E
Z
f+
E∼E0
Z
f=
E0
f=
E∼E0
Z
g=
E∼E0
Z
g+
E∼E0
Z
g=
E0
g
E
where the first and last equality follow from Corollary 6, the second and fourth equalities follow because
R
R
f = E0 g = 0 by Problem 9, and the third equality follows because f = g on E ∼ E0 .
E0
13. See the counter-example on page 78.
14. If {fn } converges uniformly to f on E, then f is bounded (see the solution to Problem 8). Thus there
exists a real number M such that |f | ≤ M on E. We also know there exists an index N such that
|fn − f | < 1 on E
for all n ≥ N . But then |fn | − |f | ≤ |fn − f | < 1, so |fn | ≤ M + 1 for all n ≥ N . Since each
fn is bounded, for each natural number n there exists a real number Mn such that |fn | ≤ Mn on
E. Define M 0 = max{M1 , M2 , · · · , MN , M + 1}. Then |fn | ≤ M 0 for all n, so {fn } is a uniformly
pointwise bounded sequence on E that converges pointwise to f . The Bounded Convergence Theorem
R
R
then implies limn→∞ E fn = E f .
15. The assertions can be verified by thoroughly reviewing the proofs in the text.
16. For any natural number n, define
En =
1
x ∈ E f (x) >
n
Since f is measurable, En is a measurable set. Because f is non-negative, f ≥ f · χEn ≥
all n. By Theorem 5, we have
Z
Z
f≥
0=
E
Z
f · χEn ≥
E
En
63
1
1
· χEn = · m(En ) ≥ 0
n
n
1
n
· χEn for
We can therefore conclude m(En ) = 0 for all n. Now define
E0 = {x ∈ E : f (x) > 0}
and observe that {En } converges upwards to E0 . By continuity of measure, we have
m(E0 ) = lim m(En ) = 0
n→∞
Since f is non-negative, f = 0 on E ∼ E0 , Therefore f = 0 a.e.
4.3
The Lebesgue Integral of a Measurable Nonnegative Function
17. Let h be a bounded, measurable, non-negative function on E. Then 0 ≤ h ≤ f on E and
R
Problem 9. Therefore E f = 0.
R
E
h = 0 by
18. Let f be a bounded, measurable function on a set E. Suppose E0 and E1 are measurable subsets of E
with finite measure such that f ≡ 0 on E ∼ E0 and E ∼ E1 . Then f ≡ 0 on E0 ∼ E1 and E1 ∼ E0 .
R
R
Since E0 ∼ E1 and E1 ∼ E0 have finite measure, we know that E0 ∼E1 f = E1 ∼E0 f = 0 from the
definition of the Lebesgue integral for bounded, measurable functions over sets of finite measure. We
therefore have
Z
Z
Z
f=
E0
Z
f+
E0 ∼E1
Z
f=
E0 ∩E1
Z
f+
E1 ∼E0
f=
E0 ∩E1
f
E1
where the first and last equality follow by Corollary 6. Since the value of the integral does not depend
on the choice of set of finite measure outside of which f vanishes, the integral of a bounded measurable
function of finite support is well-defined.
19. Let E = [0, 1] and D = (0, 1]. Since f is a monotonic function defined on D, f is measurable on E by
Propositions 4 and 5 of Chapter 3.
Let {fn } denote the sequence of step functions defined by
fn =
n
n2
−1
X
−n
2−αk2
χ(2−(k+1)2−n ,2−k2−n ]
k=0
If α ≥ 0, then {fn } is a uniformly bounded sequence of measurable functions on E converging pointwise
to f . If α < 0, then {fn } is an increasing sequence of non-negative measurable functions on E
converaging pointwise to f . Therefore by the Bounded and Monotone Convergence Theorems, we
R
R
know that E fn → E f . Since
i −n
−n
−n
−n
m 2−(k+1)2 , 2−k2
= 1 − 2−2
2−k2
we have
Z
−2−n
fn = 1 − 2
E
n
−1 n2X
k=0
64
−n
2−(1+α)2
k
If α 6= −1, we can use the geometric summation formula to obtain
1 − 2−(1+α)n
−n
fn = 1 − 2−2
·
1 − 2−(1+α)2−n
E
Z
(See Problem 15(iii) of Chapter 1.) Using L’Hôpital’s rule, it is straight-forward to show that
Z
lim
n→∞
fn =
E


1
1+α
if α > −1
∞
if α < −1
If α = −1, we have
Z
−n
fn = 1 − 2−2
n2n → ∞
E
We can therefore conclude

∞
f=
 1
E
if α ≤ −1
Z
if α > −1
1+α
R
20. It is necessary and sufficient to show that
E
h ≤ M for all bounded, measurable functions of finite
support h that satisfy 0 ≤ h ≤ f on E. Let h denote such a function. For each natural number n,
define
hn = min{h, fn } on E
R
R
R
Since hn ≤ fn on E, we know that E hn ≤ E fn ≤ M for all n. Therefore E hn = limn→∞ E hn ≤ M ,
R
where the first equality is proven in the text.
To see that this result implies Fatou’s Lemma, suppose lim inf
R
E
fn < ∞. Let {fnk } denote a subse-
quence of {fn } that satisfies
Z
Z
fnk = lim inf
lim
k→∞
fn
E
E
R
R
For any > 0, there exists an index K such that E fnk ≤ lim inf E fn + for all k ≥ K. The above
R
R
result then implies E f ≤ lim inf E fn + . Since can be made arbitrarily small, we can conclude
R
R
that E f ≤ lim inf E fn .
R
To see that Fatou’s Lemma implies this result, suppose there exists a real number M such that fn ≤
R
R
R
M for all n. Then lim inf fn ≤ M , so by Fatou’s Lemma we know that f ≤ lim inf fn ≤ M .
21. By the definition of the Lebesgue integral, we can find a bounded, measurable function of finite support
g that satisfies 0 ≤ g ≤ f on E and
Z
Z
f−
g>
E
E
2
Let E0 = {x ∈ E : g(x) 6= 0} and note that m(E0 ) < ∞. By the definition of the upper Lebesgue
integral for bounded functions on a set of finite measure, we can find a simple function η that satisfies
η ≤ g on E0 and
Z
Z
g−
η>
E0
E0
2
Since it is always possible to replace η with max{0, η}, we can assume without loss of generality that
65
η ≥ 0. Extend η to E by defining η = 0 on E ∼ E0 . Then 0 ≤ η ≤ f on E, η has finite support, and
Z
Z
|f − η| =
E
Z
f−
E
Z
Z
g−
η<
E
η+
E
E
=
2
Z
Z
g−
E0
η+
E0
<
2
Now suppose E is a closed and bounded interval. Since g is bounded, there exists a positive real
number M such that |g| ≤ M on E. By Lusin’s Theorem, we can find a continuous function g̃ defined
on E and a closed set F ⊆ E that satisfy
g̃ = g on F and m(E ∼ F ) <
8M
The function g̃ may be chosen such that |g̃| ≤ M (see the construction in Problem 47 of Chapter 1).
A continuous function on a closed and bounded interval is Riemann integrable (see Problem 6), so by
the definition of the lower Riemann integral we can find a step function h ≤ g̃ on E that satisfies
Z
Z
g̃ −
h>
E
E
4
Define h(x) ≡ 0 for x ∈
/ E. Since E has finite measure, the step function h has finite support and
Z
Z
Z
Z
|f − g| +
|g − g̃| +
|g̃ − h|
E
E
E
Z
< +
|g − g̃| +
2
4
ZE∼F
Z
3
≤
+
|g| +
|g̃|
4
E∼F
E∼F
3
≤
+ 2 · m(E ∼ F ) · M
4
3
<
+
4
4
=
|f − h| ≤
E
R
R
22. Let E be a measurable set. Because of Fatou’s Lemma, it suffices to show that lim sup E fn ≤ E f .
R
R
R
Since lim sup R fn ≤ R f < ∞, we can assume without loss that R fn is finite for all n. Since fn is
R
R
non-negative, we know that fn · χE ≤ fn and fn · χR∼E ≤ fn . Therefore E fn and R∼E fn are finite
R
R
R
and E fn = R fn − R∼E fn . We can thus conclude
Z
Z
lim sup
fn −
fn = lim sup
E
R
Z
fn
Z
R∼E
Z
≤ lim sup
fn − lim inf
R
Z
Z
≤
f−
f
R∼E
ZR
=
f
fn
R∼E
E
where the third line follows from Fatou’s Lemma.
Pn
23. Let fn = k=1 ak χ[k,k+1) . Then {fn } is an increasing sequence of measurable simple functions on E
66
that converges upward to f . By Lemma 1 and the Monotone Convergence Theorem, we have
Z
Z
f = lim
E
24.
fn = lim
n→∞
n→∞
E
n
X
ak =
k=1
∞
X
ak
k=1
(i) By the Simple Approximation Theorem, we can find an increasing sequence of simple functions
{ψn } on E that converges pointwise on E to f . Since we can always replace ψn with max{0, ψn },
we can assume without loss that each ψn is non-negative. Define ϕn = ψn · χ[−n,n] . Then each
ϕn is a non-negative simple function on E of finite support and {ϕn } converges pointwise on E
to f .
(ii) Let
A = {ϕ | ϕ simple, of finite support and 0 ≤ ϕ ≤ f on E}
and let
Z
ϕ ϕ∈A
x = sup
E
R
f ≥ E ϕ by
R
the definition of the Lebesgue integral for nonnegative measurable functions. Therefore E f ≥ x.
Suppose ϕ ∈ A. Since simple functions are bounded and measurable, we know that
R
E
Applying Part (i) of this problem, we can construct an increasing sequence of simple functions
{ϕn } in A that converges pointwise on E to f . By the Monotone Convergence Theorem, we know
R
R
that E f = limn→∞ E ϕn . Thus for any > 0, we can find an natural number n such that
Z
Z
f≤
ϕn + ≤ x + E
R
Since this expression holds for all , we must have E f ≤ x.
R
R
R
R
25. Since fn ≤ f on E for all n, we know that E fn ≤ E f . Thus lim sup E fn ≤ E f , which along with
R
R
Fatou’s Lemma implies limn→∞ E fn = E f .
R
R
26. Let fn = χ(−∞,−n] . Then fn converges downward to f ≡ 0 but limn→∞ fn = ∞ =
6 0 = f.
27. Let gn = inf k≥n fk . Then {gn } is an increasing sequence of non-negative, measurable functions on E
that converges pointwise on E to lim inf fn . By the Monotone Convergence Theorem, we know that
R
R
R
R
lim inf fn = limn→∞ E gn . But since gn ≤ fn , we also know that E gn ≤ E fn for all n. Therefore
E
R
R
limn→∞ E gn ≤ lim inf E fn . Combining these results, we can conclude
Z
Z
lim inf fn ≤ lim inf
4.4
fn
The General Lebesgue Integral
28. Suppose f is non-negative. By Problem 24(i), we can construct an increasing sequence of non-negative
simple functions {ϕn } on E, each of finite support, which converges pointwise on E to f . Then ϕn · χC
is an increasing sequence of non-negative measurable functions converging upwards to f · χC . We
67
therefore have
Z
Z
Z
f · χC = lim
ϕn · χC = lim
n→∞
E
Z
ϕn =
n→∞
E
C
f
C
where the first and last equalities follow from the Monotone Convergence Theorem and the middle
equality follows from Problem 10.
Now suppose f is integrable. By the previous result and the monotonicity of integration, we have
Z
f+ =
Z
C
and
f
Since f · χC = [f · χC ] and f
Z
Z
f=
C
f+ −
C
Z
f− =
−
Z
C
f+ < ∞
E
Z
−
=
f
C
+
Z
E
Z
+
f + · χC <
−
Z
f− < ∞
· χC <
E
E
−
· χC = [f · χC ] , we have
Z
f + · χC −
E
f − · χC =
Z
E
[f · χC ]+ −
Z
E
[f · χC ]− =
E
Z
f · χC
E
29. Let E = [1, ∞) and define
E0 =
∞ [
n=1
1
n − ,n
2
Consider the function
f = χE 0 − χE∼E 0
P∞
Then an = 0 for all n, so the series
n=1
an converges absolutely and therefore also converges.
P∞
n=1 an does not imply
However, f is not integrable. Thus convergence and absolute convergence of
integrability of f .
Now suppose f is integrable. Then
n
X
|ai | ≤
n Z
X
i=1
i=1
i+1
Z
|f | =
i
n+1
Z
|f | =
1
|f | · χ[1,n+1)
E
But |f | · χ[1,n+1) is an increasing sequence of non-negative, measurable functions converging pointwise to |f | on E. We can therefore conclude
∞
X
n=1
Z
|an | ≤ lim
n→∞
Z
|f | · χ[1,n+1) =
E
|f | < ∞
E
by the Monotone Convergence Theorem. Thus integrability of f implies that
P∞
n=1
an converges and
converges absolutely.
30. Let gn = inf k≥n fk . Then gn converges pointwise on E to lim inf fn , |gn | ≤ g on E, and gn ≤ fn for all
68
n. By the Lebesgue Dominated Convergence Theorem and monotonicity of integration, we have
Z
Z
lim inf fn = lim
n→∞
E
gn
E
Z
≤ lim inf
fn
E
Applying the above inequality, we can also conclude
Z
Z
Z
lim inf(−fn ) ≥ − lim inf −
fn = lim sup
fn
Z
lim sup fn = −
E
E
E
E
31. Let g 0 be a finite, integrable function and h0 be a nonnegative function that together satisfy g + h =
g 0 + h0 . Then
g + − g − + h = [g 0 ]+ − [g 0 ]− + h0
which implies
g + + [g 0 ]− + h = [g 0 ]+ + g − + h0
By the linearity of integration for non-negative functions, we have
Z
g+ +
E
Z
Z
[g 0 ]− +
Z
E
E
Z
[g 0 ]+ +
h=
E
g− +
E
Z
h0
E
which implies
Z
E
g+ −
Z
g− +
Z
E
Z
[g 0 ]+ −
h=
E
Z
E
[g 0 ]− +
E
Z
h0
E
We can conclude from this expression that
Z
Z
g+
Z
h=
E
E
E
g0 +
Z
h0
E
32. I first prove the following preliminary result:
Claim: Let {an } and {bn } be two sequences of real numbers. Suppose {an } converges to a ∈ R. Then
lim inf(an + bn ) = a + lim inf bn and lim sup(an + bn ) = a + lim sup bn .
Proof: For any n, we have inf k≥n (ak + bk ) ≥ inf k≥n ak + inf k≥n bk . Therefore
lim inf(an + bn ) ≥ lim inf an + lim inf bn = a + lim inf bn
Now let {bnk } be a subsequence of {bn } that converges to lim inf bn . Then
lim inf(an + bn ) ≤ lim (ank + bnk ) = a + lim inf bn
k→∞
where the inequality follows because lim inf(an + bn ) is the smallest subsequential limit of the sequence
{bn }. We can therefore conclude lim inf(an + bn ) = a + lim inf bn . This result then implies
lim sup(an + bn ) = − lim inf(−(an + bn )) = −(−a + lim inf(−bn )) = a + lim sup bn
69
Let {fn } be a sequence of measurable functions on E that converges pointwise a.e. on E to f . Let {gn }
be a sequence of nonnegative measurable functions on E that converges pointwise a.e. on E to g and
R
R
that satisfies |fn | ≤ gn on E for all n. Suppose limn→∞ E gn = E g < ∞. We may assume without
R
R
R
loss that E gn < ∞ for all n. By the monotonicity of integration, we know that E |fn | ≤ E gn < ∞.
We also have
Z
Z
Z
|f | ≤ lim inf
Z
|fn | ≤ lim inf
E
g<∞
gn =
E
E
E
where the first inequality follows from Fatou’s Lemma. By possibly excising from E a countable
collection of sets of measure zero and applying Proposition 15, we may assume that fn , gn , f and g
are finite on E. The functions g − f and gn − fn for each n are then properly defined, non-negative
and measurable. Moreover, the sequence {gn − fn } converges a.e. on E to g − f . We therefore have
Z
Z
Z
g−
E
(g − f )
Z
≤ lim inf (gn − fn )
Z E
Z
gn − lim sup
fn
= lim
n→∞ E
E
Z
Z
=
g − lim sup
fn
f=
E
E
E
E
where the second line follows from Fatou’s Lemma and the third line follows from linearity of integration
and the preliminary result. We can thus conclude
Z
Z
fn ≤
lim sup
f
E
E
An analogous argument applied to the sequence {fn − gn } can be used to show
Z
Z
f ≤ lim inf
E
33. Suppose
R
E
fn
E
|f − fn | → 0. Since |fn | − |f | ≤ |f − fn | on E for all n and |fn | − |f | → 0 a.e. on E, we
have
Z
(|fn | − |f |) = 0
lim
n→∞
E
by the General Lebesgue Dominated Convergence Theorem. Since f is integrable, this expression
implies
Z
Z
|fn | =
lim
n→∞
E
|f |
E
by linearity of integration.
Conversely, suppose limn→∞
R
E
|fn | =
R
E
|f |. Then limn→∞
R
E
(|fn | + |f |) = 2
|fn − f | ≤ |fn | + |f | on E for all n and |f − fn | → 0 a.e. on E, we have
Z
|f − fn | = 0
lim
n→∞
E
70
R
E
|f | < ∞. Since
by the General Lebesgue Dominated Convergence Theorem.
34. Let fn = f · χ[−n,n] . Then {fn } is an increasing sequence of non-negative measurable functions on E
converging pointwise to f on R. By the Monotone Convergence Theorem and Problem 28, we have
Z
Z
Z
f = lim
n→∞
R
Z
n→∞
R
n
f · χ[−n,n] = lim
fn = lim
n→∞
R
f
−n
35. Let {yn } be a sequence in [0, 1] that converges to 0. Define fn (x) = f (x, yn ). Then |fn | ≤ g on [0, 1]
and fn converges pointwise to f on [0, 1]. By the Lebesgue Dominated Convergence Theorem, we have
Z
1
1
Z
n→∞
0
Z
1
f (x, yn )dx
fn (x)dx = lim
f (x)dx = lim
n→∞
0
0
Since this expression holds for any sequence {yn } in [0, 1] converging to 0, we can conclude
1
Z
1
Z
f (x)dx = lim
f (x, y)dx
y→0
0
0
Now suppose f (x, y) is continuous in y for each x. Fix y ∈ [0, 1] and let {yn } denote a sequence in [0, 1]
converges to y. Let fn (x) = f (x, yn ) and let f (x) = f (x, y). Then |fn | ≤ g on [0, 1] and fn converges
pointwise to f on [0, 1]. By the Lebesgue Dominated Convergence Theorem, we have
1
Z
h(y) =
f (x, y)dx
0
1
Z
=
f (x)dx
0
Z
1
= lim
n→∞
fn (x)dx
0
Z
= lim
n→∞
1
f (x, yn )dx
0
= lim h(yn )
n→∞
Since this expression holds for any sequence {yn } in [0, 1] converging to y, we can conclude
h(y) = lim
h (y 0 )
0
y →y
Thus h is continuous at y.
36. Notice that the left-hand side of the expression we are asked to verify may not be well-defined. For
example, suppose we define f (x, y) on Q as

1

 +y
x
f (x, y) =

0
if 0 < x ≤ 1
if x = 0
Although this function satisfies the conditions given in the problem, the derivative of
with respect to y is not defined.
71
R1
0
f (x, y)dx = ∞
To avoid this problem, assume there exists y0 ∈ [0, 1] such that f (x, y0 ) is an integrable function of x
on [0, 1]. For any (x, y) ∈ [0, 1]2 , we have
y
Z
|f (x, y)| = f (x, y0 ) +
y0
y
∂f
(x, t)dt
∂y
Z
≤ |f (x, y0 )| +
y0
∂f
(x, t) dt
∂y
≤ |f (x, y0 )| + g(x)
By monotonicity and linearity of integration, we can conclude from this expression that f (x, y) is an
integrable function of x on [0, 1] for any y ∈ [0, 1].
Now fix y ∈ [0, 1]. Let {hn } denote a sequence that converges to zero and assume y + hn ∈ [0, 1] for all
n. Let f (x) =
∂f
∂y (x, y)
and define
f (x, y + hn ) − f (x, y)
hn
fn (x) =
Then fn (x) converges pointwise to f (x) on [0, 1] and |fn (x)| ≤ g(x) by the Mean Value Theorem. We
can therefore conclude
d
dy
Z
0
1
R1
R1
f (x, y + hn )dx − 0 f (x, y)dx
0
f (x, y)dx = lim
n→∞
hn
Z 1
= lim
fn (x)dx
n→∞
Z
0
1
=
f (x)dx
0
Z
1
=
0
∂f
(x, y)dx
∂y
where the second equality follows from linearity of integration and the third equality follows from the
Lebesgue Dominated Convergence Theorem.
4.5
Countable Additivity and Continuity of Integration
37. Since {En } is a descending countable collection of measurable subsets of E and m (∩∞
n=1 En ) = m(∅) =
0, we have
Z
Z
0=
f = lim
∩∞
n=1 En
n→∞
f
En
by continuity of integration. The desired result then follows from the formal definition of a limit.
38.
(i) For any natural number n ≥ 2, we have
Z
n
Z
f=
1
1
∞
nX
k=1
n−1
X (−1)k
(−1)k
· χ[k,k+1) =
k
k
k=1
This series converges by the alternating series test, so limn→∞
72
Rn
1
f is well-defined. However, we
also have
∞
Z
n
Z
|f | = lim
|f | = lim
n→∞
1
∞
nX
Z
n→∞
1
1
k=1
n−1
X1
1
· χ[k,k+1) = lim
=∞
n→∞
k
k
k=1
where the first equality follows from the Monotone Convergence Theorem and the last equality
follows from the divergence of the harmonic series. Therefore f is not integrable over [1, ∞).
(ii) Extend the function to [0, ∞) by defining

 sin x
x
f (x) =
0
if x > 0
if x = 0
For any natural number n, we have
Z
2πn
Z
(2n−1)π
2π(n−1)
sin x
dx +
x
Z
2nπ
sin x
dx
x
(2n−2)π
(2n−1)π
Z π
Z π
sin(x + (2n − 1)π)
sin(x + (2n − 2)π)
dx +
dx
=
x
+
(2n
−
2)π
x + (2n − 1)π
Z0 π
Z π 0
sin x
sin x
=
dx −
dx
x
+
(2n
−
2)π
x
+
(2n
− 1)π
0
0
f=
Thus
Z
2πn
f=
0
n Z
X
2n−1
X
(−1)i
i=0
But
f
2π(k−1)
k=1
=
2πk
Z
0
π
sin x
dx
x + iπ
Rπ
sin x
0 x+iπ
is positive and decreasing in i, so the above series converges by the alternating
R 2πn
R1
series test. Therefore limn→∞ 0 f is well-defined. Since f is continuous, 0 f is finite and
Rn
R 2πn
R1
limn→∞ 1 f = limn→∞ 0 f − 0 f is well-defined.
We also see that
Z π
sin(x + (2n − 1)π)
sin(x + (2n − 2)π)
dx +
dx
|f | =
x
+
(2n
−
2)π
x + (2n − 1)π
0
2π(n−1)
0
Z π
Z π
sin x
sin x
=
dx +
dx
0 x + (2n − 2)π
0 x + (2n − 1)π
Z π
Z π
1
1
≥
sin x dx +
sin x dx
(2n − 1)π 0
2nπ 0
2
2
=
+
(2n − 1)π 2nπ
Z
2πn
Z
π
73
which means
∞
Z
2πn
Z
|f |
|f | = lim
n→∞
0
0
2n
2X1
n→∞ π
i
i=1
≥ lim
=∞
by the Monotone Convergence Theorem and the divergence of the harmonic series. Since
R∞
is finite, this result implies 1 |f | is infinite. Thus f is not integrable over [1, ∞).
R1
0
|f |
The above results do not contradict continuity of integration because continuity only applies to integrable functions.
39. Let f be integrable over E and suppose {En }∞
n=1 is an ascending countable collection of measurable
subsets of E. Define E0 = ∅ and Ck = Ek ∼ Ek−1 . Then
{Ck }∞
k=1 is disjoint,
n
[
Ck = En for all n, and
k=1
∞
[
Ek =
k=1
∞
[
Ck
k=1
We therefore have
Z
∪∞
n=1 En
n Z
X
Z
f=
f = lim
∪∞
n=1 Cn
n→∞
k=1
Z
Z
f = lim
n→∞
Ck
f = lim
∪n
i=1 Ck
n→∞
f
En
where the second equality follows from countable additivity of integration and the third equality from
the additivity over domains property.
Now suppose {En }∞
n=1 is a descending countable collection of measurable subsets of E. We then have
Z
Z
Z
f−
f=
∩∞
n=1 En
f
E∼∩∞
n=1 En
E
Z
Z
f−
=
E
f
∪∞
n=1 (E∼En )
Z
Z
f − lim
f
n→∞ E∼E
E
n
Z
Z
= lim
f−
f
n→∞
E
E∼En
Z
= lim
f
=
n→∞
En
where the first and last lines follow from the additivity over domains property and the third line follows
from continuity of integration for an ascending countable collection of measurable sets.
74
4.6
Uniform Integrability: The Vitali Convergence Theorem
40. For any x ∈ R, we have
Z
x
Z
|f | ≤
|f | < ∞
−∞
R
by monotonicity of integration. Therefore F (x) =
Rx
−∞
f is well-defined.
Now fix > 0. By Proposition 23, there exists δ > 0 such that if A ⊆ E is measurable and m(A) < δ,
R
then A |f | < . Suppose |x0 − x| < δ and without loss of generality assume x0 > x. Then
Z
|F (x0 ) − F (x)| =
x0
Z
x0
f ≤
|f | < x
x
Therefore F is continuous.
Now suppose

1

 √
f (x) = 2 x

0
Then



0


√
F (x) =
x



1
if 0 ≤ x < 1
otherwise
if x < 0
if 0 ≤ x < 1
if x ≥ 1
But F is not Lipschitz (see the solution to Problem 50 of Chapter 1).
41. Let fn = χ[0,n) . Fix > 0 and let δ = . Then for any measurable set A ⊆ R with m(A) < δ, we have
Z
Z
|fn | ≤
1 = m(A) < A
A
Therefore the sequence of functions {fn } is uniformly integrable. But {fn } → χ[0,∞) on R, but χ[0,∞)
is not integrable.
42. Let E = [0, 1] and define hn = nχ[0,1/n) − nχ[1−1/n,1] . Then E has finite measure, each hn is integrable
R
and {hn } converges pointwise a.e. to h ≡ 0 on E. We also have E hn = 0 for all n, so
Z
lim
n→∞
hn = 0
E
Let = 1/2. Fix δ > 0 and choose N > 1/δ. Let A = [0, 1/N ). Then A ⊆ E is measurable and m(A) < δ,
but
Z
hn = 1 ≥ A
for all n ≥ N . Therefore {hn } is not uniformly integrable.
43. Fix > 0 and choose α ∈ R. Suppose α = 0. Then for any measurable set A ⊆ E,
R
A
|αfn | = 0 < for all n. Thus {αfn } is uniformly integrable. If α 6= 0, choose δ to satisfy the uniform integrability
75
challenge for /|α|. If A ⊆ E is measurable and m(A) < δ, then
Z
Z
|αfn | = |α|
=
|α|
|fn | < |α| ·
A
A
for all n. Therefore {αfn } is uniformly integrable for all α.
Now choose δf and δg to satisfy the /2 uniform integrability challenge for for {fn } and {gn }, respectively. Let δ = min{δf , δg }. If A ⊆ E is measurable and m(A) < δ, then
Z
Z
Z
|fn + gn | ≤
A
|fn | +
+ =
2 2
|gn | <
A
A
for all n. Therefore {fn + gn } is uniformly integrable.
Combining the above results, we conclude that {αfn + βgn } is uniformly integrable for all α and β.
44.
(i) Suppose f is non-negative. By Problem 24, we can find an increasing sequence of simple functions
{ϕn }, each of finite support, that converges pointwise to f . By the Monotone Convergence
Theorem, there exists an index N such that
Z
Z
|f − ϕn | =
Z
f−
R
ϕn < R
for all n ≥ N .
In general, we can write f = f + − f − where f + and f − denote the positive and negative parts of
f . By the previous result, we can find simple functions of finite support η + and η − that satisfy
Z
Z
|f − η | <
2
R
+
+
|f − − η − | <
R
2
Let η = η + − η − . Then η is a simple function of finite support that satisfies
Z
Z
|f − η| =
R
f + − η + − (f − − η − ) ≤
R
Z
f + − η+ +
R
Z
f − − η− < R
(ii) Suppose f is non-negative. Let fn = min{f, n} · χ[−n,n] . Then {fn } converges upward to f , so
by the Monotone Convergence Theorem
Z
Z
lim
n→∞
Pick n such that
fn =
R
Z
f
R
Z
f−
R
fn <
R
2
fn is a bounded, measurable function on the closed and bounded interval I = [−n, n]. By Problem
18 of Chapter 3, there exists a step function s on I and a measurable subset F of I for which
fn = s on F and m(I ∼ F ) <
4n
Without loss, we can assume 0 ≤ s ≤ n (if not, replace s with max{0, min{s, n}}). By the
76
additivity over domains and monotonicity properties of integration, we have
Z
Z
|fn − s| =
Z
Z
|fn − s| +
I
F
Z
|fn − s| ≤
|fn | +
I∼F
I∼F
|s| ≤ 2n · m(I ∼ F ) <
I∼F
2
Extend s to R by defining s = 0 on R ∼ I. Then
Z
Z
Z
|f − s| ≤
Z
|f − fn | +
R
R
Z
|fn − s| =
R
|f − fn | +
R
|fn − s| < I
The result for general f can be proven using the same argument employed in part (i).
(iii) We can proceed as in part (ii), replacing s with a continuous function g and the result from
Problem 18 of Chapter 3 with Lusin’s Theorem.
45. By the additive domains property, we have
Z
|fˆ| =
Z
R
|fˆ| +
Z
E
|fˆ| =
R∼E
Z
|f | < ∞
E
Therefore fˆ is integrable. The additive domains property likewise implies
R
R
fˆ =
R
E
f.
By Problem 45, we can find a simple function η̂ on R that satisfies
Z
|fˆ − η̂| < R
Let η denote the restriction of η̂ to E. Then
Z
Z
|f − η| =
E
|fˆ − η̂| ≤
Z
E
|fˆ − η̂| < R
An analogous argument can be used to find a continuous function satisfying the desired property.
46. Since |f (x) cos nx| < |f (x)|, f (x) cos nx is integrable over R. By Problem 44(ii), we can find a step
R∞
function s that vanishes outside a closed, bounded interval and satisfies −∞ |f − s| < /2. Since s is a
step function, there exists a partition P = {x0 , x1 , · · · , xN } and numbers c1 , · · · , cN such that
s(x) =

c
i
0
if xi−1 < x < xi , 1 ≤ i ≤ N
otherwise
We therefore have
Z
∞
s(x) cos(nx)dx =
−∞
N
X
i=1
Z
xi
ci
cos(nx)dx
xi−1
=
N
1 X
ci (sin(nxi ) − sin(nxi−1 ))
n i=1
≤
2 · N · maxi |ci |
n
77
For all n >
4·N ·maxi |ci |
,
we have
Z ∞
s(x) cos(nx)dx
f (x) − s(x) cos(nx)dx +
f (x) cos(nx)dx ≤
−∞
−∞
−∞
Z ∞
Z ∞
s(x) cos(nx)dx
|f (x) − s(x)| dx +
≤
Z
∞
Z
∞
−∞
−∞
<
Since can be made arbitrarily small, the result follows.
47.
(i) Suppose f = χE for a measurable set E. For any real number t, we have
χE (x + t) = χE−t (x) for all x ∈ R
Translation invariance of Lebesgue measure then implies
∞
Z
Z
∞
f (x)dx =
−∞
−∞
χE (x)dx = m(E) = m(E − t) =
Z ∞
Z
χE−t (x)dx =
−∞
∞
Z
∞
χE (x + t)dx =
−∞
f (x + t)dx
−∞
The result extends to simple functions by linearity of integration. If f is a measurable non-negative
function, then by the Simple Approximation Theorem there exists an increasing sequence of simple
functions {ϕn } that converges pointwise to f . For each x, {ϕn (x + t)} is an increasing sequence
converging to f (x + t). The Monotone Convergence Theorem therefore implies
Z
∞
Z
∞
f (x)dx = lim
−∞
n→∞
Z
∞
ϕn (x)dx = lim
n→∞
−∞
Z
∞
ϕn (x + t)dx =
−∞
f (x + t)dx
−∞
The result can be extended to a general integrable function f by applying the result for nonnegative functions to the positive and negative parts of f .
(ii) Let {tn } denotes a sequence of real numbers converging to 0. Since g is bounded, there exists
a number M such that |g| < M on R. We can use the result from Problem 44(iii) to find a
continuous function fˆ on R that vanishes outside a closed and bounded set E and that satisfies
R
|f − fˆ| < 2M
. We therefore have
R
Z
∞
g(x)(f (x) − f (x + t))dx
−∞
Z
∞
≤M
−∞
Z ∞
<M
fˆ(x) − fˆ(x + tn ) dx + M
Z
∞
f (x) − fˆ(x) dx + M
−∞
Z
∞
f (x + tn ) − fˆ(x + tn ) dx
−∞
fˆ(x) − fˆ(x + tn ) dx + −∞
(1)
where the last line follows from part (i). By The Extreme Value Theorem, fˆ is bounded above
by some constant M̂ on E. Thus fˆ(x + tn ) − fˆ(x) converges pointwise to 0 on a bounded set
Ê ⊇ E, vanishes outside Ê, and is bounded above by 2M̂ . The Dominated Convergence Theorem
78
then implies
Z
n→∞
n→∞
Ê
∞
Z
fˆ(x + tn ) − fˆ(x) = lim
0 = lim
fˆ(x + tn ) − fˆ(x)
−∞
Taking the lim sup of both sides of (1), we obtain
Z
∞
g(x)(f (x) − f (x + tn ))dx ≤ lim sup
n
−∞
Since can be made arbitrarily small, we have
Z
∞
g(x)(f (x) − f (x + tn ))dx = 0
lim
n→∞
−∞
Since this expression holds for all sequences {tn } converging to zero, the desired result follows.
48. Since g is bounded, there exists a real number M such that |g| ≤ M on E. Monotonicity of integration
then implies
Z
Z
Z
|f · g| =
|f | · |g| ≤ M
E
|f | < ∞
E
E
49.
(i) → (ii): If f is 0 a.e. on R and g is a bounded function, then f g = 0 a.e. on R. But this implies
R
R
fg = 0
by Proposition 9.
(ii) → (iii): Fix a measurable set A. If (ii) is true, then
R
f=
A
R
R
f · χA = 0 by Problem 28.
(iii) → (iv): (iv) follows from (iii) because every open set is measurable.
(iv) → (i): Let fn = |f | · χ(−n,n) . Then {fn } converges upward to |f |. If (iv) is true, then
Z
Z
Z
|f | = lim
R
n→∞
n
|f | = 0
fn = lim
n→∞
R
−n
where the first equality follows from the Monotone Convergence Theorem. (i) then follows from
Proposition 9.
50. Suppose F is uniformly integrable. Fix > 0 and let δ > 0 correspond to the challenge in the uniform
integrability criteria. For any measurable set A ⊆ E satisfying m(A) < δ, we have
Z
Z
f ≤
A
|f | < A
for all f ∈ F.
To show the converse, choose δ > 0 such that if A ⊆ E is measurable and m(A) < δ, then
R
A
f < 2 .
Fix an A ⊆ E that satisfies these conditions. Let B = {x ∈ E : f (x) ≥ 0}. Then f = f + on B and
f = −f − on E ∼ B, where f + and f − denote the positive and negative parts of f . Moreover, B is a
79
measurable set and both m(A ∩ B) and m(A ∩ B C ) are less than δ. We therefore have
Z
Z
Z
|f | =
|f | +
A
|f |
A∩B C
A∩B
Z
f+ +
=
Z
f−
A∩B C
A∩B
Z
Z
f
f +
=
A∩B C
A∩B
<
for all f ∈ F. Thus F is uniformly integrable.
51. If U is an open set, then E ∩ U is a measurable set contained in E. Uniform integrability then
immediately implies the condition given in the problem.
To show the converse, fix > 0 and let δ correspond to the challenge given in the problem. Let
A ⊆ E be a measurable set that satisfies m(A) < δ. By Theorem 11 of Chapter 2, we can find an open
set U containing A for which m(U ∼ A) < δ − m(A). Then
m(E ∩ U) = m(A) + m((E ∩ U) ∼ A) ≤ m(A) + m(U ∼ A) < δ
Therefore
Z
Z
|f | ≤
|f | < A
E∩U
for all f ∈ F. Thus F is uniformly integrable.
52. (a) Define fn = n · χ[0,1/n] and F = {fn }. Then
Choose n >
1/δ
and define A =
[0, 1/n].
R1
|fn | = 1 for all n. Let = 1/2 and fix any δ > 0.
R
Then m(A) = 1/n < δ but A |fn | = 1 > . Therefore F is
0
not uniformly integrable.
(b) Let F be a family of functions on [0, 1] that satisfies the given properties. Fix > 0 and let δ = /2.
Choose f ∈ F. Since f is bounded and measurable, it is integrable over [0, 1] by Theorem 4. If
A ⊆ [0, 1] is a measurable set satisfying m(A) < δ, then
Z
Z
|f | ≤
A
1 = m(A) < δ < A
by monotonicity of integration. Therefore F is uniformly integrable.
(c) Let F be a family of functions on [0, 1] that satisfies the given properties. Extend each f ∈ F to
Rb
the real line by defining f ≡ 0 on R ∼ [0, 1]. Then a |f | ≤ b − a for all (a, b) ⊆ R. Fix > 0 and
choose a measurable set A ⊆ [0, 1] such that m(A) < . By Theorem 11 of Chapter 2, we can find
an open set O ⊇ A such that m(O ∼ A) < − m(A). By Proposition 9 of Chapter 1, O can be
written as
O=
∞
[
(ai , bi )
i=1
80
where a1 < b1 < a2 < · · · . Since m(O) = m(A) + m(O ∼ A) < , we have
Z
Z
|f | ≤
|f | =
O
A
∞ Z
X
n=1
bn
|f | ≤
an
∞
X
(bn − an ) = m(O) < n=1
by monotonicity and countable additivity of integration. Therefore F is uniformly integrable.
5
Lebesgue Integration: Further Topics
5.1 Uniform Integrability and Tightness: A General Vitali Convergence
Theorem
1. If {hn } is uniformly integrable and tight over E, then limn→∞
R
E
hn = 0 by the Vitali Convergence
Theorem.
Conversely, suppose limn→∞
R
E
hn = 0. Pick N such that
R
E
hn < for all n ≥ N . By Problem 2,
we know that the finite collection of functions {hn }N
n=1 is tight. We can therefore find a set of finite
measure E0 ⊆ E such that
Z
|hn | < E∼E0
for all n < N . Since
Z
Z
|hn | ≤
hn < E∼E0
E
for all n ≥ N , we can conclude that {hn } is tight over E. The proof that {hn } is uniformly integrable
remains unchanged from the proof of uniform integrability in Theorem 26 of Chapter 4.
2. For each fk , we can apply Proposition 1 to find a set of finite measure Ek ⊆ E that satisfies
Z
|fk | < E∼Ek
Let E0 =
Sn
k=1
Ek . Then m(E0 ) ≤
Pn
k=1
m(Ek ) < ∞ and
Z
Z
|fk | ≤
E∼E0
|fk | < E∼Ek
for k = 1, · · · , n. Therefore {fk }nk=1 is tight.
{fk }nk=1 is uniformly integrable by Proposition 24 of Chapter 4.
R
3. Fix > 0 and choose α ∈ R. Suppose α = 0. Then E∼∅ |αfn | = 0 < , so {αfn } is tight. If α 6= 0, let
E0 ⊆ E denote a set finite measure that satisfies
Z
|fn | <
E∼E0
for all n. Then
Z
|α|
Z
|αfn | = |α|
E∼E0
|fn | < E∼E0
81
for all n. Therefore {αfn } is tight.
Now choose Ef and Eg to satisfy the /2 tightness challenge for {fn } and {gn }, respectively. Let
E0 = Ef ∪ Eg . Then m(E0 ) ≤ m(Ef ) + m(Eg ) < ∞ and
Z
Z
Z
|fn + gn | ≤
E∼E0
Z
|fn | +
E∼E0
Z
|gn | ≤
|fn | +
E∼E0
E∼Ef
|gn | < E∼Eg
for all n. Therefore {fn + gn } is tight.
Combining the above results, we conclude that {αfn + βgn } is tight for all α and β.
{αfn + βgn } is uniformly integrable from Problem 43 of Chapter 4.
4. Suppose {fn } is uniformly integrable and tight over E. Let E0 ⊆ E denote a set of finite measure that
satisfies
Z
2
|fn | <
E∼E0
for all n. Let δ correspond to the /2 uniform integrability challenge. Choose a measurable set A ⊆ E
R
that satisfies m(A ∩ E0 ) < δ. Then A∩E0 |fn | < /2 for all n, so
Z
Z
Z
|fn | =
A
Z
|fn | +
A∩E0
Z
|fn | ≤
A∼E0
|fn | +
|fn | < A∩E0
E∼E0
for all n.
Conversely, suppose the given statement is true. Let δ and E0 correspond to the given challenge.
R
Pick a measurable set A ⊆ E that satisfies m(A) < δ. Then m(A ∩ E0 ) < δ, so A |fn | < for all n.
Therefore {fn } is uniformly integrable. Now define A = E ∼ E0 . Then m(A ∩ E0 ) = m(∅) = 0 < δ, so
R
|f | < for all n. Therefore {fn } is tight.
E∼E0 n
5. Suppose {fn } is uniformly integrable and tight. Fix > 0. Let δ correspond to the /3 uniform
integrability challenge. Since {fn } is tight, we can find a set of finite measure E0 that satisfies
Z
|fn | <
R∼E0
3
for all n. Let Er = (−r, r) ∩ E0 . Then Er ↑ E0 , so limr→∞ m(Er ) = m(E0 ) by continuity of
measure. Choose r such that m(E0 ) < m(Er ) + δ. Then m(E0 ∼ Er ) = m(E0 ∼ (−r, r)) < δ, so
R
|f | < /3 for all n by uniform integrability. Let O denote an open subset of R that satisfies
E0 ∼(−r,r) n
R
m(O ∩ (−r, r)) < δ. Then O∩(−r,r) |fn | < /3 for all n by uniform integrability, so we have
Z
Z
|fn | =
Z
|fn | +
O
O∩(−r,r)
Z
≤
|fn |
O∼(−r,r)
Z
|fn | +
O∩(−r,r)
Z
≤
|fn |
R∼(−r,r)
Z
|fn | +
O∩(−r,r)
E0 ∼(−r,r)
<
82
Z
|fn | +
|fn |
R∼E0
for all n.
Conversely, suppose the statement is true. Let δ and r correspond to the given challenge. Let A
denote a measurable subset of R that satisfies m (A ∩ (−r, r)) < δ. By Theorem 11 of Chapter 2, we
can find an open subset O of R that contains A and that satisfies m(O ∼ A) < δ − m(A ∩ (−r, r)).
Then
m(O ∩ (−r, r)) ≤ m(A ∩ (−r, r)) + m(O ∼ A) < δ
so that
Z
Z
|fn | ≤
|fn | < O
A
for all n. But then {fn } is uniformly integrable and tight by Problem 4.
5.2
Convergence in Measure
6. Suppose {fn } → g in measure on E. Let
Ek =
Then
1
x ∈ E : |f (x) − g(x)| >
k
1
1
+ m |fn − g| >
→ 0 as n → ∞
m(Ek ) ≤ m |fn − f | >
2k
2k
S∞
S∞
Therefore m(Ek ) = 0 for all k, so m ( k=1 Ek ) = 0. Since f = g on E ∼ k=1 Ek , f = g a.e. on E.
Conversely, suppose f = g a.e. on E. Then there exists a set E0 with m(E0 ) = 0 such that f = g on
E ∼ E0 . We then have
m (|fn − g| > η) = m (x ∈ E ∼ E0 : |fn (x) − g(x)| > η)
= m (x ∈ E ∼ E0 : |fn (x) − f (x)| > η)
= m (|fn − f | > η)
→0
Thus {fn } → g in measure on E.
7. Fix η > 0. First suppose that g is a simple function. Then g can be written as
PK
k=1 ck χAk ,
where
each ck is non-zero scalar, each Ak ⊆ E is a measurable set, and {Ak }K
k=1 are disjoint. We have
K
X
η
m (|fn · g − f · g| > η) =
m
x ∈ Ak : |fn (x) − f (x)| >
|ck |
k=1
K
X
η
≤
m |fn − f | >
|ck |
k=1
→ 0 as n → ∞
where the last line follows because {fn } → f in measure on E. Thus {fn · g} → f · g in measure on E.
Now let g be a measurable function on E that is finite a.e. on E. By the Simple Approximation
83
Theorem, we can construct a sequence {φk } of simple functions on E that converges pointwise on E
to g. By Egoroff’s Theorem, for any > 0 we can find a closed set F contained in E for which
{φk } → g uniformly on F and m(E ∼ F ) < Choose K such that |g − φk | < 1 on F for all k ≥ K. Then for any k ≥ K, we have
m ({|fn · g − f · g| > η}) ≤
n
η o
+
m x ∈ F : |fn (x) − f (x)| · |g(x) − φk (x)| >
2
n
η o
η
m x ∈ E ∼ F : |fn (x) − f (x)| · |g(x) − φk (x)| >
+ m |fn · φk − f · φk | >
2
2
η
η
+ + m |fn · φk − f · φk | >
< m |fn − f | >
2
2
→
where the last follows because {fn } → f in measure on E and from the result for simple functions.
This expression implies lim supn m ({|fn · g − f · g| > η}) ≤ for all > 0, so we can conclude that
lim m ({|fn · g − f · g| > η}) = 0
n→∞
Thus {fn · g} → f · g in measure on E.
Now suppose {gn } → g in measure on E. Applying our previous result, we obtain
r r η
η
|fn − f | >
+m
|gn − g| >
+
3
3
n
n
η o
η o
+ m |gn · f − g · f | >
m |fn · g − f · g| >
3
3
→0
m ({|fn · gn − f · g| > η}) ≤ m
Therefore {fn · gn } → f · g in measure on E. From this result, we can immediately infer that {fn2 } → f 2
in measure on E.
8. Let {fn } denote a sequence of nonnegative measurable functions on E. Suppose {fn } → f in measure
on E. Let {nk } denote an increasing sequence of natural numbers that satisfies
Z
Z
fnk → lim inf
n→∞
E
fn
E
Since {fnk } → f in measure on E, we can use Theorem 4 to find a further subsequence {fnkj } that
converges a.e. on E to f . By applying Fatou’s Lemma for pointwise convergence to this subsequence,
we can conclude
Z
Z
f ≤ lim inf
E
j→∞
E
Z
fnkj = lim inf
n→∞
fn
E
Fatou’s Lemma therefore holds if “pointwise convergence a.e.” is replaced with “convergence in measure”. Since the Monotone and Dominated Convergence Theorems follow directly from Fatou’s Lemma,
“pointwise convergence a.e.” can be replaced with “convergence in measure” for these results as well.
Now suppose {fn } is uniformly integrable and tight over E. By Theorem 4, there exists a subsequence
84
{fnj } that converges a.e. on E to f . We can infer that f is integrable by applying the Vitali Convergence
Theorem to this subsequence. Now fix > 0. By uniform integrability of {fn } and Proposition 23 of
Chapter 4, |fn − f | is uniformly integrable. Furthermore, |fn − f | is tight by tightness of {fn } and
Proposition 1. Since |fn − f | → 0 in measure on E, Corollary 5 implies
Z
|fn − f | = 0
lim
n→∞
Since
Z
E
Z
−
Z
|fn − f | ≤
E
fn −
f≤
E
we can conclude that
Z
E
Z
lim
Z
fn =
n→∞
|fn − f |
E
E
f
E
Thus the Vitali Convergence Theorem also remains valid if “pointwise convergence a.e.” is replaced
with “convergence in measure”.
9. Let E = R, fn = χ[n,∞) and f = 0. Then {fn } → f pointwise on E. However, for any ∈ (0, 1) we
have
m({|fn − f | > ) = m([n, ∞)) = ∞ 6→ 0
Therefore {fn } does not converge in measure to f on E.
10. Suppose {fn } → f in measure on E. Fix a scalar α ∈ R and choose any η > 0. If α = 0, then
lim m(|αfn | > η) = lim m(∅) = 0
n→∞
n→∞
so that {αfn } → 0 in measure on E. If α 6= 0, then
η
lim m (|αfn − αf | > η) = lim m |fn − f | >
=0
n→∞
n→∞
|α|
so that {αfn } → αf in measure on E.
Now suppose {gn } → g in measure on E. Then
η
η
m (|fn + gn − f − g| > η) ≤ m |fn − f | >
+ m |gn − g| >
→0
2
2
so that {fn + gn } → f + g converges in measure on E.
Combining the above results, we can conclude that {αfn + βgn } → αf + βg in measure on E for all α
and β.
11. Suppose {fn } → f in measure on E. Let {fnj } denote a subsequence of {fn }. Then {fnj } → f
in measure on E, so there exists a further subsequence that converges pointwise a.e. on E to f by
Theorem 4.
Conversely, suppose {fn } does not converge in measure on E to f . Then there exists some η > 0 and
> 0 such that
m(|fn − f | > η) > infinitely often
85
We can therefore find a subsequence {fnj } that satisfies
m(|fnj − f | > η) > n
o
fnjk , we have
for all j. Since m(E) < ∞, for any further subsequence
m
n
x ∈ E : fnjk (x) − f (x) > η infinitely often
o
∞ [
∞
\
=m
!
{|fnjk − f | > η}
l=1 k=l
∞
[
= lim m
l→∞
!
{|fnjk − f | > η}
k=l
≥
by continuity of measure. But since
o n
x ∈ E : fnjk (x) − f (x) > η infinitely often ⊆ x ∈ E : {fnjk (x)} does not converge to fn (x)
we cannot have
n
o
fnjk converging pointwise a.e. to f on E.
12. Let {aj } denote a sequence of real numbers that satisfies |aj+1 − aj | ≤ 2−j for all j. Fix > 0 and let
N denote an index satisfying 2−N +1 < . Pick m, n ≥ N . Without loss of generality, assume m > n.
We then have
|am − an | ≤
m−1
X
|aj+1 − aj |
j=n
≤ 2−n
m−n−1
X
2−j
j=0
−n+1
≤2
<
Therefore {aj } is Cauchy and converges by Theorem 17 of Chapter 1.
13. Suppose {fn } is Cauchy in measure. Then for each natural number j, there exists an index Nj such
that
m |fm − fn | > 2−j < 2−j
for all m, n ≥ Nj . Let {nj } denote a strictly increasing sequence of natural numbers defined recursively
by
n 1 = N1 ,
nj+1 = max{nj + 1, Nj+1 }
and let
Ej = x ∈ E : fnj+1 (x) − fnj (x) > 2−j ,
E0 =
∞
[
j=1
Since nj , nj+1 ≥ Nj , we must have
m(Ej ) < 2−j
86
Ej
for all j. Countable subadditivity of measure then implies
m(E0 ) ≤
∞
X
m(Ej ) < 1
j=1
If x ∈ E ∼ E0 , then the sequence {fnj (x)} converges by Problem 12. We can therefore define the
function
f (x) =

 lim fnj (x)
if x ∈ E ∼ E0

0
if x ∈ E0
j→∞
Since fnj+1 (x) − fnj (x) is a measurable function of x for all j, each Ej is a measurable set. Therefore
E0 is also measurable, so χE∼E0 is a measurable function. Thus gj := fnj · χE∼E0 is a measurable
function for all j. Since {gj } → f pointwise, f is measurable.
Now fix η > 0 and > 0. For any natural number j, we have
η
η
+ m |fnj − f | >
m (|fn − f | > η) ≤ m |fn − fnj | >
2
2
Since {fn } is Cauchy in measure, there exists an index N such that
η
m (|fn − f | > η) < + m |fnj − f | >
2
(1)
for all n, nj ≥ N . Pick J that satisfies 2−J < η/4. If x ∈ E ∼ E0 , then for any j ≥ J we have
fnj (x) − f (x) ≤
≤
j+m−1
X
fni+1 (x) − fni (x) + fnj+m (x) − f (x)
i=j
∞
X
2−i + fnj+m (x) − f (x)
i=j
= 2−j+1 + fnj+m (x) − f (x)
η
≤ + fnj+m (x) − f (x)
2
for any natural number m. Since {fnj } → f on E ∼ E0 , by letting m go to infinity we can conclude
that
fnj (x) − f (x) ≤
for all j ≥ J. Therefore
η
2
n
ηo
E ∼ E0 ⊆ x ∈ E : fnj (x) − f (x) ≤
2
which implies
n
ηo
Ej0 := x ∈ E : fnj (x) − f (x) >
⊆ E0
2
for all j ≥ J. Since χEj0 ≤ χE0 for j ≥ J, Fatou’s Lemma implies
Z
lim inf χE0 − χ
E j→∞
Ej0
87
≤ lim inf
j→∞
Z E
χE0 − χEj0
Upon rearrangement, this expression can be written as
Z
m(E0 ) −
E
Since
lim sup χEj0 ≤ m(E0 ) − lim sup m Ej0
j→∞
(2)
j→∞

1
lim sup χEj0 (x) =
0
j→∞
if x ∈ Ej0 infinitely often
otherwise
and m(E0 ) < ∞, (2) implies
η
≤ m x ∈ Ej0 infinitely often
lim sup m |fnj − f | >
2
j→∞
But since Ej0 ⊆ E0 for all j ≥ J, we must have

m
∞
[

Ej0  ≤ m(E0 ) < ∞
j=J
The Borel-Cantelli Lemma then implies m x ∈ Ej0 infinitely often = 0. Therefore
η
lim sup m |fnj − f | >
=0
2
j→∞
We can therefore conclude
m(|fn − f | > η) ≤ for all n ≥ N by taking the lim sup of (1) over j. Since can be chosen to be arbitrarily small,
{fn } → f in measure on E.
14. Suppose {fn } → f in measure on E. Fix > 0 and let η = /m(E). Let
E 0 := {x ∈ E : |fn (x) − f (x)| > η}
Then
Z
ρ(fn , f ) =
E
Z
=
E0
|fn − f |
1 + |fn − f |
|fn − f |
+
1 + |fn − f |
Z
E∼E 0
|fn − f |
1 + |fn − f |
≤ m(E 0 ) + η · m(E)
< m({x ∈ E : |fn (x) − f (x)| > η) + where the third line follows because
|fn −f |
1+|fn −f |
≤ 1 on E 0 and
convergence in measure, this expression implies
lim sup ρ(fn , f ) ≤ n→∞
88
|fn −f |
1+|fn −f |
≤ η on E ∼ E 0 . But by
Since can be chosen to be arbitrarily small and ρ(fn , f ) ≥ 0 for all n, we must have limn→∞ ρ(fn , f ) =
0.
Conversely, suppose limn→∞ ρ(fn , f ) = 0. Fix η > 0. Then
η
|fn − f |
m (|fn − f | > η) = m
>
1 + |fn − f |
1+η
Z
|fn − f |
1+η
≤
η
1
+
|fn − f |
E
1+η
=
· ρ(fn , f )
η
→ 0 as n → ∞
where the first line follows because the function
x
1+x
is strictly increasing over [0, ∞) and the second
line follows from Markov’s inequality. Therefore {fn } → f in measure on E.
5.3
Characterizations of Riemann and Lebesgue Integrability
15. Let Ef , Eg , and Ef ·g denote the sets of points in [a, b] at which f , g and f · g fail to be continuous,
respectively. By Theorem 8, we know that m(Ef ) and m(Eg ) have measure 0.
Now suppose x ∈ [a, b] ∼ (Ef ∪ Eg ). Fix > 0. There exists a δ > 0 such that for all x0 ∈ [a, b]
satisfying |x0 − x| < δ, we have
r 3
r |g(x0 ) − g(x)| < min
,
3(|f (x)| + 1)
3
|f (x0 ) − f (x)| < min
,
3(|g(x)| + 1)
But then for all x0 ∈ [a, b] satisfying |x0 − x| < δ, we have
|f (x0 )g(x0 ) − f (x)g(x)| ≤ |f (x0 ) − f (x)||g(x0 ) − g(x)| + |f (x0 ) − f (x)||g(x)| + |g(x0 ) − g(x)||f (x)| < Thus f · g is continuous at x, so x ∈ [a, b] ∼ Ef ·g . We can conclude that
Ef ·g ⊆ Ef ∪ Eg
But this implies m(Ef ·g ) ≤ m(Ef ) + m(Eg ) = 0. Since f · g is bounded, f · g is continuous by Theorem
8.
16. If f is a bounded function, then f is Riemann integrable over [a, b] by Theorem 8. But then f is
Lebesgue integrable over [a, b] by Theorem 3 of Chapter 4. Theorem 7 then implies f is measurable.
Let E0 denote the set of discontinuity points of f . Since f is continuous on E ∼ E0 , the restriction of
f to E ∼ E0 is measurable by Proposition 3 of Chapter 3. But since m(E0 ) = 0, the restriction of f
to E0 is also measurable. Therefore f is measurable by Proposition 5 of Chapter 3.
89
17. For each natural number n, define
4n + 1
= 2n(n + 1)
(4n + 4)n
4n + 2
f
=0
(4n + 4)n
4n + 3
f
= −2n(n + 1)
(4n + 4)n
4n + 4
f
=0
(4n + 4)n
f
Extend f to all of (0, 1] by linear interpolation. Set f (0) = 0. By construction, f is continuous on
(0, 1] and
Z
f =0
[1/n,1]
for all n. Therefore the sequence
nR
f
[1/n,1]
o
converges. However,
Z
Z
|f | = lim
[0,1]
n→∞
|f | = lim n = ∞
[1/n,1]
n→∞
Therefore f is not Lebesgue integrable over [0, 1].
R
R
If f is non-negative, then limn→∞ [1/n,1] f = [0,1] f by the Monotone Convergence Theorem. In this
R
R
case, [0,1] f must be integrable if limn→∞ [1/n,1] f converges.
6
Differentiation and Integration
6.1
Continuity of Monotone Functions
1. By Proposition 2, we can find an increasing function f on (a, b) that is continuous only at (a, b) ∼ C
and that satisfies 0 ≤ f ≤ 1. If b ∈ C, define f (b) = 2. Then f is not continuous at b because
|f (b) − f (x)| ≥ 1 for all x ∈ (a, b). If b ∈
/ C, define f (b) = sup{f (x) : x ∈ (a, b)}. Pick a sequence
{xn } in (a, b) that converges upward to b. Since f is bounded and increasing on (a, b), the sequence
{f (xn )} converges to some real number y. Since f (xn ) ≤ f (b) for all n, we must have y ≤ f (b). For
any x ∈ (a, b), there is an index n such that x < xn < b. But then f (x) ≤ f (xn ) ≤ y because f is
increasing. Therefore y must also be an upper bound of {f (x) : x ∈ (a, b)}, which implies y ≥ f (b).
We conclude that limn→∞ f (xn ) = f (b), so f is continuous at b.
If we also define f (a) = −1 if a ∈ C and f (a) = inf{f (x) : x ∈ (a, b)} if a ∈
/ C, then f satisfies the
desired property.
2. Since the rational numbers are countable, we can use Problem 1 to construct an increasing function f˜
on [0, 1] that is continuous only at the irrational numbers in [0, 1]. Define f (x) = f˜(x) + x. Then f
is strictly increasing, continuous at the irrational numbers in [0, 1], and discontinuous at each rational
number in [0, 1].
90
3. Extend f to R by defining



inf{f (y) : y ∈ E}


g(x) = f (x)



sup{f (y) : y ∈ E and y < x}
if x ≤ inf E
if x > inf E and x ∈ E
if x > inf E and x ∈
/E
Now suppose f is not continuous at some x0 ∈ E. Then there exists > 0 such that for all δ > 0,
there exists x ∈ E satisfying |x − x0 | < δ and |f (x) − f (x0 )| > . But this implies that for any δ > 0,
there exists x ∈ R satisfying |x − x0 | < δ and |g(x) − g(x0 )| > . Therefore g is not continuous at
x0 , implying that the set of points at which f fails to be continuous must be contained in the set of
points at which g fails to be continuous. But since g is monotone, g is continuous except possibly at a
countable number of points by Theorem 1. We conclude that the set of points at which f fails to be
continuous must be countable, since it is contained in a countable set.
4. Let E = {0, 1} and C = {0}. Any function defined on E is continuous on E, so no function on E can
only be continuous at E ∼ C = {1}.
6.2
Differentiability of Monotone Functions: Lebesgue’s Theorem
5. Let E = [0, 1] and let F denote the collection of all degenerate intervals of the form [x, x] for x ∈ E.
Then for each x ∈ E and > 0, there is an interval I ∈ F that contains x and has `(I) < . However,
for any finite disjoint subcollection {Ik }nk=1 of F we have
∗
m
E∼
n
[
!
Ik
=1
k=1
6. Let E denote a set of finite outer measure and let F denote a collection of general, non-degenerate
intervals. Suppose for each point x ∈ E and > 0, there is an interval I ∈ F that contains x and
satisfies `(I) < . Without loss of generality, assume all the intervals in F are bounded. Define F̄ as
F̄ = {I¯ : I ∈ F}
Then F̄ is a collection of closed, bounded, nondegenerate intervals that covers E in the sense of Vitali.
By the Vitali Covering Lemma, for any > 0 there is a finite disjoint subcollection {I¯k }n of intervals
k=1
in F̄ for which
∗
m
E∼
n
[
!
I¯k
<
k=1
For each k, there exists Ik ∈ F such that I¯k ⊇ Ik and m∗ (I¯k ∼ Ik ) = 0. By the monotonicity and
91
subadditivity of outer measure, we have
∗
m
E∼
n
[
!
Ik
∗
≤m
E∼
k=1
n
[
!
I¯k
+m
k=1
n
[
n
[
∗
I¯k ∼
k=1
n
[
!
Ik
k=1
!
I¯k ∼ Ik
∗
<+m
k=1
≤+
n
X
m∗ I¯k ∼ Ik
k=1
=
7. Consider the Weierstrass function
f (x) =
∞
X
1
sin(2n x)
n
2
n=0
It can be shown that f is continuous everywhere but nowhere differentiable. But if f were monotone
on an open interval (a, b), then f would be differentiable almost everywhere on (a, b) by Lebesgue’s
Theorem. Therefore f cannot be monotone on any open interval.
8. Let z denote the midpoint of I. Fix x ∈ I ∩ J and pick any y ∈ J. Then
|y − x| ≤ `(J) ≤
`(I)
,
γ
|x − z| ≤
`(I)
2
By the Triangle Inequality, we have
|y − z| ≤ |y − x| + |x − z| ≤
1 1
+
2 γ
· `(I)
If γ ≥ 1/2, then |y − z| ≤ 5/2 · `(I). Thus y ∈ 5 ∗ I.
The result does not hold if 0 < γ < 1/2. For example, if J = [0, 10] and I = [−1, 1] then 5 ∗ I =
[−5, 5] 6⊇ J.
9. Suppose E has measure zero. Then for any index n, there exists a countable collection of nonempty
open, bounded intervals In = {In,k }∞
k=1 that covers E and that satisfies
∞
X
`(In,k ) < 2−n
(1)
k=1
The collection I =
S∞
n=1
In is a union of countable sets and is therefore countable. Now fix x ∈ E.
Since I1 covers E, there exists some I1 ∈ I1 that contains x. Now suppose we have found intervals
I1 , · · · , IJ in I that satisfy
x ∈ Ij for j = 1, · · · , J
`(I1 ) > · · · > `(IJ )
(2)
Choose n such that 2−n < `(IJ ). Since In covers E, there exists some IJ+1 ∈ In that contains x.
Moreover, `(IJ+1 ) < 2−n by (1). Continuing in this way, we can recursively construct an infinite series
92
of distinct intervals in I that contain x. Moreover, (1) implies
∞ X
∞
X
`(In,k ) < 1 < ∞
n=1 k=1
Conversely, suppose {Ik }∞
k=1 is a countable collection of open intervals for which each point in E
belongs to infinitely many Ik ’s and
∞
X
`(Ik ) < ∞
k=1
By the Borel-Cantelli Lemma, there exists a set E0 of measure zero such that all points not in E0 belong
to finitely many of the Ik ’s. This implies E ⊆ E0 , so E must have measure zero by monotonicity of
measure.
10. Fix x, x0 ∈ E and assume x < x0 . Then
`((ck , dk ) ∩ (−∞, x)) ≤ `((ck , dk ) ∩ (−∞, x0 ))
for all k. Therefore f is increasing.
Since x is in infinitely many of the intervals, we can assume without loss that ck < x < dk for all k.
For each index k, pick k > 0 such that
(x, x + k ) ⊆ (ck , dk )
For each index n, define ˜n = min{1 , 2 , · · · , n }. Then for all 0 < h < ˜n , we have
n 1X
f (x + h) − f (x)
≥
`((ck , dk ) ∩ (−∞, x + h)) − `((ck , dk ) ∩ (−∞, x))
h
h
k=1
n 1X
`(ck , x + h) − `(ck , x)
=
h
k=1
=n
Thus for any M > 0, we can find an > 0 such that
f (x + h) − f (x)
>M
h
for all 0 < h < . Therefore D̄(x) = ∞. Since D̄(x) is not finite, f is not differentiable at x.
93
11. Applying Problem 47(i) of Chapter 4, we have
b
Z
Z
g(t + γ)dt =
g(t + γ)χ(a,b) (t)dt
a
R
Z
g(t)χ(a,b) (t − γ)dt
=
R
Z
=
g(t)χ(a+γ,b+γ) (t)dt
R
Z
b+γ
g(t)dt
=
a+γ
Suppose f is integrable over the closed, bounded interval [a, b]. Extend f to take the value f (b) on
(b, b + 1]. Then for any a ≤ u < v ≤ b and 0 < h ≤ 1, we have
Z
v
u
Z v
f (x + h)
f (x)
dx −
dx
h
h
u
u
Z v+h
Z v
f (x)
f (x)
=
dx −
dx
h
h
u+h
u
! Z
Z v
Z v+h
Z u+h
v
f (x)
f (x)
f (x)
f (x)
dx +
dx +
dx −
dx
= −
h
h
h
h
u
v
u
u
Z v+h
Z u+h
f (x)
f (x)
=
dx −
dx
h
h
v
u
f (x + h) − f (x)
dx =
h
Z
v
12. Fix x ∈ Q and h > 0. Choose q ∈ Q satisfying x < q < x + h and define t = q − x. Then 0 < |t| ≤ h
and Difft χQ (x) = 0. Since Difft χQ (x) ≤ 0 for any t 6= 0, we must have sup0<|t|≤h Difft χQ (x) = 0 for
all h. Therefore D̄χQ (x) = 0.
1
Now suppose x ∈
/ Q. Fix M > 0 and h > 0 and define = min h, M
. Pick q ∈ Q satisfying
x < q < x + and define t = q − x. Then 0 < |t| ≤ h and
Difft χQ (x) =
1
1
> ≥M
t
Since M was arbitrary, sup0<|t|≤h Difft χQ (x) = ∞ for all h. Therefore D̄χQ (x) = ∞.
Putting these results together, we conclude

0 if x ∈ Q
D̄χQ (x) =
∞ if x ∈
/Q
By instead choosing q ∈
/ Q in the above proof, it is straight-forward to show

−∞ if x ∈ Q
DχQ (x) =
0
if x ∈
/Q
94
13. Consider the sequence of disjoint intervals {In }∞
k=1 constructed recursively in the proof of the Vitali
Covering Lemma. From the proof, we know that for any > 0 there exists an n such that
∞
[
"
∗
0≤m
E∼
#
"
∗
Ik ≤ m
E∼
k=1
n
[
#
Ik < k=1
Since can be made arbitrarily small, we have
∞
[
"
m
∗
E∼
#
Ik = 0
k=1
14. Suppose I := {Iα } is a collection of closed, bounded, nondegenerate intervals indexed by α. Define F
as
F = {[a, b] : −∞ < a < b < ∞ and [a, b] ⊆ Iα for some Iα ∈ I}
Then F is a collection of closed, bounded, nondegenerate intervals that covers
S
α Iα
in the sense of
Vitali. By the Vitali Covering Lemma, for each > 0 there exists a finite disjoint subcollection {I˜k }nk=1
of F for which
"
m∗ E ∼
n
[
#
I˜k < k=1
Then
Sn
˜ is a closed set contained in
k=1 Ik
S
α Iα .
We conclude
S
α Iα
is measurable by Theorem 11(iii)
of Chapter 2.
15. For any t 6= 0, we have
1
Difft f (0) = sin
t
Choose h > 0. Pick a natural number k such that
1
π
+ 2πk ≥
2
h
Set t = 1/( π2 + 2πk). Then 0 < |t| ≤ h and Difft f (0) = 1. Since h was arbitrary, we must have
sup Difft f (0) ≥ 1
D̄f (0) = lim
h→0 0<|t|≤h
Since sin(·) is bounded above by 1, we must also have D̄f (0) ≤ 1. Thus D̄f (0) = 1.
An analogous argument can be used to show Df (0) = −1.
16. Suppose g is non-negative. Then f is increasing on [a, b], so f is differentiable almost everywhere on
(a, b).
In the general case, write g = g + − g − where g + and g − denote the positive and negative parts of g. By
Rx
Rx
the above result, the functions h1 (x) = a g + and h2 (x) = a g − are differentiable almost everywhere
on (a, b). Thus there exists sets E1 and E2 of measure zero such that h1 and h2 are differentiable at all
points in (a, b) ∼ E1 and (a, b) ∼ E2 , respectively. By Problem 20 and the linearity of differentiation,
95
we have
h01 − h02 = D(h1 ) + D(−h2 ) ≤ D(h1 − h2 ) ≤ D(h1 − h2 ) ≤ D(h1 ) + D(−h2 ) = h01 − h02
on (a, b) ∼ (E1 ∪ E2 ). Since f = h1 − h2 , this expression implies that f is differentiable on (a, b) ∼
(E1 ∪ E2 ). But E1 ∪ E2 has measure zero, so f is differentiable almost everywhere on (a, b).
17. Extend f to the closed interval [a, b] by setting f (a) = inf x∈(a,b) f (x) and f (b) = supx∈(a,b) f (x). Since
f is bounded, −∞ < f (a) ≤ f (b) < ∞. The extended function is increasing on [a, b], so we can apply
Corollary 4 to conclude
Z
b
f 0 ≤ sup f (x) − inf f (x)
x∈(a,b)
x∈(a,b)
a
18. Suppose c ∈ (a, b) is a local minimizer for f . Then there exists some h∗ > 0 such that f (c) ≤ f (x) for
all x ∈ (c − h∗ , c + h∗ ). For any h ∈ (0, h∗ ), we have
f (c + h) − f (c)
≥0
h
which implies
Df (c) = lim
sup
h→0 0<|t|≤h
We also have
f (c + t) − f (c)
≥0
t
f (c + h) − f (c)
≤0
h
for any h ∈ (−h∗ , 0). Thus
Df (c) = lim
inf
h→0 0<|t|≤h
f (c + t) − f (c)
≤0
t
19. Corrected exercise: let f be continuous on [a, b] with Df ≥ 0. Show that f is increasing
on [a, b].
Choose u, v, and such that a < u < v < b and > 0. Let
S = {x ∈ [u, v] : f (x) + x ≥ f (u) + u}.
S is not empty as u ∈ S and S is bounded above by v. Therefore we can define s∗ = sup S. S is also
closed because f is continuous. Therefore s∗ ∈ S.
We will show s∗ = v by way of contradiction. Suppose s∗ < v. For each h > 0, define
g(h) =
f (s∗ + t) − f (s∗ )
t
0<|t|≤h
inf
Since limh→0 g(h) = Df (s∗ ) ≥ 0, there exists h0 ∈ (0, v − s∗ ) such that g(h0 ) > −. But this means
f (s∗ + t) − f (s∗ )
> −
t
96
for all 0 < |t| ≤ h0
(3)
Pick t0 ∈ (0, h0 ) and define s0 = s∗ + t0 . By (3), we have
f (s0 ) − f (s∗ )
> −
s0 − s∗
Rearranging, we obtain
f (s0 ) + s0 > f (s∗ ) + s∗ ≥ f (u) + u
where the second inequality follows because s∗ ∈ S. But this implies s0 ∈ S and s0 > s∗ , a contradiction.
We can conclude that
f (v) + v ≥ f (u) + u
Since this expression holds for all > 0, we have f (v) ≥ f (u). Thus f is increasing on (a, b). Since
f is continuous, this result can be extended to all of [a, b] by keeping v fixed and letting u → a and
keeping u fixed and letting v → b.
20. Fix h > 0. Then
f (t + h) − f (t)
g(t + h) − g(t)
f (t + h) + g(t + h) − (f (t) + g(t))
+ inf
≤ inf
h
h
h
0<|t|≤h
0<|t|≤h
0<|t|≤h
f (t + h) + g(t + h) − (f (t) + g(t))
≤ sup
h
0<|t|≤h
inf
f (t + h) − f (t)
g(t + h) − g(t)
+ sup
h
h
0<|t|≤h
0<|t|≤h
≤ sup
Taking limits as h → 0 yields the desired result.
21.
(i) We first show that D(f ◦ g)(γ) ≤ Df (c) · g 0 (γ). If Df (c) = ∞, this inequality holds trivially. So
suppose Df (c) < ∞. Then for any > 0, there exists δ > 0 such that
f (y) − f (c)
≤ Df (c) + y−c
(4)
for all y satisfying 0 < |y − c| ≤ δ. Since g 0 (γ) > 0, there also exists an open interval I containing
γ such that
g(x) − g(γ)
>0
x−γ
(5)
for all x ∈ I ∼ {γ}. Let {xn } denote a sequence in I ∼ {γ} that converges to γ. By continuity
of g, |g(xn ) − g(γ)| < δ for all n sufficiently large. From (4), we therefore have
(f ◦ g)(xn ) − (f ◦ g)(γ)
≤ Df (c) + g(xn ) − g(γ)
for all n sufficiently large. But this implies
g(xn ) − g(γ)
(f ◦ g)(xn ) − (f ◦ g)(γ)
(f ◦ g)(xn ) − (f ◦ g)(γ) g(xn ) − g(γ)
=
·
≤ Df (c) + xn − γ
g(xn ) − g(γ)
xn − γ
xn − γ
97
for all n sufficiently large. Taking the lim sup of both sides, we obtain
lim sup
n→∞
(f ◦ g)(xn ) − (f ◦ g)(γ)
≤ Df (c) + g 0 (γ)
xn − γ
Since this expression holds for arbitrarily small , we can conclude
lim sup
n→∞
(f ◦ g)(xn ) − (f ◦ g)(γ)
≤ Df (c) · g 0 (γ)
xn − γ
But since {xn } was an arbitrary sequence converging to γ, we must have D(f ◦g)(γ) ≤ Df (c)·g 0 (γ).
We now prove D(f ◦ g)(γ) ≥ Df (c) · g 0 (γ). If D(f ◦ g)(γ) = ∞, then the inequality holds trivially.
So suppose D(f ◦ g)(γ) < ∞. For any > 0, there exists δ > 0 such that
(f ◦ g)(x) − (f ◦ g)(γ)
≤ D(f ◦ g)(γ) + x−γ
for all x satisfying 0 < |x − γ| ≤ δ. Now g is a strictly increasing on I by (5). By Problem 45 of
Chapter 2, g has a continuous inverse on g(I). Thus for any sequence {yn } in g(I) converging to
c, there exists a sequence {xn } in I converging to γ such that yn = g(xn ) for all n. We therefore
have
f (yn ) − f (c) g(xn ) − g(γ)
(f ◦ g)(xn ) − (f ◦ g)(γ)
·
=
≤ D(f ◦ g)(c) + yn − c
xn − γ
xn − γ
for n large enough. Rearranging, we obtain
f (yn ) − f (c)
≤ (D(f ◦ g)(c) + ) ·
yn − c
g(xn ) − g(γ)
xn − γ
−1
for n large enough. Taking the lim sup of both sides, we obtain
lim sup
n→∞
f (yn ) − f (c)
D(f ◦ g)(c) + ·≤
yn − c
g 0 (γ)
Since this expression holds for arbitrarily small , we can conclude
lim sup
n→∞
D(f ◦ g)(c)
f (yn ) − f (c)
·≤
yn − c
g 0 (γ)
But since this result holds for any sequence {yn } converging to c, we have Df (c) · g 0 (γ) ≤
D(f ◦ g)(γ).
(ii) Since Df (c) and Df (c) are finite, there exists δ0 > 0 and finite positive number M such that
f (y) − f (c)
<M
y−c
for all y satisfying |y − c| < δ0 . Since g 0 (γ) = 0, for any > 0 there exists δ1 > 0 such that
g(x) − g(γ)
<
x−γ
M
98
if |x − γ| < δ1 . Now let {xn } denote a sequence converging to γ. Then there exists N1 such that
|xn − γ| < δ1 for all n ≥ N1 . By continuity of g, there also exists N0 such that |g(xn ) − g(γ)| < δ0
for n ≥ N0 . We therefore have
(f ◦ g)(xn ) − (f ◦ g)(γ)
g(xn ) − g(γ)
(f ◦ g)(xn ) − (f ◦ g)(γ)
≤
·
<M·
=
xn − γ
g(xn ) − g(γ)
xn − γ
M
for all n ≥ max{N0 , N1 }. But since can be made arbitrarily small, this expression implies
lim
n→∞
(f ◦ g)(xn ) − (f ◦ g)(γ)
=0
xn − γ
for any sequence {xn } converging to γ. We can therefore conclude D(f ◦ g)(γ) = 0.
22. Suppose f is an increasing function on an interval I. Fix a real number y and consider the set
A = {x ∈ I : f (x) ≤ y}
If A is empty, it is trivially measurable. Now suppose A is non-empty. Let x0 and x1 denote two points
in A that satisfy x0 ≤ x1 . Fix λ ∈ [0, 1] and define
xλ = λx0 + (1 − λ)x1
Since f is increasing and xλ ≤ x1 , we must have
f (xλ ) ≤ f (x1 ) ≤ y
Therefore xλ ∈ A. We conclude that A is an interval and therefore measurable by Proposition 8 of
Chapter 2. Since A is measurable for any choice of y, f is a measurable function.
If f is decreasing, then −f is increasing and therefore measurable. f is then measurable by Theorem
6 of Chapter 3.
23. Let M ≥ 0 be a real number that satisfies
−M ≤ Df (x) ≤ Df (x) ≤ M
for all x ∈ (a, b). Fix x0 , x1 ∈ (a, b) and assume without loss that x0 < x1 . Define the function
g(x) = f (x) −
f (x1 ) − f (x0 )
(x − x0 )
x1 − x0
(6)
Since g is a continuous function defined on the closed and bounded interval [x0 , x1 ], the Extreme Value
Theorem implies the existence of some c ∈ [x0 , x1 ] such that g(c) ≤ g(x) for all x ∈ [x0 , x1 ]. Suppose
c ∈ (x0 , x1 ). Then Dg(c) ≥ 0 by Problem 18. Using the result from Problem 20, we have
0 ≤ Dg(c) ≤ Df (c) −
99
f (x1 ) − f (x0 )
x1 − x0
Upon rearrangement, this expression yields
f (x1 ) − f (x0 ) ≤ Df (c) · (x1 − x0 ) ≤ M · (x1 − x0 )
Now suppose c ∈ {x0 , x1 }. Then g(c) = f (x0 ), so we have
f (x1 ) − f (x0 )
f (x) − f (x0 )
≤
x1 − x0
x − x0
for all x ∈ (x0 , x1 ). Thus for all t ∈ (0, x1 − x0 ), we have
f (x1 ) − f (x0 )
f (x0 + t) − f (x0 )
≤
x1 − x0
t
which implies
f (x1 ) − f (x0 )
≤ lim
h→0
x1 − x0
f (x0 + t) − f (x0 )
= Df (x0 ) ≤ M
t
0<|t|≤h
sup
We conclude that
f (x1 ) − f (x0 ) ≤ M · (x1 − x0 )
An analogous argument using lower derivatives in place of upper derivatives can be used to show
f (x0 ) − f (x1 ) ≥ M · (x1 − x0 )
Therefore |f (x1 ) − f (x0 )| ≤ M |x1 − x0 |, so f is Lipschitz on (a, b).
It remains to show that the Lipschitz condition is also satisfied when x0 = a or x1 = b. Suppose x0 = a
but continue to assume x1 ∈ (a, b). Fix > 0. Since f is continuous at a, there exists 0 < δ < b − a
such that
|f (x) − f (a)| < for all x ∈ [a, a + δ). Let {an } denote a sequence in (a, a + δ) that converges to a. Then
|f (x1 ) − f (a)| ≤ |f (x1 ) − f (an )| + |f (an ) − f (a)| ≤ M · |x1 − an | + for all n. Taking the limit as n → ∞, we obtain
|f (x1 ) − f (a)| ≤ M · |x1 − a| + Since can be made arbitrarily small, we have
|f (x1 ) − f (a)| ≤ M · |x1 − a|
An analogous argument can be applied to the other end point to show that f is Lipschitz on [a, b].
24. The derivative of f on (0,1] is given by
2
f (x) = − · cos
x
0
100
1
x2
+ 2x · sin
1
x2
For each natural number k, the function f 0 satisfies
f 0 (x) ≥
on the interval
1
x




1
1

Ek = 

 √π + 2π · k , r 3π
+ 2π · k
4
The integral of |f 0 | over Ek therefore satisfies
Z
|f 0 | ≥
Z
Ek
Ek
where M is any constant larger than
1
2
1
1
= · log
x
2
· log
12
11
≈
1
23 .
1+2·k
3
4 +2·k
≥
M
k
Since {Ek }∞
k=1 is a sequence of disjoint sets in
(0, 1], we must have
Z
1
|f 0 | ≥
0
∞ Z
X
k=1
|f 0 | ≥ M
Ek
∞
X
1
=∞
k
k=1
0
Thus |f | is not integrable.
6.3
Functions of Bounded Variation: Jordan’s Theorem
25. The Weierstrass function is continuous everywhere on [0, 1] but is nowhere differentiable. Therefore by
the contrapositive to Corollary 6, f must not be of bounded variation on any subinterval [a, b] of [0, 1].
26. Fix a natural number n. Define cn =
Pn =
1
√
n· 2
and consider the partition
1 1
2 2
n−1 n−1
0, cn , , + cn , , + cn , · · · ,
,
+ cn , 1
n n
n n
n
n
Then V (f, Pn ) = 2 · n. Since 2 · n diverges, f is not of bounded variation.
27. The total variation of f on [0, x] is given by



sin x


T V (f[0,x] ) = 2 − sin x



4 + sin x
0≤x<
π
2
<x≤
3·π
2
π
2
3·π
2
<x≤2·π
We can write f (x) = h(x) − g(x) where



2 · sin x


h(x) = f (x) + T V (f[0,x] ) = 2



4 + 2 · sin x
g(x) = T V (f[0,x] )
By Lemma 5, h and g are increasing.
101
0≤x<
π
2
<x≤
3·π
2
π
2
3·π
2
<x≤2·π
28. Since f is a step function on [a, b], there exists a partition P = {x0 , x1 , · · · , xn } of [a, b] and numbers
c1 , · · · , cn such that
f (x) = ci if xi−1 < x < xi
Now consider the partition
0
P =
x0 + x1
xn−1 + xn
x0 ,
, x1 , · · · , xn−1 ,
, xn
2
2
Any refinement of P 0 does not change the variation of f , so we have
T V (f ) = V (f, P 0 ) =
n
X
(|f (xk−1 ) − ck | + |f (xk ) − ck |)
k=1
29. (a) Fix a natural number n and consider the partition
( r
Pn =
We then have
0,
2
,
2·n·π
s
2
,··· ,
(2 · n − 1) · π
r
2
,
2·π
r )
2
π
1
1
1
2
+
V
= · 1 + + ···
0, π
π
2
n−1 n
h
i
√
Since the harmonic series diverges and V f 0, 2 , Pn ≤ V f[−1,1] , Pn , f is not of bounded
fh
√ 2 i , Pn
π
variation on [−1, 1].
(b) Observe that
g 0 (x) = − sin(1/x) + 2 · x · cos(1/x)
for x 6= 0. We also have
Diff h g(0) = h · cos(1/h)
Taking the limit as h goes to zero, we can conclude
g 0 (0) = 0
Therefore g(·) is differentiable on [−1, 1] and its derivative satisfies |g 0 (x)| ≤ 3. This means g(·) is
Lipschitz on [−1, 1], which implies g(·) is of bounded variation on [−1, 1].
30. That linear combinations of functions of bounded variation are also of bounded variation follows immediately from Problem 34.
Suppose f and g are real-valued functions of bounded variation on the closed, bounded interval [a, b].
Then both f and g must be bounded on [a, b]. Thus there exists a real number M > 0 such that
|f (x)| < M and |g(x)| < M for all x ∈ [a, b]. Now let P = {x0 , · · · , xk } denote a partition of [a, b].
102
Then
V (f · g, P ) =
k
X
|f (xi ) · g(xi ) − f (xi−1 ) · g(xi−1 )|
i=1
≤
k
X
(|f (xi ) − f (xi−1 )| · |g(xi )| + |g(xi ) − g(xi−1 )| · |f (xi−1 |)
i=1
≤ (V (f, P ) + V (g, P )) · M
≤ (T V (f, P ) + T V (g, P )) · M
Since V (f · g, P ) is bounded above, f · g is of bounded variation on [a, b].
31. Let P 0 = {x0 , · · · , xk } denote a partition of [a, b]. Suppose x ∈ (a, b) is not in P 0 . Let P denote the
partition obtained from adjoining x to P 0 . Then there exists n ∈ {1, · · · , k} such that xn−1 < x < xn .
By the Triangle Inequality, we have
V (f, P ) =
X
|f (xi ) − f (xi−1 )| + |f (x) − f (xn−1 )| + |f (xn ) − f (x)|
i6=n
≥
X
|f (xi ) − f (xi−1 )| + |f (xn ) − f (xn−1 )|
i6=n
= V (f, P 0 )
But this proves the desired result, since any refinement of P 0 can be formed by sequentially adjoining
a finite number of additional partition points.
32. From the definition of total variation, we have
T V (f ) = sup{V (f, P ) | P a partition of [a, b]}
Now fix a partition P 0 of [a, b] and define
s = sup{V (f, P ) | P a refinement of P 0 }
For any partition P of [a, b], we can form a refinement P 00 of P 0 by combining the partition points
of P and P 0 . By Problem 31, we have V (f, P ) ≤ V (f, P 00 ) ≤ s. Since P was arbitrary, s ≥ T V (f ).
Furthermore, we must have T V (f ) ≥ s since all refinements of P 0 are partitions of [a, b]. We can
therefore conclude T V (f ) = s.
Now set P0 = {a, b}. Since T V (f ) = sup{V (f, P ) | P a refinement of P0 }, we can find a refinement P1
of P0 that satisfies
|T V (f ) − V (f, P1 )| < 1
Moreover, by Problem 31 we have V (f, P0 ) ≤ V (f, P1 ). Continuing in this manner, we can construct
a sequence of partitions {Pn } that satisfies
|T V (f ) − V (f, Pn )| <
and V (f, Pn−1 ) ≤ V (f, Pn ) for all natural numbers n.
103
1
n
33. Suppose T V (f ) < ∞. Fix > 0 and choose a partition P = {x0 , · · · , xk } of [a, b] such that
V (f, P ) ≥ T V (f ) −
2
By pointwise convergence, there exists an index N such that
|fn (xi ) − f (xi )| <
,
4·k
i = 0, 1, · · · , k
for all n ≥ N . We therefore have
T V (f ) ≤ V (f, P ) +
=
k
X
2
|f (xi ) − f (xi−1 )| +
i=1
≤
k
X
|f (xi ) − fn (xi )| +
i=1
2
k
X
|fn (xi ) − fn (xi−1 )| +
i=1
k
X
|f (xi−1 ) − fn (xi−1 )| +
i=1
2
< V (fn , P ) + ≤ T V (fn ) + for all n ≥ N . But this implies
T V (f ) ≤ lim inf T V (fn )
Now suppose T V (f ) = ∞. Fix a real number M > 0 and choose a partition P = {x0 , · · · , xk } of [a, b]
such that
V (f, P ) ≥ M
By pointwise convergence, for any > 0 there exists an index N such that
|fn (xi ) − f (xi )| <
,
2·k
i = 0, 1, · · · , k
for all n ≥ N . Using the same steps as before, we can conclude
M ≤ V (f, P ) ≤ T V (fn )
for all n ≥ N . Since M can be arbitrarily large, we must have lim inf T V (fn ) = ∞.
34. By the triangle inequality, we have
V (f + g, P ) =
k
X
|f (xi ) − f (xi−1 ) + g(xi ) − g(xi−1 )|
i=1
≤
k
X
|f (xi ) − f (xi−1 )| +
i=1
k
X
i=1
= V (f, P ) + V (g, P )
104
|g(xi ) − g(xi−1 )|
for any partition P of [a, b]. But this implies
T V (f + g) ≤ sup{V (f, P ) + V (g, P ) | P a partition of [a, b]}
≤ sup{V (f, P ) | P a partition of [a, b]} + sup{V (g, P ) | P a partition of [a, b]}
= T V (f ) + T V (g)
We also have
V (αf, P ) =
k
X
|αf (xi ) − αf (xi−1 )|
i=1
= |α|
k
X
|f (xi ) − f (xi−1 )|
i=1
= |α| · V (f, P )
which implies
T V (αf ) = sup{|α| · V (f, P ) | P a partition of [a, b]}
= |α| · sup{V (f, P ) | P a partition of [a, b]}
= |α| · T V (f )
35. The derivative of f on (0, 1] is given by
f 0 (x) = −β · xα−β−1 · cos(1/xβ ) + α · xα−1 · sin(1/xβ )
Suppose α > β > 0. Then f 0 is integrable:
Z
1
0
Z
|f | ≤
0
1
(β · xα−β−1 + α · xα−1 )dx =
0
By the Fundamental Theorem of Calculus, we have
Z
x
f 0 = f (x) − f (y)
y
for any x > y > 0. Taking the limit as y → 0, we obtain
Z
f (x) =
0
105
x
f0
β
+1<∞
α−β
by continuity of integration. Now let P = {x0 , · · · , xk } be a partition of [0, 1]. We then have
V (f, P ) =
k
X
|f (xi ) − f (xi−1 )|
i=1
=
k Z
X
≤
k
X
Z
xi−1
f0
0
xi−1
Z
i=1
xi
k Z
X
xi
f0
|f 0 |
xi−1
i=1
1
Z
f0 −
0
i=1
=
xi
|f 0 |
=
0
<∞
Since P was arbitrary, f is of bounded variation.
Now suppose β ≥ α > 0. Fix an index n and consider the following partition:
( 1/β 1/β
1/β )
2
2
2
Pn = 0,
,
,··· ,
,1
2nπ
(2n − 1)π
π
It can be shown
V (f, Pn ) = c
n−1
X
k=0
≥
1
+ sin(1) − c
(1 + 2k)α/β
n
X
c
2α/β
1
+ sin(1) − c
α/β
k
k=1
where
c=2
α/β
2
π
Since 0 < α/β ≤ 1, the series on the right-hand side diverges. Therefore f is not of bounded variation.
36. Corrected exercise: Let f fail to be of bounded variation on [0, 1]. Show that there is a
point x0 in [0, 1] such that f fails to be of bounded variation on each subinterval of [0, 1]
that contains x0 in its interior.
Suppose f is not of bounded variation on an interval [an , bn ] ⊆ [0, 1]. Define
an+1 =
bn+1

a
if T V (f[an ,(an +bn )/2] ) = ∞
(a + b )/2
n
n
otherwise
n

(a + b )/2
n
n
=
b
n
if T V (f[an ,(an +bn )/2] ) = ∞
otherwise
106
Since T V (f[an ,bn ] ) = T V (f[an ,(an +bn )/2] )+T V (f[(an +bn )/2,bn ] ) = ∞, f must not be of bounded variation
on [an+1 , bn+1 ]. Starting at a1 = 0 and b1 = 1, we can use this recursion to define a descending,
countable collection of nonempty closed intervals {[an , bn ]}∞
n=1 that satisfy T V (f[an ,bn ] ) = ∞ for all n.
T∞
By the Nested Set Theorem, there exists an x0 ∈ n=1 [an , bn ]. Choose a subinterval [a, b] of [0, 1] that
contains x0 in its interior. Then there exists n large enough such that [an , bn ] ⊆ [a, b]. But this implies
T V (f[a,b] ) ≥ T V (f[an ,bn ] ) = ∞, so f is not of bounded variation on [a, b].
6.4
37.
Absolutely Continuous Functions
(i) Let
f (x) =

x cos(π/2x)
if 0 < x ≤ 1
0
if x = 0
It was shown in Section 6.3 that f is not of bounded variation on [0, 1]. Therefore by Theorem
8, f is not absolutely continuous on [0, 1]. But for each > 0, f is continuous with bounded first
derivative on [, 1]. Therefore f is Lipschitz on [, 1] by the Mean Value Theorem, implying that
f is absolute continuous on [, 1] by Proposition 7.
(ii) Fix > 0. Let δ1 correspond to the /2 challenge for continuity at 0. Let δ2 correspond to the
/2 challenge of absolute continuity on [δ1 , 1]. Define δ = min(δ1 , δ2 ). Choose a finite, disjoint
Pn
collection of open intervals {(ak , bk )}nk=1 contained in [0, 1] that satisfy
k=1 (bk − ak ) < δ.
Assume the intervals are ordered so that
a1 < b1 < a2 < b2 < · · · < an < bn
n
Also assume there exists an index m such that {(ak , bk )}m
k=1 is contained in [0, δ1 ) and {(ak , bk )}k=m+1
is contained in [δ1 , 1] (if δ1 ∈ (ak , bk ) for some k, replace (ak , bk ) with (ak , δ1 ) and (δ1 , bk )). Since
bm < δ1 , we have
m
X
|f (bk ) − f (ak )| = f (bm ) − f (a1 ) −
k=1
m
X
(f (ak ) − f (bk−1 ))
k=2
≤ f (bm ) − f (0)
< /2
where the first two lines follow because f is increasing and the third line follows from continuity
Pn
at 0. Since m+1 (bk − ak ) < δ2 , we also have
n
X
|f (bk ) − f (ak )| < /2
k=m+1
by absolute continuity of f on [δ1 , 1]. Combining these results, we can conclude
n
X
|f (bk ) − f (ak )| < k=1
107
which implies that f is absolutely continuous on [0, 1].
(iii) Fix ∈ (0, 1). Then |f 0 (x)| ≤
1
√
2 for all x ∈ [, 1], implying that f is Lipschitz and thus
absolutely continuous on [, 1]. Since f is also increasing on [0, 1], we can conclude that f is
absolutely continuous on [0, 1] by part (ii).
To see that f is not Lipschitz on [0, 1], fix a constant c ≥ 0. If c = 0, then
|f (x1 ) − f (x0 )| > 0 = c · |x1 − x0 |
for x0 , x1 ∈ [0, 1] satisfying x0 6= x1 . If c > 0, let x0 = 0 and choose x1 ∈ (0, min(1/c2 , 1)). We
then have
|f (x1 ) − f (x0 )| =
√
x1 > c · x1 = c · |x1 − x0 |
Therefore there can exist no Lipschitz constant for f on [0, 1].
38. Suppose f is absolutely continuous on [a, b]. Fix > 0 and let δ > 0 correspond to the /2-challenge
for absolute continuity. Let {(ak , bk )}∞
k=1 be a countable disjoint collection of open intervals in [a, b]
P∞
satisfying k=1 (bk − ak ) < δ. Then for each index n, we have
n
X
|f (bk ) − f (ak )| < /2
k=1
by absolute continuity. But since this expression holds for all n, we have
∞
X
|f (bk ) − f (ak )| ≤ /2 < k=1
as desired.
Since every finite collection is countable, the converse follows immediately from the definition of absolute continuity.
39. Suppose f is absolutely continuous on [a, b]. Fix > 0 and let δ correspond to the -challenge in
Problem 38. Pick a measurable subset E of [a, b] that satisfies m(E) < δ. By Theorem 11 of Chapter
S∞
2, there exists a countable collection of disjoint open intervals {(ak , bk )}∞
k=1 such that E ⊆
k=1 (ak , bk )
S∞
and m ( k=1 (ak , bk ) ∼ E) < δ − m(E). We then have
∞
X
k=1
(bk − ak ) = m
∞
[
!
(ak , bk )
=m
k=1
∞
[
!
(ak , bk ) ∼ E
k=1
108
+ m(E) < δ
But this implies
∗
∗
m (f (E)) ≤ m
∞
[
f
!!
(ak , bk )
k=1
∞
[
∗
=m
!
f ((ak , bk ))
k=1
∞
[
≤m
!
[f (ak ), f (bk )]
k=1
=
∞
X
(f (bk ) − f (ak ))
k=1
<
where the first line follows from monotonicity of outer measure, the second line follows because the
image of a union equals the union of images, the third line follows from the Intermediate Value Theorem
and monotoncity of f , the fourth line follows from countable subadditivity, and the final line follows
from Problem 38.
Correction: The converse requires f to be continuous. Suppose the condition in the problem
is true. Fix > 0 and let δ correspond to the -challenge in the problem. Let {(ak , bk )}nk=1 denote
Pn
a collection of disjoint open intervals contained in [a, b] that satisfy k=1 (bk − ak ) < δ. Since f is
increasing, {(f (ak ), f (bk ))}nk=1 is also a collection of disjoint open intervals. We therefore have
n
X
|f (bk ) − f (ak )| =
k=1
n
X
m ((f (ak ), f (bk )))
k=1
=m
≤m
n
[
!
(f (ak ), f (bk ))
k=1
n
[
!
f ((ak , bk ))
k=1
=m f
n
[
!!
(bk − ak )
k=1
<
where the first line follows by definition of the measure of an interval, the second line follows by
additivity of measure, the third line follows from the Intermediate Value Theorem and monotonicity
of f , the fourth line follows because the image of a union equals the union of images, and the last line
follows from the condition in the problem.
40. Suppose f is an increasing, absolutely continuous function on [a, b]. Let E ⊆ [a, b] denote a set of
measure zero. By Problem 39, for any > 0 we have m∗ (f (E)) < . Since can be chosen to be
arbitrarily small, we must have m∗ (f (E)) = 0.
The above result implies ψ(x) is not absolutely continuous. But since ψ(x) = ϕ(x) − x and x is
absolutely continuous, ϕ must not be absolutely continuous by Problem 42.
109
41. Suppose E is measurable. By Theorem 11 of Chapter 2, there exists an Fσ set F such that F ⊆ E
and m∗ (E ∼ F ) = 0. By (i), f (F ) is an Fσ set and is also therefore measurable. By (ii), f (E ∼ F )
has measure 0 and is therefore measureable. Since f (E) = f (F ∪ (E ∼ F )) = f (F ) ∪ f (E ∼ F ), we
can infer that f (E) is measurable as well.
(i) Let E denote a closed subset of [a, b]. Since E is closed and bounded, E is compact. The
continuous image of a compact set is compact, so f (E) is also closed and bounded.
Now suppose F is an Fσ set. Then there exists a countable collection of closed sets {En }∞
n=1 such
S∞
S∞
that F = n=1 En . But this implies f (F ) = n=1 f (En ). By the above result, we know that
f (En ) is closed for all n; therefore f (F ) is also an Fσ set.
(ii) Follows from Problem 40.
42. Sum: Suppose f and g are absolutely continuous on [a, b]. Fix > 0. Let δ1 and δ2 correspond
to the /2-challenge of absolute continuity for f and g on [a, b], respectively. Define δ = min(δ1 , δ2 )
and let {(ak , bk )}nk=1 denote a finite collection of disjoint open intervals contained in [a, b] that satisfy
Pn
k=1 (bk − ak ) < δ. Then
n
X
|(f (bk ) + g(bk )) − (f (ak ) + g(ak ))| ≤
k=1
n
X
|f (bk ) − f (ak )| +
k=1
n
X
|g(bk ) − g(ak )|
k=1
<
where the first line follows from the Triangle Inequality and the second line follows from the definition
of absolute continuity. We can therefore conclude that f + g is absolutely continuous on [a, b].
Product: Suppose f and g are absolutely continuous on [a, b]. Fix > 0. Since f and g are continuous
on [a, b], they must be uniformly bounded on [a, b] by the Extreme Value Theorem. Thus there exists
2·M -challenge of
{(ak , bk )}nk=1 denote
M ≥ 0 such that |f | ≤ M and |g| ≤ M on [a, b]. Let δ1 and δ2 correspond to the
absolutely continuity for f and g on [a, b], respectively. Let δ = min(δ1 , δ2 ) and let
Pn
a finite collection of disjoint open intervals contained in [a, b] that satisfy k=1 (bk − ak ) < δ. Then
n
X
|f (bk ) · g(bk ) − f (ak ) · g(ak )| =
k=1
≤
≤
n
X
k=1
n
X
k=1
n
X
|(f (bk ) − f (ak )) · g(bk ) + (g(bk ) − g(ak )) · f (ak )|
|f (bk ) − f (ak )| · |g(bk )| +
n
X
|g(bk ) − g(ak )| · |f (ak )|
k=1
|f (bk ) − f (ak )| · M +
k=1
n
X
|g(bk ) − g(ak )| · M
k=1
<
Therefore f · g is absolutely continuous on [a, b].
43.
(i) By Problem 37, we know that f is continuous on [0, 1]. An analogous argument can be used
to show that f is absolutely continuous on [−1, 0]. But if f is absolutely continuous on [a, b]
and on [b, c], then f is absolutely continuous on [a, c]. To show this, fix > 0. Let δ1 and
δ2 correspond to the /2 challenge of absolutely continuity for f on [a, b] and [b, c] respectively.
Define δ = min(δ1 , δ2 ) and choose a finite disjoint collection {(ak , ck )}nk=1 of open intervals in
110
(a, c) that satisfy
Pn
k=1 (ck
− ak ) < δ. Without loss, assume there exists an index m such that
(ak , ck ) ⊆ [a, b] for k = 1, · · · , m and (ak , ck ) ⊆ [b, c] for k = m + 1, · · · , n. Then
n
X
|f (ck ) − f (ak )| =
k=1
m
X
n
X
|f (ck ) − f (ak )| +
k=1
|f (ck ) − f (ak )| < k=m+1
The first derivative of g is given by

sin
g 0 (x) =
0
π
2x
·
π
2
+ 2 · x · cos
π
2x
if x 6= 0
if x = 0
Therefore |g 0 (x)| ≤ 4 for all x ∈ [−1, 1], which implies that g is Lipschitz and thus absolutely
continuous on [−1, 1].
(ii)
2
2
2
+
+ · · · 2/3
2/3
2/3
(2n)
(2n − 2)
2
1
1
+
+
·
·
·
+
1
= 21/3
n2/3
(n − 1)2/3
n
X
1
= 21/3
2/3
k
k=1
V (f ◦ g, Pn ) =
(iii) Since
Pn
1
k=1 kp
diverges for 0 < p ≤ 1, V (f ◦ g, Pn ) is unbounded above. Therefore f ◦ g is not of
bounded variation and thus not absolutely continuous on [−1, 1].
44. Fix > 0 and let c ≥ 0 denote a Lipschitz constant for f . If c = 0, then f ◦ g is constant and the result
is trivial. If c > 0, let δ correspond to the /c-challenge for absolutely continuity of g on [a, b]. Let
Pn
{(ak , bk )}nk=1 be a finite disjoint collection of open intervals in (a, b) that satisfy k=1 (bk − ak ) < δ.
Then
n
X
|(f ◦ g)(bk ) − (f ◦ g)(ak )| ≤ c
k=1
n
X
|g(bk ) − g(ak )| < k=1
Therefore f ◦ g is absolutely continuous on [a, b].
45. Assume without loss that g is strictly increasing on [a, b]. Fix > 0 and let δ 0 correspond to the
-challenge for absolutely continuity of f on R. Let δ correspond to the δ 0 -challenge for absolutely
continuity of g on [a, b]. Let {(ak , bk )}nk=1 be a finite disjoint collection of open intervals in (a, b)
Pn
that satisfy k=1 [bk − ak ] < δ. Because g is strictly increasing, {(g(ak ), g(bk ))}nk=1 is a finite disjoint
Pn
0
collection of open intervals in R. Moreover,
k=1 [g(bk ) − g(ak )] < δ by absolute continuity of g.
Pn
Therefore k=1 |(f ◦ g)(bk ) − (f ◦ g)(ak )| < by absolute continuity of f .
46.
• Assertion 1: FLip , FAC and FBV are closed under linear combinations:
That FBV and FAC are closed under linear combinations follows directly from Problems 34 and
42, respectively. Suppose f and g are Lipschitz on E. Let c1 and c2 denote their respective
111
Lipschitz constants. Then for any x, x0 ∈ E, we have
|(f (x) + g(x)) − (f (x0 ) + g(x0 ))| ≤ |f (x) − f (x0 )| + |g(x) − g(x0 )| ≤ (c1 + c2 ) · |x − x0 |
Therefore f + g is also Lipschitz, so FLip is closed under addition. Now let α ∈ R denote an
arbitrary scalar. Then
|α · f (x) − α · f (x0 )| = |α| · |f (x) − f (x0 )| ≤ |α| · c · |x − x0 |
Therefore α · f is also Lipschitz and FLip is closed under scalar multiplication. We conclude that
FLip is closed under linear combinations.
• Assertion 2: A function in one of these collections has its total variation function in the same
collection:
This claim was shown for FAC in the proof of Theorem 8.
Suppose f is bounded variation on [a, b]. Since g(x) = T V (f[a,x] ) is increasing and g(a) = 0,
T V (g[a,b] ) = T V (f[a,b] ) < ∞. Thus T V (f[a,x] ) is also of bounded variation.
Now suppose f is Lipschitz on [a, b] with Lipschitz constant c. Pick x, x0 ∈ [a, b] and assume
without loss that x ≤ x0 . Then
|T V (f[a,x] ) − T V (f[a,x0 ] )| = |T V (fx,x0 )|
≤ c · |x − x0 |
where the first line follows from (20) on page 117 in the textbook, and the second line follows
from the example on the top of page 117. Therefore T V (f[a,x] ) is also Lipschitz.
47. If f is absolutely continuous, then the condition in the problem follows immediately from the triangle
inequality.
Conversely, suppose the condition in the problem holds. Fix > 0 and let δ correspond to the /2challenge in the prompt. Let {(ak , bk )}nk=1 be a finite disjoint collection of open intervals in (a, b) that
Pn
satisfy k=1 (bk − ak ) < δ. Then
X
[bk − ak ] < δ
k:f (bk )≥f (ak )
which implies
X
X
|f (bk ) − f (ak )| =
k:f (bk )≥f (ak )
[f (bk ) − f (ak )] < /2
k:f (bk )≥f (ak )
Likewise,
X
[bk − ak ] < δ
k:f (bk )<f (ak )
112
which implies
X
X
|f (bk ) − f (ak )| = −
[f (bk ) − f (ak )]
k:f (bk )<f (ak )
k:f (bk )<f (ak )
X
=
[f (bk ) − f (ak )] < /2
k:f (bk )<f (ak )
Therefore
n
X
X
|f (bk ) − f (ak )| =
k=1
X
|f (bk ) − f (ak )| +
k:f (bk )≥f (ak )
|f (bk ) − f (ak )| < k:f (bk )<f (ak )
Thus δ satisfies the -challenge for absolute continuity, so f is absolutely continuous on [a, b].
6.5
Integrating Derivatives: Differentiating Indefinite Integrals
48. ϕ is singular, but ϕ(0) = 0 and ϕ(1) = 1. We therefore have
Z
1
ϕ0 = 0 6= 1 = ϕ(1) − ϕ(0)
0
Theorem 10 then implies that ϕ is not absolutely continuous.
In contrast with the reasoning in Problem 40, the above argument exploits a property of the derivative
of ϕ.
49. By Problem 11, we know that
Z
b
Diff 1/n f = Av1/n f (b) − Av1/n f (a)
a
for all n. Since f is continuous at a, for any > 0, there exists an index N such that
if |x − a| <
1
then |f (x) − f (a)| < N
Thus for all n ≥ N , we have
a+1/n
Z
f − f (a)
Av1/n f (a) − f (a) = n
a
a+1/n
Z
(f − f (a))
= n
a
≤
≤
1
n
Z
1
n
Z
=
113
a+1/n
|f (x) − f (a)|
a
a+1/n
a
(1)
Since > 0 can be made arbitrarily small, we conclude that limn→∞ Av1/n f (a) = f (a). Likewise,
limn→∞ Av1/n f (b) = f (b). Taking limits of (1), we obtain
b
Z
Diff 1/n f = f (b) − f (a)
lim
n→∞
a
Since f 0 = limn→∞ Diff 1/n f a.e. on [a, b], we also know that
b
Z
Z
0
f =
a
Therefore if limn→∞
Rb
a
b
Z
Rb
a
a
lim Diff 1/n f
a n→∞
limn→∞ Diff 1/n f , then
b
Z
f 0 = lim
Z
Diff 1/n f =
n→∞
a
Conversely, if
Rb
Diff 1/n f =
b
a
b
lim Diff 1/n f = f (b) − f (a)
a n→∞
f 0 = f (b) − f (a) then,
b
Z
lim
n→∞
b
Z
f 0 = f (b) − f (a) = lim Diff 1/n f
Diff 1/n f =
a
n→∞
a
50. By the Vitali Convergence Theorem, we have
b
Z
f0 =
Z
Rb
a
b
Z
lim Diff 1/n f = lim
a n→∞
a
Therefore
b
Diff 1/n f
n→∞
a
f 0 = f (b) − f (a) by Problem 49.
51. By the Dominated Convergence Theorem, we know that
b
Z
Z
0
f =
a
Therefore
Rb
a
b
b
Z
lim Diff 1/n f = lim
a n→∞
Diff 1/n f
n→∞
a
f 0 = f (b) − f (a) by Problem 49.
52. f · g is absolutely continuous by Problem 42. Theorem 10 then implies
Z
a
(f · g)0 = f (b) · g(b) − f (a) · g(a)
b
By the Product Rule, (f · g)0 = f 0 · g + g 0 · f a.e. [a, b]. Therefore
a
Z
f0 · g +
b
Z
a
g 0 · f = f (b) · g(b) − f (a) · g(a)
b
by linearity of integration. Since g is continuous on [a, b], there exists M > 0 such that |g| ≤ M on
[a, b]. We also know that f 0 is integrable by Theorem 10. Therefore
Z
b
a
|f 0 · g| =
Z
a
|f 0 | · |g| ≤ M
b
Z
b
114
a
|f 0 | < ∞
by monotonicity of integration. We can thus conclude
Z
a
f 0 · g = f (b) · g(b) − f (a) · g(a) −
Z
b
a
f0 · g
b
53. Suppose f is Lipschitz. Then there exists c ≥ 0 such that
f (x + t)
≤c
t
for any x and t that satisfy x + t ∈ [a, b]. Thus if f is differentiable at x, then
f (x + t)
≤c
t
|f 0 (x)| = lim
t→0
Since f is differentiable a.e. by Theorem 10, we must have |f 0 | ≤ c a.e. on [a, b].
Conversely, suppose there exists c ≥ 0 such that |f 0 | ≤ c a.e. on [a, b]. If x, x0 are in [a, b] and x ≤ x0 ,
then f is absolutely continuous on [x, x0 ]. We therefore have
0
Z
x0
0
|f (x) − f (x )| =
Z
x0
f ≤
x
|f 0 | ≤ c|x − x0 |
x
by Theorem 10 and monotonicity of integration. Therefore f is Lipschitz on [a, b].
54.
(i) Fix > 0 and δ > 0. Define the collection of closed, bounded, non-degenerate intervals
F=
d−c
[c, d] ⊆ (a, b) : c < d and f (d) − f (c) < ·
b−a
Since f is singular, there exists a set E0 with m(E0 ) = 0 such that f 0 (x) = 0 for all x ∈ (a, b) ∼ E0 .
f (x+h)−f (x)
h
Let E = (a, b) ∼ E0 . Because f is increasing and limh→0
= 0 for all x ∈ E, F covers
E in the sense of Vitali. By the Vitali Covering Lemma, there exists a finite, disjoint subcollection
{(ck , dk )}N
k=1 of F for which
m E∼
N
[
!
[ck , dk ]
<δ
k=1
Without loss, assume the subcollection is ordered so that
c1 < d1 < c2 < d2 < · · · < cN < dN
Then
(a, b) ∼
N
[
[ck , dk ] =
k=1
115
n
[
(ak , bk )
k=1
where
n=N +1

a
ak =
d
if k = 1
k−1

c
k
bk =
b
if k = 2, · · · , n
if k = 1, · · · , n − 1
if k = n
Then
m
n
[
!
(ak , bk )
=m E∼
k=1
N
[
!
(ck , dk ) ∼ E0
+m E ∼
k=1
≤m E∼
N
[
N
[
!
(ck , dk ) ∩ E0
k=1
!
(ck , dk )
+ m (E0 )
k=1
<δ
Since [ck , dk ] ∈ F for k = 1, · · · , N , we also have
n
X
[f (bk ) − f (ak )] = f (b) − f (a) −
k=1
N
X
[f (dk ) − f (ck )]
k=1
PN
> f (b) − f (a) − ·
− ck )
b−a
k=1 (dk
≥ f (b) − f (a) − (ii) Fix > 0. By Corollary 4, f 0 is integrable over [a, b]. By Proposition 23 of Chapter 4, there
exists δ > 0 such that
Z
if A ⊆ [a, b] is measurable and m(A) < δ, then
f0 <
A
2
Choose a finite disjoint collection {(ak , bk )}nk=1 of open intervals in (a, b) for which
n
X
[bk − ak ] < δ and f (b) − f (a) −
k=1
Then
n
X
[f (bk ) − f (ak )] <
k=1
Z
f0 <
Sn
k=1 (ak ,bk )
2
2
by (2). Assume without loss that the intervals are ordered so that
a = b0 ≤ a1 < b1 < a2 < b2 < · · · < an < bn ≤ an+1 = b
116
(2)
We then have
f (b) − f (a) −
n
X
[f (bk ) − f (ak )] =
k=1
n+1
X
[f (ak ) − f (bk−1 )] ≥
k=1
n+1
X Z ak
k=1
0
Z
f0
f =
bk−1
[a,b]∼
Sn
k=1 (ak ,bk )
where the inequality follows from Corollary 4. Thus
b
Z
f0 =
Z
f0 +
Z
Sn
a
k=1 (ak ,bk )
f0 < [a,b]∼
Rb
Since can be made arbitrarily small, we must have
Sn
k=1 (ak ,bk )
f 0 = 0. But f 0 ≥ 0 because f is increasing,
a
so f 0 = 0 a.e. by Proposition 9 of Chapter 4.
PN
(iii) Fix > 0 and δ > 0. Since limN →∞ n=1 (fn (b) − fn (a)) = f (b) − f (a), there exists an index N
for which
N
X
<
[fn (b) − fn (a)]
2 n=1
f (b) − f (a) −
The function
PN
n=1 [fn (x)
(3)
− fn (x)] is increasing and singular on [a, b], so by part (i) there exists
a finite disjoint collection {(ak , bk )}K
k=1 of open intervals in (a, b) for which
K
X
[bk − ak ] < δ
k=1
and
K X
N
X
[fn (bk ) − fn (ak )] >
k=1 n=1
N
X
[fn (b) − fn (a)] −
n=1
2
(4)
for k = 1, · · · , K. Combining (3) and (4), we obtain
N
K X
X
f (b) − f (a) − <
[fn (bk ) − fn (ak )]
(5)
k=1 n=1
Since each fn is increasing, we also know that
N
X
n=1
[fn (bk ) − fn (ak )] ≤
∞
X
[fn (bk ) − fn (ak )] = f (bk ) − f (ak )
n=1
for k = 1, · · · , K. Substituting into (5), we obtain
f (b) − f (a) − <
K
X
[f (bk ) − f (ak )]
k=1
But then f satisfies the property in part (i), so we can conclude that f is singular by part (ii).
55.
(i) Since f is of bounded variation, v is of bounded variation by the remark at the end of Section 6.4.
Therefore both f and v are differentiable almost everywhere on [a, b] by Corollary 6. But then
there exists a set E ⊆ [a, b] of measure zero such that f and v are differentiable on [a, b] ∼ E.
117
Now for any interval [c, d] ⊆ [a, b], we have
f (d) − f (c) ≤ |f (d) − f (c)| ≤ T V (f[c,d] ) = T V (f[a,d] ) − T V (f[a,c] ) = v(d) − v(c)
Therefore
f (d) − v(d) ≤ f (c) − v(c) for all a ≤ c < d ≤ b
Thus f (x) − v(x) is decreasing on [a, b]. We also know that f (x) + v(x) is increasing on [a, b] by
Lemma 5. Therefore for all x ∈ [a, b] ∼ E, we have
f 0 (x) − v 0 (x) ≤ 0 and f 0 (x) + v 0 (x) ≥ 0
which implies |f 0 | ≤ v 0 almost everywhere on [a, b]. By monotonicity of integration and Corollary
4, we have
b
Z
|f 0 | ≤
b
Z
a
v 0 ≤ v(b) − v(a) = T V (f[a,b] )
a
(ii) Suppose f is absolutely continuous. Fix > 0 and choose a partition P = {x0 , x1 , · · · , xk } that
satisfies
k
X
|f (xi ) − f (xi−1 )| > T V (f[a,b] ) − i=1
Since f is absolutely continuous, we have
k
X
|f (xi ) − f (xi−1 )| =
k Z
X
i=1
xi
0
f ≤
k Z
X
xi−1
i=1
i=1
xi
0
Z
|f | ≤
xi−1
b
|f 0 |
a
by Theorem 10. We can therefore conclude
Z
T V (f[a,b] ) − ≤
b
|f 0 |
a
Rb
for any > 0, which implies T V (f[a,b] ) ≤ a |f 0 |. As the reverse inequality was proven in part
Rb
(i), we must have T V (f[a,b] ) = a |f 0 |.
Rb
Conversely, suppose T V (f[a,b] ) = a |f 0 |. Then
b
Z
|f 0 | ≤
T V (f[a,b] ) =
a
Z
b
v 0 ≤ T V (f[a,b] )
a
by part (i). But this expression implies
Z
b
(v 0 − |f 0 |) = 0
a
Since v 0 − |f 0 | ≥ 0, v 0 = |f 0 | a.e. on [a, b] by Proposition 9 of Chapter 4. Therefore for any
a ≤ c < d ≤ b, we have
Z
d
0
Z
|f | =
c
d
v 0 = T V (f[c,d] ) ≥ |f (d) − f (c)|
c
118
Now fix > 0. Since f 0 is integrable, there exists a δ > 0 such that for any finite, disjoint
collection {(ak , bk )}nk=1 of open intervals contained in [a, b],
n
X
if
Z
[bk − ak ] < δ
∪n
k=1 (ak ,bk )
k=1
But since |f (bk ) − f (ak )| ≤
n
X
R bk
ak
|f 0 | < then
|f 0 | for k = 1, · · · , n, we have
|f (bk ) − f (ak )| ≤
k=1
n Z
X
bk
Z
0
k=1
|f 0 | < |f | =
∪n
k=1 (ak ,bk )
ak
for any finite, disjoint collection {(ak , bk )}nk=1 of open intervals contained in [a, b] satisfying
Pn
k=1 [bk − ak ] < δ. Therefore f is absolutely continuous on [a, b].
(iii) Corollaries 4 and 12 can be viewed as an application of the results in parts (i) and (ii) to increasing
functions.
56.
(i) By Proposition 9 of Chapter 1, O =
intervals
{(ak , bk )}∞
k=1 .
S∞
k=1 (ak , bk )
for some countable collection of disjoint open
The Intermediate Value Theorem and Theorem 10 then imply
Z
bk
m(g((ak , bk ))) = g(bk ) − g(ak ) =
g0
ak
for k = 1, 2, · · · . By countable additivity of measure and integration, we have
∞
[
m(g(O)) = m g
!!
(ak , bk )
k=1
∞
[
=m
!
g((ak , bk ))
k=1
=
=
∞
X
m (g((ak , bk )))
k=1
∞ Z bk
X
k=1
Z
=
g0
ak
g0
O
(ii) Since E is a Gδ -set, there exists a countable collection of open sets {Gk }∞
k=1 such that E =
T∞
G
.
Because
g
is
injective,
we
have
k=1 k
g(E) = g
∞
\
k=1
!
Gk
=
∞
\
g(Gk )
k=1
Since g is continuous and strictly increasing, g maps open intervals to open intervals. Therefore g
maps open sets to open sets, which implies that g(E) is a Gδ -set. Therefore g(E) is measurable.
Tn
Now assume without loss that G1 ⊆ (a, b). Define On = k=1 Gk and let En = On ∼ E. By
119
Problem 39, for any > 0 there exists a δ > 0 such that
if m(En ) < δ, then m(g(En )) < Since En ↓ ∅ and m(E1 ) ≤ b − a, limn→∞ m(En ) = 0 by continuity of measure. Therefore
there exists an index N such that m(En ) < δ for all n ≥ N . The above expression then implies that m(g(En )) < for all n ≥ N . Since can be made arbitrarily small, we must have
limn→∞ m(g(En )) = 0.
Now monotonicity and subadditivity of outer measure imply
m(g(E)) ≤ m(g(On ))
= m(g(E ∪ En ))
= m(g(E) ∪ g(En ))
≤ m(g(E)) + m(g(En ))
By part (i), we also know that
Z
g 0 = m(g(On ))
On
for n = 1, 2, · · · . Combining these expressions, we obtain
Z
g 0 ≤ m(g(E)) + m(g(En ))
m(g(E)) ≤
On
Taking limits of both sides and applying continuity of integration, we obtain m(g(E)) =
R
E
g0 .
(iii) Suppose E has measure zero. Then for any > 0, m∗ (g(E)) < by Problem 39. Therefore
R
g(E) has measure zero. We also know that E g 0 = 0 by Problem 9 of Chapter 4. Therefore
R
m(g(E)) = 0 = E g 0 , as desired.
(iv) By Theorem 11 of Chapter 2, there exists a Gδ set E containing A for which m(E ∼ A) = 0.
Corollary 18 of Chapter 4 and parts (ii) and (iii) then imply
Z
g0 =
A
Z
g0 +
Z
E
Z
=
g0
E∼A
g0
E
= m(g(E))
But monotonicity and subadditivity of measure imply
m(g(A)) ≤ m(g(E)) ≤ m(g(A)) + m(g(E ∼ A)) = m(g(A))
R
Therefore m(g(A)) = m(g(E)). Combining these results, we have A g 0 = m(g(A)).
Pn
(v) Since ϕ is simple, it can be written as ϕ = k=1 ck · χEk where c1 , · · · , ck are distinct values
and Ek = {y ∈ [c, d] : ϕ(y) = ck }. Since g : [a, b] → [c, d] is surjective by the Intermediate Value
Theorem, g g −1 (Ek ) = Ek . We can therefore apply linearity of integration and part (iv) to
120
conclude
b
Z
ϕ(g(x))g 0 (x)dx =
a
=
=
=
=
n
X
Z
b
χEk (g(x))g 0 (x)dx
ck
a
k=1
n
X
Z
g0
ck ·
g −1 (Ek )
k=1
n
X
ck · m g g −1 (Ek )
k=1
n
X
ck · m(Ek )
k=1
n
X
Z
d
ck ·
χEk
c
k=1
Z d
ϕ(y)dy
=
c
(vi) By the Simple Approximation Theorem, there exists an increasing sequence {ϕn } of simple functions on [c, d] that converges pointwise to f on [c, d]. The Monotone Convergence Theorem and
part (v) then imply
d
Z
d
Z
f (y)dy = lim
ϕn (y)dy
n→∞
c
c
b
Z
ϕn (g(x))g 0 (x)dx
= lim
n→∞
a
b
Z
f (g(x))g 0 (x)dx
=
a
(vii) Since g is injective, O = g −1 (g(O)). Therefore χg(O) (g(x)) = χO (x). Applying part (vi), we
conclude
Z
m(g(O)) =
d
Z
χg(O) (y)d(y) =
c
b
χg(O) (g(x))g 0 (x)dx =
a
Z
b
χO (x)g 0 (x)dx =
a
Z
g 0 (x)dx
O
57. Let a = 0 and b = 1. Let g denote the Cantor-Lebesgue function and let f (y) = 1 on [0, 1]. Then
g 0 = 0 a.e., so
Z
d
Z
f (y)dy 6=
1=
c
b
f (g(x))g 0 (x)dx = 0
a
58. Let F denote a generalized Cantor set of measure 1 − α for some α ∈ (0, 1). Let E = [0, 1] ∼ F and let
Z
f (x) =
x
χE
0
Then f is absolutely continuous by Theorem 11. Now pick any x, x0 ∈ [0, 1] satisfying x < x0 . By
Problem 39 of Chapter 2, E is an open set that is dense in [0, 1]. Therefore there exists y ∈ (x, x0 ) and
121
r > 0 such that (y − r/2, y + r/2) ⊆ E ∩ (x, x0 ). But then
f (x0 ) − f (x) =
Z
x0
χE = m(E ∩ [x, x0 ]) ≥ r > 0
x
by monotonicity of measure. Therefore f (x0 ) > f (x), so f is strictly increasing on [0, 1]. By Theorem
14, there exists a set E0 of measure zero such that f 0 = χE on [0, 1] ∼ E0 . But then f 0 = 0 on F ∼ E0 ,
a set of positive measure.
R g(x)
Rx
59. Define h(x) = g(a) f (s)ds − a f (g(t))g 0 (t)dt. The claim is that h(b) − h(a) = 0. However showing
that h0 = 0 a.e. [a, b] does not prove this claim (the Cantor-Lebesgue function is a counter-example).
Rx
60. By Theorems 10 and 11, f (x) = f (a) + a f 0 = f (a) for all x ∈ [a, b].
Suppose g0 + h0 = g1 + h1 on [a, b], where g0 and g1 are absolutely continuous and h0 and h1 are
singular. Furthermore, assume h0 (a) = h1 (a) = 0. Then g00 = g10 on [a, b] and g0 (a) = g1 (a), which
implies
x
Z
g00 = g1 (a) +
g0 (x) = g0 (a) +
a
Z
x
g10 = g1 (x)
a
for x ∈ [a, b]. Since h0 − h1 = g1 − g0 , the above expression also implies h0 = h1 on [a, b].
6.6
Convex Functions
61. Suppose ϕ is convex. The statement in the prompt holds trivially for n = 1. Now suppose the statement
is true for some index n. Let x1 , · · · , xn , xn+1 denote points in (a, b) and let λ1 , · · · , λn , λn+1 denote
Pn+1
Pn
Pn
nonnegative numbers that satisfy k=1 λk = 1. Define λ = k=1 λk and x = k=1 λk xk /λ. By the
induction hypothesis, we know that
ϕ(x) ≤
n
X
λk ϕ(xk )/λ
k=1
We therefore have
ϕ
n+1
X
!
λ k xk
= ϕ (λx + (1 − λ)xn+1 )
k=1
≤ λϕ(x) + (1 − λ)ϕ(xn+1 )
≤
n
X
λk ϕ(xk ) + λn+1 ϕ(xn+1 )
k=1
=
n+1
X
λk ϕ(xk )
k=1
where the second line follows from convexity of ϕ. Therefore the statement is true for n + 1, so we
conclude by induction that the statement holds for all natural numbers n.
The converse follows immediately because the definition of convexity is a special case of the statement
in the prompt.
122
Pn
Now suppose f is a simple function over [0, 1]. Let k=1 ck χEk denote the canonical representation of
Pn
f . Then m(Ek ) ≥ 0 for all k and k=1 m(Ek ) = 1, so
!
n
X
f (x)dx = ϕ
ck m(Ek )
1
Z
ϕ
0
k=1
n
X
≤
m(Ek )ϕ (ck )
k=1
Z 1
(ϕ ◦ f )(x)dx
=
0
where the second line applies the statement in the prompt and the first and third lines follow from the
definition of the integral of a simple function.
62. If ϕ is convex, then the prompt statement follows immediately from the definition of convexity.
Conversely, suppose the prompt statement is true. Fix x1 , x2 ∈ (a, b) and λ ∈ [0, 1]. Write the binary
representation of λ as
λ=
∞
X
ck 2−k
k=1
where ck ∈ {0, 1} for all k. Since 1 =
P∞
−k
k=1
2
, we also have
∞
X
1−λ=
(1 − ck )2−k
k=1
We can now derive the expression
∞
X
ϕ (λx0 + (1 − λ)x1 ) = ϕ
−k
ck 2
x0 +
k=1
=ϕ
∞
X
!
(1 − ck )2
−k
x1
k=1
c1 x1 + (1 − c1 )x2
+
2
P∞
−k
x0
k=1 ck+1 2
+
P∞
− ck+1 )2−k x1
k=1 (1
2
P∞
P∞
x0 + k=1 (1 − ck+1 )2−k x1
ϕ(c1 x1 + (1 − c1 )x2 ) ϕ
k=1 ck+1 2
+
2
2 P
P∞
∞
−k
−k
c
2
x
x1
c1 ϕ(x1 ) + (1 − c1 )ϕ(x2 ) ϕ
0+
k=1 k+1
k=1 (1 − ck+1 )2
=
+
2
2
−k
≤
where the inequality follows from the prompt statement and the last line follows because c1 ∈ {0, 1}.
If we continue recursively, we obtain the inequality
ϕ (λx0 + (1 − λ)x1 ) ≤
n
X
−k
ck 2
ϕ(x1 )+
k=1
n
X
(1−ck )2
−k
k=1
−n
ϕ(x2 )+2
ϕ
∞
X
k=1
−k
ck+n 2
x0 +
∞
X
!
−k
(1 − ck+n )2
k=1
for any index n. Since ϕ is continuous, it is uniformly bounded on the interval [x1 , x2 ]. Therefore
the last term can be made arbitrarily small by taking n large enough. We can therefore take limit to
obtain
ϕ (λx0 + (1 − λ)x1 ) ≤
∞
X
k=1
ck 2−k ϕ(x1 ) +
∞
X
k=1
123
(1 − ck )2−k ϕ(x2 ) = λϕ(x1 ) + (1 − λ)ϕ(x2 )
x1
√
63. Consider the function ϕ(x) = − 1 − x2 over the interval [0, 1]. Then
ϕ0 (x) = √
x
1 − x2
ϕ00 (x) =
1
(1 − x2 )3/2
on (0, 1). Since ϕ00 ≥ 0 on (0, 1), ϕ is convex on (0, 1) by Proposition 15. Since ϕ is continuous on
[0, 1], ϕ must also be convex on [0, 1]. However ϕ0 is unbounded on (0, 1) and therefore ϕ cannot be
Lipschitz on [0, 1].
64. If ϕ00 ≥ 0 on (a, b), then ϕ is convex on (a, b) by Proposition 15.
Conversely, suppose ϕ is convex. Fix x ∈ (a, b). By Lemma 16, we know that
ϕ0 (x + h) − ϕ0 (x)
≥0
h
for all h 6= 0 satisfying x + h ∈ (a, b). Taking the limit as h → 0, we conclude that ϕ00 (x) ≥ 0.
65. Taking derivatives, we obtain the expressions
ϕ0 (t) = bp(a + bt)p−1
ϕ00 (t) = b2 p(p − 1)(a + bt)p−2
for t ∈ (0, ∞). If p ≥ 1, ϕ00 ≥ 0 on (0, ∞). Therefore ϕ is convex on (0, ∞) by Proposition 15. Since ϕ
is continuous at 0, ϕ is convex on [0, ∞).
66. Claim: Suppose ϕ is a convex function on (−∞, ∞) that satisfies
1
Z
ϕ
Z 1
f (x)dx =
(ϕ ◦ f )(x)dx
0
0
for all integrable functions f over [0, 1]. Then there exist constants c0 and c1 such that ϕ(x) = c0 + c1 x.
Proof: Let c0 = ϕ(0) and c1 = ϕ(1) − ϕ(0). Then for x ∈ [0, 1], we have
Z
1
χ[0,x] (t)dt
ϕ(x) = ϕ
0
1
Z
(ϕ ◦ χ[0,x] )(t)dt
=
0
= xϕ(1) + (1 − x)ϕ(0)
= c0 + c1 x
For x ∈ (1, ∞), we have
Z
ϕ(1) = ϕ
1
x · χ[0,1/x] (t)dt
0
Z
1
ϕ(x · χ[0,1/x] (t))dt
1
1
ϕ(0)
= ϕ(x) + 1 −
x
x
=
0
⇒
ϕ(x) = c0 + c1 x
124
For x ∈ (−∞, 0), we have
1
Z
x · χ[0,1/(1−x)] (t) + ·χ(1/(1−x),1] (t) dt
ϕ(0) = ϕ
0
1
Z
ϕ x · χ[0,1/(1−x)] (t) + ·χ(1/(1−x),1] (t) dt
=
0
=
⇒
−x
1
ϕ(x) +
ϕ(1)
1−x
1−x
ϕ(x) = c0 + c1 x
67. Claim: Let ϕ be a convex function on (−∞, ∞) and let f be an integrable function over the closed,
bounded, non-degenerate interval [a, b] such that ϕ ◦ f is also integrable over [a, b]. Then
1
b−a
ϕ
Z
!
b
f (x)dx
a
1
≤
b−a
Proof: Define
α=
1
b−a
Z
Z
b
(ϕ ◦ f )(x)dx
a
b
f (x)dx
a
Choose m to lie between ϕ(α− ) and ϕ(α+ ). Since ϕ is convex, we know that
ϕ (f (x)) ≥ m (f (x) − α) + ϕ̃(α)
Thus by monotonicity and linearity of integration, we have
Z
b
(ϕ ◦ f )(x)dx ≥ m ·
a
!
b
Z
f (x)dx − (b − a) · α
+ ϕ(α) · (b − a)
a
Dividing both sides by b − a yields the desired result.
68. Since
d2
dx2
exp(x) = exp(x) ≥ 0 for all x ∈ R, exp(·) is convex by Proposition 15. The desired result
then follows immediately from Jensen’s Inequality.
Qk
69. Correction: limk→∞ n=1 ζnαn might not exist. For example, if αn = 2−n and ζn = exp ((−2)n ),
Qk
Qk
P∞
then n=1 ζnαn = exp((1 + (−1)k+1 )/2). We therefore prove lim supk→∞ n=1 ζnαn ≤ n=1 αn ζn .
Let α0 = 0 and define En = [αn−1 , αn−1 + αn ). Then m(En ) = αn for all n. For each k, define
Pk
fk = n=1 ζn · χEn . Then fk is a positive measurable function over [0, 1], so by Problem 70 we have
log
k
X
n=1
!
αn · ζn
Z
= log
1
Z
fk (x)dx ≥
0
1
log (fk (x)) dx =
0
k
X
αn · log(ζn ) = log
n=1
k
Y
!
ζnαn
n=1
Taking exponentials, we obtain the expression
k
X
αn · ζn ≥
n=1
k
Y
ζnαn
n=1
for all k. Since {αn · ζn } is a sequence of positive numbers, the series
125
Pk
n=1
αn · ζn converges. We
therefore have
∞
X
k
Y
αn · ζn ≥ lim sup
k→∞ n=1
n=1
ζnαn
70. The function − log(·) is convex on (0, ∞) by Proposition 15. Since g is positive on [0, 1], we can use
the Jensen’s Inequality proof in the text to show
1
Z
− log
1
Z
g(x)dx ≤ −
0
log(g(x))dx
0
The desired result then follows by multiplying both sides of the inequality by −1.
71. Suppose there exist constants c1 and c2 for which
|ϕ(t)| ≤ c1 + c2 |t|
for all t ∈ R. If f is integrable over [0, 1], then f is finite a.e. on [0, 1]. Therefore
|(ϕ ◦ f )(x)| ≤ c1 + c2 |f (x)|
for almost x ∈ [0, 1]. Monotonicity of measure then implies
Z
1
Z
|(ϕ ◦ f )(x)|dx ≤ c1 + c2
1
|f (x)|dx < ∞
0
0
Thus ϕ ◦ f is integrable whenever f is.
The following preliminary result will be used in the proof of the converse.
Claim: Let {bn } be an increasing sequence of real numbers and suppose limn→∞ bn = ∞. There exists
a non-negative sequence {an } of real numbers such that
∞
X
an < ∞ and
n=1
∞
X
an · bn = ∞
n=1
Proof: Assume without loss of generality that b1 > 0 (otherwise just set an = 0 until bn > 0). Define
kn = blog2 bn c
For any fixed index n, let mn denote the total number of indicies n0 satisfying kn0 = kn . Now define
1
mn 2kn
an =
Then an > 0 for all n, and
∞
X
n=1
an ≤
∞
X
1
<∞
2n
n=1
We also have
2kn ≤ bn
126
which implies
∞
X
an · bn ≥
n=1
∞
X
1=∞
n=1
as desired.
Now suppose ϕ ◦ f is integrable over [0, 1] whenever f is. Let g denote the function
g(x) =
ϕ(x)
x
We first show there must exist x0 ≥ 0 and c1 ≥ 0 such that
|g(x)| ≤ c1 if |x| ≥ x0
To prove this claim, suppose g is unbounded on every set of the form (∞, −x0 ] ∪ [x0 , ∞). Since ϕ is
continuous, we can define the sequence
yn = max g(x)
1≤|x|≤n
As g is unbounded on (∞, −1] ∪ [1, ∞), we must have yn ↑ ∞. We can therefore apply the preliminary
lemma to construct a non-negative sequence {an } such that
0<
∞
X
∞
X
an < ∞ and
n=1
an · yn = ∞
n=1
Let {xn } denote a sequence satisfying g(xn ) = yn and 1 ≤ |xn | ≤ n. Let a =
dn =
Define
E1 = [0, d1 ) ,
En =
∞
X
"n−1
X
dn ≤
n=1
n=1
an and define
an
|xn | · a
dk ,
k=1
(If dn = 0, set En = ∅.) Since
P∞
n
X
!
dk
n = 2, 3, ...
k=1
∞
X
an
=1
a
n=1
we must have En ⊆ [0, 1] for all n. Let f denote the function
f=
∞
X
|xn | · χEn
n=1
Then
1
Z
|f (x)|dx =
0
but
Z
1
|(ϕ ◦ f )(x)|dx =
0
∞
X
∞
X
|xn | · dn = 1
n=1
|ϕ(xn )| · dn =
n=1
∞
X
n=1
127
yn · |xn | · dn =
∞
1X
an · yn = ∞
a n=1
Thus ϕ ◦ f is not integrable over [0, 1] even though f is.
We conclude that there must exist x0 ≥ 0 and c1 ≥ 0 such that
|ϕ(x)| ≤ c1 · |x| if |x| ≥ x0
Let c0 = max|x|≤x0 |ϕ(x)|. Since c0 ≥ 0 and c1 · |x| ≥ 0, we must have
|ϕ(x)| ≤ c0 + c1 · |x|
for all x ∈ R, as desired.
7
The Lp spaces: Completeness and Approximation
7.1
Normed Linear Spaces
1. Let f and g belong to C[a, b]. Then by monotonicity and linearity of integration, we have
Z
||f + g||1 =
b
b
Z
|f + g| ≤
Z
b
(|f | + |g|) =
a
Z
|f | +
a
a
b
|g| = ||f ||1 + ||g||1
a
Therefore || · ||1 satisfies the triangle inequality. For any real number α, we have
b
Z
||αf ||1 =
Z
|αf | =
a
b
Z
|α| · |f | = |α|
a
b
|f | = |α| · ||f ||1
a
by linearity of integration. Therefore || · ||1 satisfies positive homogeneity. Since |f | ≥ 0, ||f ||1 ≥ 0 by
monotonicity of integration. Moreover, ||f ||1 = 0 if and only if f = 0 by Proposition 9 of Chapter 4.
Thus || · ||1 satisfies nonnegativity, so || · ||1 is a norm on C[a, b].
Now fix c ≥ 0. Define xn =
1
nb
+
n−1
n a
and consider the following sequence of functions {fn }:

x−a

 2n
1−
xn − a
fn (x) = b − a

0
for x ∈ [a, xn ]
for x ∈ (xn , b]
Then fn ∈ C[a, b] and
||fn ||max =
2n
,
b−a
for all n. Therefore ||fn ||max > c||fn ||1 for all n >
||fn ||1 = 1
c(b−a)
.
2
Thus there is no c ≥ 0 for which ||f ||max ≤
c||f ||1 for all f ∈ C[a, b].
Now let c = b − a. Since |f | ≤ ||f ||max , we must have
Z
||f ||1 =
b
Z
|f | ≤
a
b
||f ||max = (b − a) · ||f ||max = c||f ||max
a
for all f ∈ C[a, b].
128
Pn
2. Let f (x) =
i=0
Pm
ai xk and g(x) =
i
i=0 bi x ,
where {ai }ni=0 and {bi }m
i=0 are sets of real numbers and
n and m are non-negative integers. Assume without loss of generality that n ≥ m and define bi = 0
for i = m + 1, m + 2, · · · , n. Then for any scalars α and β, we have
αf (x) + βg(x) =
n
X
(αai + βbi )xi
i=0
Thus αf + βg is in X, so X is a linear space. The above expression also implies
||f + g|| =
n
X
|ai + bi | ≤
i=0
n
X
|ai | +
i=0
m
X
|bi | = ||f || + ||g||
i=1
so that the triangle inequality holds. We also have
||αf || =
n
X
|αai | = |α|
i=0
n
X
|ai | = |α| · ||f ||
i=0
so that || · || satisfies positive homogeneity. Non-negativity of || · || follows directly from non-negativity
of the absolute value function. Therefore || · || is a norm on X.
3. Let f and g denote functions in L1 [a, b]. We then have
Z
||f +g|| =
b
b
Z
x2 |f (x)+g(x)|dx ≤
a
x2 (|f (x)|+|g(x)|)dx =
Z
a
b
x2 |f (x)|dx+
a
Z
b
x2 |g(x)|dx = ||f ||+||g||
a
so the triangle inequality holds. For any real number α, we have
Z
||αf || =
b
x2 |αf (x)|dx = |α|
a
Z
b
x2 |f (x)|dx = |α| · ||f ||
a
Therefore positive homogeneity also holds. Since x2 |f (x)| ≥ 0 on [a, b], ||f || ≥ 0. Moreover, ||f || = 0
if and only if x2 |f | = 0 a.e. [a, b] by Proposition 9 of Chapter 4. But since x2 6= 0 a.e. [a, b], x2 |f | = 0
if and only if f = 0 a.e. [a, b]. Thus || · || satisfies non-negativity.
4. We first show that the set
U = M : m{x ∈ [a, b] |f (x)| > M } = 0
is equal to the set of essential upper bounds of f . Suppose M is an essential upper bound of f . Then
there exists a set E0 of measure zero such that |f | ≤ M on [a, b] ∼ E0 . But then
{x ∈ [a, b] |f (x)| > M } ⊆ E0
which implies
m
x ∈ [a, b] |f (x)| > M
=0
by monotonicity and non-negativity of measure. That each element of M is an essential upper bound
of f follows immediately from the definition of an essential upper bound.
We therefore conclude that ||f ||∞ = inf U . To show that ||f ||∞ ∈ U , let Mn denote a sequence in U
129
converging downward to ||f ||∞ . Then
∞
\
x ∈ [a, b] |f (x)| ≤ Mn = x ∈ [a, b] |f (x)| ≤ ||f ||∞
n=1
Since
b−a=m
x ∈ [a, b] |f (x)| ≤ Mn
=m
x ∈ [a, b] |f (x)| ≤ Mn
+m
x ∈ [a, b] |f (x)| > Mn
for all n, continuity of measure implies
b−a=m
x ∈ [a, b] |f (x)| ≤ ||f ||∞
The excision property of measure then implies
m
x ∈ [a, b] |f (x)| > ||f ||∞
= m([a, b]) − m
x ∈ [a, b] |f (x)| ≤ ||f ||∞
= (b − a) − (b − a)
=0
Therefore ||f ||∞ ∈ U , so ||f ||∞ = min U .
Now suppose f is continuous. Since f ≤ ||f ||max on [a, b], ||f ||max is an essential upper bound of f .
Therefore ||f ||max ≥ ||f ||∞ . To prove the converse, let M ≥ 0 denote an essential upper bound of f .
Then there exists a set E of measure zero such that f ≤ M on [a, b] ∼ E. Now suppose there exists
x ∈ [a, b] such that f (x) > M . Because f is continuous at x, there must exist a non-degenerate interval
I ⊆ [a, b] that contains x and that satisfies f (x0 ) > M for all x0 ∈ I. Then I ⊆ E, so 0 < m(I) ≤ m(E).
However m(E) = 0 by assumption. Therefore f (x) ≤ M for all x, so ||f ||max ≤ M . We conclude that
||f ||max is a lower bound of the set of essential upper bounds of f , so ||f ||max ≤ ||f ||∞ .
5. Let {ak } and {bk } denote two sequences in `∞ . Then |ak + bk | ≤ |ak | + |bk | for all k, so
||{ak } + {bk }||∞ = sup |ak + bk |
1≤k<∞
≤ sup (|ak | + |bk |)
1≤k<∞
≤ sup |ak | + sup |bk |
1≤k<∞
1≤k<∞
= ||{ak }||∞ + ||{bk }||∞
For any scalar α, we have
||α · {ak }||∞ = sup |α · ak | = sup |α| · |ak | = |α| sup |ak | = |α| · ||{ak }||∞
1≤k<∞
1≤k<∞
1≤k<∞
Since |ak | ≥ 0 for all k, we also must have
||{ak }||∞ = sup |ak | ≥ 0
1≤k<∞
130
Since ||{ak }||∞ ≥ |ak | for any k, |ak | = 0 for all k if and only if ||{ak }||∞ = 0.
Now let {ak } and {bk } denote two sequences in `1 . We then have
||{ak } + {bk }||1 =
≤
=
∞
X
|ak + bk |
k=1
∞
X
(|ak | + |bk |)
k=1
∞
X
|ak | +
k=1
∞
X
|bk |
k=1
= ||{ak }||1 + ||{bk }||1
For any scalar α, we have
||α · {ak }||1 =
∞
X
|α · ak | =
k=1
∞
X
|α| · |ak | = |α|
k=1
∞
X
|ak | = |α| · ||{ak }||1
k=1
Since |ak | ≥ 0 for all k, we also must have
||{ak }||1 =
∞
X
|ak | ≥ 0
k=1
Since ||{ak }||1 ≥ |ak | for any k, |ak | = 0 for all k if and only if ||{ak }||1 = 0.
7.2
The Inequalities of Young, Hölder, and Minkowski
6. Let E be a measurable set, 1 ≤ p < ∞, and q the conjugate of p. Choose f ∈ Lp (E) and g ∈ Lq (E)
and assume f 6= 0 and g 6= 0. By linearity of integration, we have
Z
E
Z
E
f
||f ||p
g
||g||q
p 1/p
q 1/q
1
=
||f ||p
Z
1
=
||g||q
Z
p
1/p
|f |
=1
E
q
|g|
1/q
=1
E
If Hölder’s Inequality is true for normalized functions, we have
Z
E
f
g
·
≤1
||f ||p ||g||q
R
|f · g| ≤ ||f ||p · ||g||q by linearity of integration.
R1
7. Example 1: By Problem 19 of Chapter 4, 0 xα·p < ∞ if and only if α · p > −1. Fix − p11 < α ≤ − p12 .
R1
R1
p
p
Then 0 |xα | 1 < ∞ but 0 |xα | 2 = ∞, so xα ∈ Lp1 (E) ∼ Lp2 (E).
which implies
E
Example 2: (The example given in the text is incorrect. Instead define f (x) =
131
x−1/2
1+| ln x|
and E =
(0, ∞)). Suppose p = 2. For any M ≥ 1, we have
M
Z
M
Z
p
|f | =
1
1
Z
1
Z
p
x−1
dx =
(1 + ln x)2
1
|f | =
1/M
1/M
Z
M
1
x−1
dx =
(1 − ln x)2
Z
d
dx
1
1/M
1
−
1 + ln x
d
dx
1
1 − ln x
dx = 1 −
1
1 + ln M
dx = 1 −
1
1 + ln M
Applying the Monotone Convergence Theorem, we obtain
Z
∞
Z
p
M
|f | = lim
0
M →∞
Z
p
1
|f | = lim
M →∞
1/M
p
Z
|f | +
1/M
!
M
p
|f |
=2<∞
1
Now suppose p < 2. Applying L’Hôpital’s Rule, it is possible to show that
x1−p/2
=∞
x→∞ (1 + ln x)p
lim
Therefore there exists an M ≥ 1 such that
x1−p/2
≥1
(1 + ln x)p
for all x ≥ M . But then for all x ≥ M , we have
|f (x)|p =
x−p/2
≥ x−1
(1 + ln x)p
R∞
R∞
Since M x−1 dx = ∞, we must have M |f |p = ∞ by monotonicity of integration. But this implies
R∞ p
|f | = ∞.
0
Now suppose p > 2. Again applying L’Hôpital’s Rule, it is possible to show that
x1−p/2
=∞
x→0 (1 − ln x)p
lim
Therefore there exists an M ≥ 1 such that
x1−p/2
≥1
(1 − ln x)p
for all x ∈ (0, 1/M ). But then for all x ∈ (0, 1/M ), we have
|f (x)|p =
R 1/M
x−1 dx = ∞, we must have
R∞
again implies 0 |f |p = ∞.
Since
0
R 1/M
0
x−p/2
≥ x−1
(1 − ln x)p
|f |p dx = ∞ by monotonicity of integration. But this
132
8. For any λ ∈ R, we have
Z
0≤
(λ|f | + |g|)2 =
Z
E
Z
(λ2 f 2 + 2λ · |f | · |g| + g 2 ) = λ2
E
f 2 + 2λ
E
Z
Z
|f · g| +
E
g2
(1)
E
by monotonicity and linearity of integration. Since L2 (E) is a linear space, we also know that
R
E
(f +
2
g) < ∞. Applying the above expression for λ = −1, we obtain
Z
|f · g| ≤
E
Z
1
2
f2 +
E
Z
g2
<∞
E
R
R
R
Thus λ2 E f 2 + 2λ E |f · g| + E g 2 is a well-defined quadratic equation whose minimum occurs at
R
R
λ∗ = − E |f · g|/ E f 2 . Plugging λ∗ into (1), we can infer
2 Z
|f · g|
+
g2
2
f
E
E
R
0≤−
ER
Rearranging this expression, we obtain
2
Z
|f · g|
E
Z
≤
E
f2
Z
g2
E
Taking square roots of both sides yields the Cauchy-Schwartz Inequality.
9. (Correction: Show that in Young’s Inequality there is equality if and only if ap = bq .)
Suppose ap = bq . Then
ap
bq
+
= ap = a · ap/q = ab
p
q
Now suppose ap > bq . Then
ap
bq
+
> ap = a · ap/q > ab
p
q
If ap < bq , then
Therefore ap 6= bq implies
ap
bq
+
> bq = b · bq/p > ab
p
q
ap
p
+
bq
q
6= ab.
10. Suppose there exists α and β, both not equal to zero, such that α|f |p = β|g|q . Without loss of
133
generality, assume β 6= 0. Then
1/q
α|f |p
β
E
1/q
Z
α
=
|f |p
β
E
1/q
Z
1/p Z
α
|f |p
=
|f |p
β E
E
Z
1/p Z
1/q
=
|f |p
|g|q
Z
Z
|f · g| =
E
|f | ·
E
E
= ||f ||p · ||g||q
If f or g is 0 the converse holds by setting β or α equal to 0, respectively. So suppose f and g are both
nonzero and assume
Z
|f · g| = ||f ||p · ||g||q
E
We can rearrange this expression to obtain
Z E
|f |
||f ||p
p
1
· +
p
|g|
||g||q
q
|f |
|g|
1
·
· −
q
||f ||p ||g||q
=0
By Young’s Inequality, the integrand is non-negative. Therefore by Proposition 9 of Chapter 4,
|f |
||f ||p
p
1
· +
p
a.e. on E. Problem 9 then implies
|f |
||f ||p
|g|
||g||q
q
p
·
=
|f |
1
|g|
=
·
q
||f ||p ||g||q
|g|
||g||q
q
a.e. on E. We conclude that α|f |p = β|g|q a.e. on E, where α = 1/||f ||pp and β = 1/||g||qq .
11. For any x ∈ Rn , we have
||x||p =

(Pn
1/p
k=1
|xk |p )
max |x |
k
k
if 1 ≤ p < ∞
if p = ∞
It is obvious that ||x||p ≥ 0 and that ||x||p = 0 if x = 0. ||x||p = 0 implies x = 0 because ||x||p ≥ |xk |
for k = 1, · · · , n. Therefore || · ||p satisfies non-negativity. Since T is linear, for any α ∈ R we have
||α · x||p = ||Tα·x ||p = ||α · Tx ||p = |α| · ||Tx ||p = |α| · ||x||p
Therefore || · ||p satisfies positive homogeneity. Now fix x, y ∈ Rn . Then
||x + y||p = ||Tx+y ||p = ||Tx + Ty ||p ≤ ||Tx ||p + ||Ty ||p = ||x||p + ||y||p
Therefore || · ||p satisfies the Triangle (Minkowski) Inequality. We conclude || · ||p is a norm on Rn .
134
Hölder’s Inequality in Rn is given by
n
X
for all x, y ∈ Rn
|xk · yk | ≤ ||x||p · ||y||q
k=1
This inequality follows from applying Hölder’s Inequality to Tx and Ty :
n
X
n+1
Z
|xk · yk | =
|Tx · Ty | ≤ ||Tx ||p · ||Ty ||q = ||x||p · ||y||q
1
k=1
12. If 1 ≤ p < ∞, we have
1/p
∞
Z
||Ta ||p =
ak χ[k,k+1)
=
1
∞
X
!1/p
p
|ak |
= ||a||p
k=1
If p = ∞, then
||Ta ||p = sup |ak | = ||a||p
1≤k<∞
Therefore if a ∈ `p , then Ta ∈ Lp [1, ∞). For any two sequences a, b ∈ `p , we have
||a + b||p = ||Ta+b ||p = ||Ta + Tb ||p ≤ ||Ta ||p + ||Tb ||p = ||a||p + ||b||p
Thus the Minkowski Inequality holds. Hölder’s Inequality in `p can be written as
∞
X
for all a ∈ `p , b ∈ `q
|ak · bk | ≤ ||a||p · ||b||q
k=1
This inequality follows from applying Hölder’s Inequality to Ta and Tb :
∞
X
∞
Z
|ak · bk | =
|Ta · Tb | ≤ ||Ta ||p · ||Tb ||q = ||a||p · ||b||q
1
k=1
Moreover, if a 6= 0, the sequence a∗ defined by
a∗k = ||a||1−p
· sgn(ak ) · |ak |p−1
p
satisfies
∞
X
ak · a∗k = ||a||1−p
=
p
k=1
∞
X
|ak |p = ||a||p
k=1
and
||a∗ ||q = ||a||1−p
·
p
∞
X
!1/q
|ak |q(p−1)
= ||a||1−p
· ||a||p/q
=1
p
p
k=1
13. Fix f ∈ Lp1 (E) and assume there exists M ≥ 0 such that |f | ≤ M on E. Define
E0 = {x ∈ E : |f (x)| ≤ 1}
135
Since |f |p1 > 1 on E ∼ E0 , we must have
Z
|f |p1 ≤
m(E ∼ E0 ) ≤
Z
E∼E0
|f |p1 < ∞
E
Since |f |p2 ≤ |f |p1 on E0 , we also have
Z
Z
|f |p2 ≤
Z
|f |
p2
Z
14. Since | ln(1/x)| =
|f |
=
E
p2
Z
|f |
+
E∼E0
d
dx (x ln(1/x)
Z
|f |p1 < ∞
E
E0
E0
Therefore
|f |p1 ≤
p2
Z
≤ M · m(E ∼ E0 ) +
E0
|f |p2 < ∞
E0
+ x) on (0, 1], we have
Z
1
| ln(1/x)|dx = 1 − (c ln(1/c) + c)
c
for any c ∈ (0, 1]. By the Monotone Convergence Theorem,
R1
| ln(1/x)|dx = limc→0
0
R1
c
| ln(1/x)|dx =
1
1. Therefore f ∈ L (0, 1].
Now fix any p ∈ (1, ∞). Using L’Hôpital’s Rule, it is straight-forward to show
| ln(1/x)|p
=0
x→0
1/x
lim
Therefore there exists c ∈ (0, 1] such that
| ln(1/x)|p ≤
1
x
for all x ∈ (0, c]. But this implies
Z
1
| ln(1/x)|p dx =
0
Z
c
| ln(1/x)|p dx +
0
Z
≤
0
≤
Z
1
| ln(1/x)|p dx
c
c
1
x1/p
Z
dx +
1
| ln(1/x)|p dx
c
p
p
+ (1 − c) · |ln(1/c)|
p−1
<∞
Therefore f ∈ Lp (0, 1].
For any M ≥ 0, ln(1/x) > M for x ∈ (0, e−M ). Therefore f is not essentially bounded on (0, 1], so
f∈
/ L∞ (0, 1].
15. Let E denote a set of real numbers. Suppose f ∈ Lp (E), g ∈q (E), and h ∈ Lr (E), where r, p, q ≥ 1
and
1 1 1
+ + =1
p q
r
136
Then
Z
|f · g · h| ≤ ||f ||p · ||g||q · ||h||r
E
Proof: Let t =
p·q
p+q
denote the conjugate of r. Then
Z
|f |t·(p+q)/q =
ZE
Z
|f |p < ∞
ZE
t·(p+q)/p
|g|
|g|q < ∞
=
E
E
Therefore |f |t ∈ L(p+q)/q and |g|t ∈ L(p+q)/p . Since (p + q)/q and (p + q)/p are conjugates, we have
Z
|f · g|t ≤
Z
E
|f |p
q/(p+q) Z
E
|g|q
p/(p+q)
= ||f ||tp · ||g||tq < ∞
E
by Hölder’s Inequality. But this implies |f · g| ∈ Lt (E) and ||f · g||t ≤ ||f ||p · ||g||q . Since t and r are
conjugates, we have
Z
Z
|f · g · h| =
E
|f · g| · |h| ≤ ||f · g||t · ||h||r ≤ ||f ||p · ||g||q · ||h||r
E
16. Let f1 (x) = 1 and let

n


2
fn (x) =
n



2(n − 1)
if 0 ≤ x <
if
1
n
1
≤x≤1
n
for n = 2, 3, · · · . Then
Z
1
|fn | = 1
0
for all n, so {fn } is bounded in L1 [0, 1].
Now fix ∈ (0, 1/2). For any δ > 0, we can find n such that
1/n
Z
1
n
< δ. But then m([0, 1/n)) < δ and
1
>
2
fn =
0
Therefore {fn } is not uniformly integrable.
17. Fix p ≥ 1 and let
fn = χ[n,n+1)
|fn |p = 1 for all n, so {fn } is bounded in Lp (R). Fix ∈ (0, 1) and let E0 denote a set of
T∞
finite measure. Let En = E0 ∼ (−n, n). Since m(E0 ) < ∞ and n=1 En = ∅, limn→∞ m(En ) = 0 by
Then
R
R
continuity of measure. Pick n such that m(En ) < 1 − . Then
Z
Z
|fn | =
E0
Z
|fn | +
E0 ∩(−n,n)
|fn | ≤ m(En ) < 1 − En
137
Therefore
Z
Z
Z
|fn | =
|fn | −
R∼E0
R
|fn | > 1 − (1 − ) = E0
We conclude that {fn } is not tight over R.
18. If m(E) = 0, ||f ||p = 0 for all p ≥ 1 and the result holds trivially. So assume m(E) > 0. By Corollary
3, we know that
||f ||p ≤ [m(E)]1/p · ||f ||∞
for all p ∈ (1, ∞). Since limp→∞ [m(E)]1/p = 1, we have
lim sup ||f ||p ≤ ||f ||∞
(2)
p→∞
Now define
En =
1
x ∈ E : |f (x)| > ||f ||∞ −
n
If m(En ) = 0, then ||f ||∞ − n1 would be an essential upper bound for f that is less than ||f ||∞ . But that
would contradict the definition of ||f ||∞ . Therefore m(En ) > 0 for all n. From Hölder’s Inequality, we
know that
Z
Z
|f | =
En
|f · χEn | ≤ ||f ||p · (m(En ))(p−1)/p
E
for all p ∈ (1, ∞). Since ||f ||∞ −
1
n
< |f | on En , we must have
Z
1
||f ||∞ −
· m(En ) ≤
|f |
n
En
Combining expressions and simplifying, we can conclude
||f ||∞ −
1
≤ ||f ||p · m(En )−1/p
n
Since m(En ) > 0, limp→∞ m(En )−1/p = 1. Therefore
||f ||∞ −
1
≤ lim inf ||f ||p
p→∞
n
for all n. Letting n → ∞, we obtain
||f ||∞ ≤ lim inf ||f ||p
p→∞
Combining (2) and (3), we conclude
lim ||f ||p = ||f ||∞
p→∞
19. If f = 0, then
R
E
f · g = 0 for all g and ||f ||p = 0 for 1 ≤ p < ∞. Thus the result holds.
Now assume f 6= 0. For any g ∈ Lq (E) with ||g||q ≤ 1, we have
Z
Z
f ·g ≤
E
|f · g| ≤ ||f ||p · ||g||q = ||f ||p
E
138
(3)
by Hölder’s Inequality. Therefore
Z
f · g ≤ ||f ||p
sup
g∈Lq (E), ||g||q ≤1
We also know that
Z
E
f · f ∗ = ||f ||p
E
where f ∗ denotes the conjugate function of f . Since ||f ∗ ||q = 1, we must have
Z
f · g ≥ ||f ||p
sup
g∈Lq (E), ||g||q ≤1
We can conclude that
E
Z
f · g = ||f ||p
sup
g∈Lq (E), ||g||q ≤1
E
Since the supremum is obtained at g = f ∗ , we may write
Z
f · g = ||f ||p
max
g∈Lq (E), ||g||q ≤1
E
R
20. Suppose f = 0. If g ∈ Lq (E), then g is finite a.e. E. Therefore f · g = 0 a.e. E, so E f · g = 0.
R
Conversely, suppose E f · g = 0 for all g ∈ Lq (E). By Problem 20, ||f ||p = 0. By non-negativity of
|| · ||p , f = 0.
21. Claim: The expression
R1
lim
→0+
0
f
for all f ∈ Lp [0, 1]
=0
λ
holds if and only if λ ≤ 1q , where q denotes the conjugate of p. (Define
(4)
1
q
= 0 if p = 1.)
Proof: Let q denote the conjugate of p and suppose λ < 1q . Then
R
f
0
≤ ||f ||p · 1/q−λ
λ
by Hölder’s Inequality. Since lim→0+ 1/q−λ = 0, (4) holds. Now suppose λ = 1q . Then
R
f
0
λ
Z
≤
|f |p
1/p
0
by Hölder’s Inequality. By continuity of integration, lim→0+
R
Now suppose λ > 1/q. Let f (x) = x−1+λ . Then
|f (x)|p = x−1+p·(λ−1/q)
139
0
|f |p = 0. Therefore (4) again holds.
Since p · (λ − 1/q) > 0, f ∈ Lp [0, 1]. However
R
f
0
λ
lim+
→0
=
1
6= 0
λ
Therefore (4) does not hold.
22. Let {x0 , · · · , xn } denote an arbitrary partition of [a, b], Since F is indefinite integral of an Lp [a, b]
function, there exists f ∈ Lp [a, b] such that
Z
xk
F (xk ) − F (xk−1 ) =
k = 1, · · · , n
f
xk−1
By Hölder’s Inequality,
Z
xk
|f | ≤ ||f ||p · |xk − xk−1 |1/q
k = 1, · · · , n
xk−1
Combining expressions and raising both sides to the p, we obtain
|F (xk ) − F (xk−1 )|p ≤ |xk − xk−1 |p−1
xk
Z
|f |p
k = 1, · · · , n
xk−1
Rearranging and summing over k, we can infer
n
X
|F (xk ) − F (xk−1 )|p
k=1
|xk − xk−1 |p−1
Therefore the result holds by setting M =
7.3
Rb
a
≤
n Z
X
k=1
xk
p
Z
|f | =
xk−1
|f |p .
Lp is Complete: The Riesz-Fischer Theorem
23. Define
an =


0



if n = 1
n−1
X 1




i2
i=1
Then
lim an =
n→∞
if n = 2, 3, · · ·
∞
X
1
<∞
i2
i=1
By Theorem 17 of Chapter 1, {an } is Cauchy.
We also have
|an+1 − an | =
1
for all n
n2
Thus if {n } is a sequence of positive numbers that satisfies
|an+1 − an | ≤ 2n for all n
140
a
b
|f |p < ∞
then
1
for all n
n
n ≥
But this implies
∞
X
n ≥
n=1
∞
X
1
=∞
n
n=1
Therefore {an } is not rapidly Cauchy.
24. That each αfn + βgn and αf + βg belong to X follows because X is a linear space. Using properties
of a norm, we can also infer
0 ≤ ||αfn + βgn − (αf + βg)|| ≤ ||α(fn − f )|| + ||β(gn − g)||
= |α| · ||fn − f || + |β| · ||gn − g||
But since limn→∞ ||fn − f || = limn→∞ ||gn − g|| = 0 by assumption, the above expression implies
lim ||αfn + βgn − (αf + βg)|| = 0
n→∞
Therefore {αfn + βgn } → αf + βg in X.
25. By Corollary 3, each fn and f are in Lp1 (E) and
0 ≤ ||fn − f ||p1 ≤ c · ||fn − f ||p2
for some c ∈ [0, ∞). Since limn→∞ ||fn − f ||p2 = 0 by assumption, the above expression implies
||fn − f ||p1 = 0. Therefore {fn } → f in Lp1 (E).
26. Since |fn | ≤ g for all n a.e. on E, we must have |f | ≤ g a.e. on E. We can infer that |fn |p ≤ g p and
|f |p ≤ g p a.e. on E, which implies
Z
E
|fn |p ≤
Z
g p < ∞ and
Z
E
|f |p ≤
Z
E
gp < ∞
E
Thus fn and f are both in Lp (E). By the Lebesgue Dominated Convergence Theorem, we have
Z
|fn |p =
lim
n→∞
E
Z
|f |p
E
which implies that {fn } → f in Lp (E) by Theorem 7.
27. By the Riesz-Fischer Theorem, {fn } contains a subsequence that converges pointwise to f a.e. on E.
By Propositions 4 and 5, this subsequence itself contains a rapidly Cauchy subsequence. To save on
notation, assume that {fn } itself satisfies these properties. Define
g = sup |fn |
1≤n<∞
If we show g ∈ Lp (E), the proof is complete since g ≥ |fn | for all n. Since
g ≤ sup |fn − f | + |f |
1≤n<∞
141
we have
p
p
p
sup |fn − f |
|g| ≤ 2
+ |f |
p
1≤n<∞
Since f ∈ Lp (E), it suffices to show sup1≤n<∞ |fn − f | ∈ Lp (E).
To do so, we first prove the statement
max |fk − f | ≤
1≤k≤n
n
X
|fk+1 − fk | + |fn+1 − f |
(1)
k=1
for all n. (1) follows immediately from the Triangle Inequality if n = 1. Now suppose (1) is true for
some n. Then
max
1≤k≤n+1
|fk − f | = max
max |fk − f |, |fn+1 − f |
1≤k≤n
(
≤ max
n
X
)
|fk+1 − fk | + |fn+1 − f |, |fn+1 − f |
k=1
=
n
X
|fk+1 − fk | + |fn+1 − f |
k=1
=
n+1
X
|fk+1 − fk | + |fn+1 − f | − |fn+2 − fn+1 |
k=1
≤
n+1
X
|fk+1 − fk | + |fn+2 − f |
k=1
Therefore (1) holds for all n by induction.
Now (1) implies
sup |fn − f | ≤
1≤n<∞
∞
X
|fn+1 − fn |
(2)
n=1
P∞
So it suffices to show that n=1 |fn+1 − fn | ∈ Lp (E). Because {fn } is rapidly Cauchy, there exists a
P∞
convergent series n=1 n such that
||fn+1 − fn ||p ≤ 2n
for all n. By Minkowski’s Inequality, we know that
n
X
p
|fk+1 − fk |
k=1
≤
n
X
p
|fk+1 − fk | ≤
k=1
n
X
!p
2k
(3)
k=1
for all n. By the Monotone Convergence Theorem, we can take limits of (3) to conclude
Z
E
∞
X
n=1
!p
|fn+1 − fn |
≤
∞
X
!p
||fn+1 − fn ||p
n=1
≤
∞
X
!p
2n
<∞
n=1
as desired.
28. By assumption, there exists a real number M such that ||fn ||p+θ ≤ M for all n. By Fatou’s Lemma,
142
we have
Z
p+θ
|f |
Z
|fn |p+θ ≤ M p+θ < ∞
≤ lim inf
n→∞
E
E
Therefore f ∈ Lp+θ (E). By Corollary 3, f and {fn } belong to Lp (E). It remains to show that
||fn − f ||p → 0. By Hölder’s Inequality, we have
Z
θ
A
|fn − f |p ≤ ||fn − f ||pp+θ · m(A) p+θ
for any measurable set A ⊆ E. By Minkowski’s Inequality,
||fn − f ||p+θ ≤ ||fn ||p+θ + ||f ||p+θ ≤ 2 · M
Therefore
Z
θ
|fn − f |p ≤ 2p · M p · m(A) p+θ
A
Thus for any > 0, we have
Z
|fn − f |p < A
as long as m(A) <
2p · M p
p+θ
θ
. Therefore {|fn − f |p } is uniformly integrable over E, which implies
Z
lim
n→∞
|fn − f |p = 0
E
by the Vitali Convergence Theorem.
29. Consider the sequence of polynomials
pn (x) =
n
X
xn
k=0
n!
on [0, 1]. By Taylor’s Theorem, we know that
pn (x) = ex − eu(x)
xn+1
(n + 1)!
where u(x) ∈ [0, 1]. Therefore
||pn − pm ||max ≤
e
e
+
(n + 1)! (m + 1)!
for all n and m. Thus {pn } is Cauchy in || · ||max . However, {pn } converges to ex in || · ||max , which is
not a polynomial. Therefore the space of polynomials on [0, log 2] is not complete.
30. By construction,
|fn+k (x) − fn (x)| ≤ ||fn+k − fn ||max
(4)
for all x ∈ [a, b]. Since || · ||max is a norm, we also have
||fn+k − fn ||max ≤
n+k−1
X
||fj+1 − fj ||max ≤
j=n
n+k−1
X
j=n
143
aj ≤
∞
X
j=n
aj
(5)
by the Triangle Inequality.
P∞
Now fix > 0. Since j=1 aj is a convergent series of positive numbers, there exists an index N such
P∞
that j=n aj < for all n ≥ N . Now fix n, m ≥ N and assume without loss that m > n. Substituting
k = m − n into (4) and (5), we have
|fm (x) − fn (x)| ≤ ||fm − fn ||max ≤
∞
X
aj < (6)
j=n
Thus {fn (x)} is a Cauchy sequence of real numbers. Since the reals are complete, {fn (x)} converges
to a real number. We can therefore define a real-valued function f (·) on [a, b] by
lim fn (x) = f (x)
(7)
n→∞
To see that {fn } → f uniformly on [a, b], observe that
|fn (x) − f (x)| ≤ |fn (x) − fm (x)| + |fm (x) − f (x)| ≤ ||fm − fn ||max + |fm (x) − f (x)|
for all x ∈ [a, b] and all indicies n and m. By (6), there exists an index N such that
|fn (x) − f (x)| < + |fm (x) − f (x)|
for m, n ≥ N . Letting m → ∞, we have
|fn (x) − f (x)| ≤ for all n ≥ N . Therefore {fn } → f uniformly on [a, b].
It remains to show that f is continuous. For any x, x0 ∈ [a, b], we have
|f (x) − f (x0 )| ≤ |fn (x) − f (x)| + |fn (x) − fn (x0 )| + |fn (x0 ) − f (x0 )|
Since {fn } → f uniformly on [a, b], there exists an index N such that |fn − f | < /3 on [a, b]. Therefore
if we fix n ≥ N , we have
|f (x) − f (x0 )| <
2·
+ |fn (x) − fn (x0 )|
3
Since fn is continuous, there exists δ > 0 such that |fn (x) − fn (x0 )| < /3 if |x − x0 | < δ. But then
|x − x0 | < δ implies |f (x) − f (x0 )| < Therefore f ∈ C[a, b].
31. Suppose {fn } is Cauchy in C[a, b]. Choose a sequence 1 ≤ n1 < n2 < · · · such that ||fm −fn ||max < 1/2k
for all m, n ≥ nk . Then
||fnk+1 − fnk ||max <
for all k. Since
P∞
k=1
1
2k
2−k is a convergent sequence of positive numbers, by Problem 30 there exists a
function f ∈ C[a, b] such that {fnk } → f uniformly on [a, b]. But this implies ||fnk − f ||max → 0. By
144
Proposition 4, {fn } → f in C[a, b]. Thus C[a, b] is complete.
32. For any n and k, there exists a set En,k ⊆ E with m(En,k ) = 0 such that |fn+k − fn | ≤ ||fn+k − fn ||∞
S∞ S∞
on E ∼ En,k . Let E0 = n=1 k=1 En,k . Then m(E0 ) = 0 and
|fn+k (x) − fn (x)| ≤ ||fn+k − fn ||∞ ≤
n+k−1
X
aj ≤
j=n
∞
X
aj
j=n
for all k, n and all x ∈ E ∼ E0 . Now following the argument in Problem 30, we can show there exists
a function f on E such that {fn } → f uniformly on E ∼ E0 . It remains to show that f ∈ L∞ (E).
Observe that
|f (x)| ≤ |fn (x) − f (x)| + |fn (x)| ≤ ||fn − f ||∞ + |fn (x)|
on E ∼ E0 for all n. Since ||fn − f ||∞ → 0, there must exist an index N such that ||fN − f ||∞ < ∞.
Since fN ∈ L∞ (E), there must exist a set EN with m(EN ) = 0 such that |fN | < ∞ on E ∼ EN . Since
|f | is bounded on E ∼ (E0 ∪ EN ) and m(E0 ∪ EN ) = 0, we conclude that f ∈ L∞ (E).
33. Suppose {fn } is Cauchy in L∞ (E). Using the same argument as in Problem 31, we can show there
exists f ∈ L∞ (E), a subsequence {fnk }, and E0 ⊆ E with m(E0 ) = 0 such that {fnk } → f uniformly
on E ∼ E0 . But this implies ||fnk − f ||∞ → 0. By Proposition 4, {fn } → f in L∞ (E). Thus L∞ (E)
is complete.
p
34. Fix 1 ≤ p ≤ ∞. Let {an }∞
n=1 denote a Cauchy sequence in ` , where an = (an,1 , an,2 , · · · ). Following
Problem 12, we can construct a linear operator T (·) mapping a sequence a ∈ `p to a function T (a) ∈
Lp [1, ∞) that satisfies ||T (a)||p = ||a||p . Let fn = T (an ). By linearity of T , we have
||fn − fm ||p = ||T (an ) − T (am )||p = ||T (an − am )||p = ||an − am ||p
p
Therefore the sequence {fn }∞
n=1 is Cauchy in L [1, ∞). By the Riesz-Fischer Theorem, there exists a
function f such that fn → f in Lp [1, ∞). Moreover, there exists a subsequence {fni }∞
i=1 that converges
pointwise to f a.e. on [1, ∞). For each k, we have
f = lim fni = lim ani ,k
i→∞
i→∞
a.e. on [k, k + 1)
(8)
Let bk = limi→∞ ani ,k and consider the sequence b = (b1 , b2 , · · · ). Equation (8) implies f = T (b) a.e. on
[1, ∞). We can therefore infer
||b||p = ||T (b)||p = ||f ||p < ∞
and
||ani − b||p = ||T (ani − b)||p = ||T (ani ) − T (b)||p = ||fni − f ||p → 0
Thus {ani } → b in `p . By Proposition 4, {an } → b in `p . We conclude that `p is complete and therefore
a Banach space.
35. We first most show that c and c0 are linear spaces. Let a = (a1 , a2 , · · · ) and b = (b1 , b2 , · · · ) denote
two sequences in c. Let a∞ = limk→∞ ak and b∞ = limk→∞ bk . For any two scalars α and β, we have
lim (α · ak + β · bk ) = α · a∞ + β · b∞
k→∞
145
Therefore the sequence α · a + β · b converges. Thus c is a linear space. Moreover if a∞ = b∞ = 0, then
α · a + β · b converges to 0. Thus c0 is also a linear space.
It remains to show that c and c0 are complete. Let {an } denote a Cauchy sequence in c, where
an = (an,1 , an,2 , · · · ). By Problem 34, there exists a sequence b = (b1 , b2 , · · · ) such that {an } → b in
`∞ . For any set of indicies k, l, and n, we have
|bk − bl | ≤ |bk − an,k | + |an,k − an,l | + |an,l − bl |
≤ 2 · ||an − b||∞ + |an,k − an,l |
Since {an } → b in `∞ , there exists an index n such that ||an − b||∞ < /3. Since an converges, there
exists an index N such that |an,k − an,l | < /3 for all k, l ≥ N . We can thus conclude
|bk − bl | < for all k, l ≥ N . But then b is a Cauchy sequence, and therefore must converge. Thus b is in c, so c is
complete. Now if {an } is in c0 , then
|bk | ≤ |bk − an,k | + |an,k | ≤ ||an − b||∞ + |an,k |
Since an converges to zero, there exists an index N such that |an,k | < 2 · /3 for all k ≥ N . Therefore
|bk | < for all k ≥ N , implying that b converges to zero. Thus b is in c0 , so c0 is also complete.
7.4
Approximation and Separability
36. Suppose S is dense in X. Fix g ∈ X. For each index n, there exists fn ∈ S such that ||fn − g|| <
1
n.
But then limn→∞ ||fn − g|| = 0, so {fn } → g in X. Conversely, suppose for any g ∈ X there exists a
sequence {fn } in S such that {fn } → g in X. Fix > 0 and choose n such that ||fn − g|| < . Since
fn ∈ S, it follows that S is dense in X.
37. Fix h ∈ H and > 0. Since G is dense in H, there exists g ∈ G such that ||h − g|| < /2. Likewise,
since F is dense in G, there exists f ∈ F such that ||f − g|| < /2. By the Triangle Inequality, we have
||h − f || ≤ ||h − g|| + ||g − f || < Therefore F is dense in H.
38. Let Pn denote the set of polynomials of degree n with rational coefficients. There is a 1-to-1 mapping
between Pn and Qn . By Problem 23 of Chapter 1, Qn is countable. Therefore each Pn is countable.
S∞
By Corollary 6 of Chapter 1, n=1 Pn is countable.
146
39. Fix g ∈ Lp (E). Suppose g 6= 0 so that ||g||p > 0. By Problem 20, there exists h ∈ Lq (E) such that
Z
g·h >0
E
Fix ∈ 0,
R
E
g·h
and choose f ∈ S such that
||h − f ||q <
||g||p
Then
Z
Z
Z
g · (h − f )
g·h −
f ·g ≥
E
E
E
> − ||g||p · ||h − f ||q
>0
40. To complete the proof, we must show S 0 [a, b] is a countable set. Let Sn0 [a, b] denote the step functions
ψ on [a, b] that take rational values and for which there is a partition P = {x0 , · · · , xn } of [a, b] with
ψ constant on (xk−1 , xk ) and xk rational for 1 ≤ k ≤ n − 1. Each step function in this set is fully
characterized by n−1 rational partition points and n rational values. Therefore there is 1-to-1 mapping
between Sn0 [a, b] and Q2n−1 . By Problem 23 of Chapter 1, Q2n−1 is countable. Therefore Sn0 [a, b] is
S∞
countable for each n. By Corollary 6 of Chapter 1, S 0 [a, b] = n=1 Sn0 [a, b] is countable.
41. Let E = (0, 1] and let
αn = −
n−1 1
·
n
p2
Define fn (x) = xαn and f (x) = x−1/p2 . Each fn ∈ Lp2 (E) and
Z
|fn |p1 =
E
Z
1
−
p1
xn p2
· n−1
n
0
=
1
1−
p1
p2
·
n−1
n
But this implies
Z
lim
n→∞
1
=
1 − pp12
|fn |p1 =
E
p1
Since {fn } → f pointwise a.e. on E, {fn } → f in L
p2
p2
in || · ||p1 . But f 6∈ L , so L
Z
|f |p1
E
by Theorem 7. By Proposition 4, {fn } is Cauchy
normed by || · ||p1 is not complete.
42. Suppose E is countable and consider the singleton class of functions S = {0}. For any f ∈ L∞ (E) and
> 0, ||f − 0||∞ = 0 < . Thus S is dense in L∞ (E), so L∞ (E) is separable.
Now suppose E contains a non-degenerate interval. Then E must contain a closed, bounded, nondegenerate interval [a, b]. Let S be dense in L∞ (E). Then the functions in S restricted to [a, b] must
be dense in L∞ ([a, b]). But it was proven in the text that L∞ ([a, b]) is not separable, so the restriction
of functions in S to [a, b] cannot be countable. But this implies S is not countable. Therefore L∞ (E)
is not separable.
43. Fix f ∈ X and > 0. For each n, there exists gn ∈ X0 such that ||f − gn || < 1/n. For all n, m > 2/,
147
we have
||gn − gm || ≤ ||f − gn || + ||f − gm ||
1
1
< +
n m
<
Therefore {gn } is a Cauchy sequence in X0 . Since X0 is complete, there exists g ∈ X0 such that
{gn } → g in X0 . But we also have
0 ≤ ||f − g|| ≤ ||f − gn || + ||g − gn ||
by the Triangle Inequality. Taking limits, we conclude that ||f − g|| = 0, which implies f = g. Thus
f ∈ X0 , so X0 ⊇ X. Since X0 ⊆ X by construction, we must have X0 = X.
44. Fix 1 ≤ p < ∞. For each natural number n, define
Fn = {q = (q1 , q2 , · · · ) ∈ `p : qk ∈ Q for all k and qk = 0 for k ≥ n}
There exists a one-to-one correspondence between Fn and Qn . Since Qn is countable by Problem 23
of Chapter 1, Fn is countable. By Corollary 6 of Chapter 1, F := ∪∞
n=1 Fn is also countable. We claim
P∞
p
p
that F is dense in ` . To show this, fix > 0 and a ∈ ` . Since k=1 |ak |p < ∞, there exists an index
n such that
∞
X
|ak |p <
k=n+1
p
2
For k = 1, · · · , n, choose qk ∈ Q so that
|ak − qk | <
2(k+1)/p
Then
||a − q||p =
n
X
|ak − qk | +
!1/p
p
|ak |
k=n+1
k=1
n
X
∞
X
p
p
p
<
2 ·
+
2
2
k=1
p
1/p
p
<
+
2
2
!1/p
−k
=
∞
is dense in `∞ . For any set
We show `∞ is not separable by contradiction. Suppose {qn }∞
n=1 ⊆ `
A ∈ 2N , define
an (A) =

1
if n ∈ A
0
otherwise
∞
and let a(A) = (a1 (A), a2 (A), · · · ) ∈ `∞ . Since {qn }∞
n=1 is dense in ` , there exists an index η(A) such
148
that
||a(A) − qη(A) ||∞ <
1
2
Now if A, A0 are in 2N with A 6= A0 , we have
1 = ||a(A) − a(A0 )||∞
≤ ||a(A) − qη(A0 ) ||∞ + ||qη(A) − qη(A0 ) ||∞ + ||a(A0 ) − qη(A0 ) ||∞
< ||qη(A) − qη(A0 ) ||∞ + 1
Therefore ||qη(A) − qη(A0 ) ||∞ > 0, which implies η(A) 6= η(A0 ). Thus η is a one-to-one mapping from 2N
to N. But this implies 2N is countable by Problem 17 of Chapter 1. However, Problem 22 of Chapter
shows that 2N is uncountable, a contradiction.
45. Suppose E is bounded. Then there exists M ≥ 0 such that E ⊆ (−M, M ). Fix f ∈ Lp (E) and > 0.
By Proposition 23 of Chapter 4, there exists δ > 0 such that
Z
if A ⊆ E is measurable and m(A) < δ, then
|f |p <
A
p
2p+1
By Lusin’s Theorem, there exists a continuous function g on R and a closed set F ⊆ E such that
g = f on F and m(E ∼ F ) < δ
Since F is closed, the set (−M, M ) ∼ F is open. Therefore by Proposition 9 of Chapter 1, (−M, M ) ∼
S∞
F = k=1 (ak , bk ) where {(ak , bk )}∞
k=1 is a collection of disjoint open intervals. Now define the sequence
of functions
gn (x) =

n · g(ak )


g(ak ) −
· (x − ak )


ck








0
n · g(bk )


g(bk ) +
· (x − bk )



dk







g(x)
if ak < x < ak +
ck
n
dk
ck
≤ x ≤ bk −
n
n
dk
< x < bk
bk −
n
if ak +
if
if
x∈F
where
ck = 2−k · min
1
, bk − ak
max{|g(ak )|p , 1}
dk = 2−k · min
We then have
Z
bk
ak
|g(ak )|p · ck
|g(bk )|p · dk
+
2·n
2·n
1
≤
n · 2k
|gn |p ≤
149
1
, bk − ak
max{|g(bk )|p , 1}
which implies
Z
|gn |p ≤
Z
E∼F
|gn |p =
(−M,M )∼F
∞ Z
X
bk
|gn |p ≤
ak
k=1
1
2·n
Now pick n > 2p /p . Then
Z
Z
p
p
|f − gn | =
p
Z
(|f |p + |gn |p ) < p
|f − gn | ≤ 2 ·
E
E∼F
E∼F
By construction, gn is continuous on (−M, M ). We extend gn to R by defining
gn (x) =


0
if x ≤ −M − 1









g(−M ) + g(−M ) · (x + M ) if − M − 1 < x ≤ −M




g(M ) − g(M ) · (x − M )









0
if M < x < M + 1
x≥M +1
if
Then gn vanishes outside the bounded set [−M − 1, M + 1], so gn ∈ Cc (E). Since ||f − gn ||p < , Cc (E)
is dense in Lp (E).
Now suppose E is unbounded. By the Dominated Convergence Theorem, there exists n > 0 such that
Z
|f |p <
E∼(−n,n)
p
2p+1
We previously showed how to construct a continuous function g that satisfies
Z
|f − g|p <
E∩(−n,n)
p
2
and for which there exists an M ≥ 0 such that g(x) = 0 if |x| ≥ M . Assume without loss that n ≥ M .
We then have
Z
p
Z
p
|f − g| =
E
Z
|f − g|p
|f − g| +
E∼(−n,n)
< 2p ·
Z
E∩(−n,n)
(|f |p + |g|p ) +
E∼(−M,M )
< p
Since g ∈ Cc (E) and ||f − g||p < , Cc (E) is dense in Lp (E).
46. If a = 0 or b = 0, the expression reduces to
max{|a|, |b|} ≤ 2p · max{|a|, |b|}
150
p
2
which is obviously true.
Now assume a and b are both non-zero. We first consider the case when sgn(a) = sgn(b). Define
f (x) = x1/p . Then
sgn(a) · |a|1/p − sgn(b) · |b|1/p = |f (|a|) − f (|b|)|
≤ f |a| − |b| − f (0)
= |a| − |b|
≤ a−b
1/p
1/p
≤2· a−b
1/p
where the first inequality follows because f is concave over non-negative numbers.
It remains to show the inequality holds when sgn(a) = − sgn(b). In this case,
max{|a|, |b|} ≤ |a − b|
which implies
max{|a|1/p , |b|1/p } ≤ |a − b|1/p
We therefore have
sgn(a) · |a|1/p − sgn(b) · |b|1/p = |a|1/p + |b|1/p
≤ 2 · max{|a|1/p , |b|1/p }
≤2· a−b
1/p
47. If a = 0 or b = 0, the expression reduces to
max{|a|p , |b|p } ≤ p · max{|a|p , |b|p }
which is obviously true. The inequality also holds trivially whenever |a| = |b|.
Now assume a and b are both non-zero with |a| =
6 |b|. We first consider the case when sgn(a) = sgn(b).
Define f (x) = xp . Then
sgn(a) · |a|p − sgn(b) · |b|p
f (|a|) − f (|b|)
=
|a| − |b|
|a| − |b|
≤ f 0 (|a| + |b|)
= p · (|a| + |b|)p−1
where the inequality follows from the convexity of f . Rearranging this expression, we obtain
|sgn(a) · |a|p − sgn(b) · |b|p | ≤ p · |a| − |b| · (|a| + |b|)p−1
≤ p · |a − b| · (|a| + |b|)p−1
151
Now assume sgn(a) = − sgn(b). Then
sgn(a) · |a|p − sgn(b) · |b|p = |a|p + |b|p
and
|a − b| · (|a| + |b|)p−1 = (|a| + |b|)p
Thus the inequality reduces to
|a|p + |b|p ≤ p · (|a| + |b|)p
If p ≥ 2, then
|a|p + |b|p ≤ 2 · max{|a|p , |b|p } ≤ 2 · (|a| + |b|)p ≤ p · (|a| + |b|)p
If 1 < p < 2, define g(x) = xp−1 . We then have
|a|p + |b|p
|a|
|b|
=
· g(|a|) +
· g(|b|)
|a| + |b|
|a| + |b|
|a| + |b|
≤g
|a|2 + |b|2
|a| + |b|
p−1
≤ (|a| + |b|)
where the first inequality follows from concavity of g and the second inequality follows because |a|2 +
|b|2 ≤ (|a| + |b|)2 . But this expression implies
p
p
|a|p + |b|p ≤ (|a| + |b|) ≤ p · (|a| + |b|)
48. Since f ∈ L1 (E), we know that
Z
p
Z
|Φ(f )| =
E
|f | < ∞
E
Therefore Φ(f ) ∈ Lp (E). For any f, g in L1 (E), we also have
||Φ(f ) − Φ(g)||pp =
Z
ZE
≤
sgn(f ) · |f |1/p − sgn(g) · |g|1/p
p
2p · |f − g|
E
p
= 2 · ||f − g||1
where the inequality follows from Problem 46. Thus if {fn } → f in L1 (E), then
1/p
0 ≤ lim ||Φ(fn ) − Φ(f )||p ≤ 2 · lim ||fn − f ||1
n→∞
n→∞
=0
We conclude that Φ(·) is a continuous map from L1 (E) to Lp (E).
Now suppose Φ(f ) = Φ(g). Then there exists a set E0 with m(E0 ) = 0 such that
sgn(f ) · |f |1/p = sgn(g) · |g|1/p
152
(1)
on E ∼ E0 . Fix x ∈ E ∼ E0 . If sgn(f (x)) = sgn(g(x)), then (1) implies |f (x)| = |g(x)| = 0 so that
f (x) = g(x). If sgn(f (x)) = − sgn(g(x)), then (1) implies |f (x)| + |g(x)| = 0 so that f (x) = 0 = g(x).
Therefore f = g on E ∼ E0 . We conclude that Φ(·) is one-to-one.
For g ∈ Lp (E), define the function Φ−1 (g) on E as
Φ−1 (g)(x) = sgn(g(x)) · |g(x)|p
Then since g ∈ Lp (E), we have
Z
|Φ
−1
Z
(g)| =
E
|g|p < ∞
E
Thus Φ−1 (g) ∈ L1 (E) and
Φ(Φ−1 (g)) = sgn(sgn(g) · |g|p ) · | sgn(g) · |g|p |1/p = sgn(g) · |g| = g
Therefore the image of Φ on L1 (E) is Lp (E) and its inverse is given by Φ−1 . To show that the inverse
mapping is continuous, observe that for all f , g in Lp (E) we have
||Φ
−1
(f ) − Φ
−1
Z
sgn(f ) · |f |p − sgn(g) · |g|p
(g)||1 =
E
Z
≤p·
|f − g| · (|f | + |g|)p−1
E
Z
≤ p · ||f − g||p ·
(|f | + |g|)p
(p−1)/p
E
≤ p · ||f − g||p · (||f ||p + ||g||p )p−1
≤ p · ||f − g||p · (||f − g||p + 2 · ||g||p )p−1
where the first inequality follows from Problem 47, the second inequality from Hölder’s Inequality, and
the third inequality by Minkowski’s Inequality. Thus if {gn } → g in Lp (E), then
0 ≤ lim ||Φ−1 (gn ) − Φ−1 (g)||1 ≤ p · lim ||gn − g||p · (||gn − g||p + 2 · ||g||p )p−1 = 0
n→∞
n→∞
49. Suppose L1 (E) is separable. Then there exists a countable subset S ⊆ L1 (E) that is dense in L1 (E).
We claim that Φ(S) := {Φ(f ) : f ∈ S} is dense in Lp (E), where Φ(·) was defined in Problem 48.
To show this, pick g ∈ Lp (E) and define f = Φ−1 (g). By Problem 36, there exists a sequence {fn }
in S that converges to f in L1 (E). By Problem 48, {Φ(fn )} is a sequence in Φ(S) that converges
to Φ(f ) = Φ(Φ−1 (g)) = g. We conclude by Problem 36 that Lp (E) is dense in Φ(S). Since Φ is
one-to-one, Φ(S) is countable. Thus Lp (E) is separable.
50. Suppose there exists a continuous mapping Φ from L1 ([a, b]) onto L∞ ([a, b]). Then we could use the
argument in Problem 49 to conclude that L∞ ([a, b]) is separable. However L∞ ([a, b]) was shown not
to be separable in the text, a contradiction.
51. See Problem 45.
153
8
The Lp spaces: Duality and Weak Convergence
8.1
The Riesz Representation for the Dual of Lp , 1 ≤ p < ∞
1. Let ||T ||∗∗ = sup{T (f ) | f ∈ X, ||f || ≤ 1}. Suppose M ≥ 0 satisfies
|T (f )| ≤ M · ||f || for all f ∈ X
Then for any f ∈ X with ||f || ≤ 1, we have T (f ) ≤ |T (f )| ≤ M · ||f || ≤ M . Therefore ||T ||∗∗ ≤ M ,
which implies ||T ||∗∗ ≤ ||T ||∗
Now fix f ∈ X and suppose f 6= 0. By nonnegativity positive homogeneity of || · || and linearity of T ,
we have
1
· T (f ) = T
||f ||
f
||f ||
≤ ||T ||∗∗
Thus T (f ) ≤ ||T ||∗∗ · ||f ||. If f = 0, then T (f ) = 0 = ||T ||∗∗ · 0 = ||T ||∗∗ · ||f ||. We conclude that
T (f ) ≤ ||T ||∗∗ · ||f || for all f ∈ X, so ||T ||∗∗ ≥ ||T ||∗ .
2. Let T1 and T2 denote two bounded linear functionals on X. For any real numbers α, β, γ, and δ and
any f and g in X, we have
(α · T1 + β · T2 )(γ · f + δ · g) = (α · T1 + β · T2 )(γ · f + δ · g)
= γ · (α · T1 (f ) + β · T2 (f )) + δ · (α · T1 (g) + β · T2 (g))
= γ · (α · T1 + β · T2 )(f ) + δ · (α · T1 + β · T2 )(g)
|α · T1 (f ) + β · T2 (f )| ≤ |α| · |T1 (f )| + |β| · |T2 (f )| ≤ (|α| · ||T1 ||∗ + |β| · ||T2 ||∗ ) · ||f ||
Thus α · T1 + β · T2 is also a bounded linear functional. We conclude that the bounded linear functionals
on X form a linear space.
To see that || · ||∗ is a norm on this space, first observe that T (0) = 0 by linearity. Since ||0|| ≤ 1,
||T ||∗ ≥ 0. If T = 0 for all f ∈ X, ||T ||∗ = 0. Conversely, if ||T ||∗ > 0 there must exist f ∈ X with
||f || ≤ 1 such that T (f ) > 0. But this implies T 6= 0. We conclude that || · ||∗ satisfies non-negativity.
Next, observe that
||T1 + T2 ||∗ = sup{T1 (f ) + T2 (f ) | f ∈ X, ||f || ≤ 1}
≤ sup{T1 (f ) | f ∈ X, ||f || ≤ 1} + sup{T2 (f ) | f ∈ X, ||f || ≤ 1}
= ||T1 ||∗ + ||T2 ||∗
Therefore || · ||∗ satisfies the triangle inequality.
Lastly, fix a real number α 6= 0. Since
|(α · T )(f )| = |α| · |T (f )| ≤ |α| · ||T ||∗ · ||f || for all f ∈ X
154
we must have ||α · T ||∗ ≤ |α| · ||T ||∗ . We also have
|α| · |T (f )| = |(α · T )(f )| ≤ ||α · T ||∗ · ||f || for all f ∈ X
which implies
||α·T ||∗
|α|
≥ ||T ||∗ . We conclude that ||α · T ||∗ = |α| · ||T ||∗ . Therefore || · ||∗ satisfies positive
homogeneity.
3. It was shown in the text that any bounded linear functional satisfies the continuity property. To prove
the converse, suppose a linear functional is unbounded. Then for any index n there exists fn ∈ X such
that
T (fn ) > n2 · ||fn ||
Since T (0) = 0, ||fn || > 0 for all n. For each n, define the function
f˜n =
fn
n · ||fn ||
n o
Then ||f˜n || = n−1 by positive homogeneity of || · ||, so f˜n → 0 in X. But by linearity of T , we have
T (fn )
T (f˜n ) =
>n
n · ||fn ||
for all n. Therefore |T (f˜n )| 6→ T (0), so T does not satisfy the continuity property.
4. Suppose T is bounded. Then
|T (g) − T (h)| = |T (g − h)| ≤ ||T ||∗ · ||g − h||
so T is Lipschitz and the Lipschitz constant c∗ must satisfy c∗ ≤ ||T ||∗ . If c is any constant that
satisfies the Lipschitz condition, then
|T (f )| = |T (f ) − T (0)| ≤ c · ||f − 0|| = c · ||f ||
Thus ||T ||∗ ≤ c, so ||T ||∗ ≤ c∗ . We conclude that ||T ||∗ = c∗ .
Conversely, suppose T is Lipschitz with Lipschitz constant c∗ . Then
|T (f )| = |T (f ) − T (0)| ≤ c∗ · ||f − 0|| = c∗ · ||f ||
Thus T is bounded and ||T ||∗ ≤ c∗ . If M is any constant that satisfies the boundedness condition, then
|T (g) − T (h)| = |T (g − h)| ≤ M · ||g − h||
Thus c∗ ≤ M , so c∗ ≤ ||T ||∗ . We conclude that ||T ||∗ = c∗ .
5. Assume 1 ≤ p < ∞ and let F p denote the functions in Lp (E) that vanish outside a bounded set. Pick
f ∈ Lp (E) and > 0. By the Monotone Convergence Theorem,
Z
|f |p →
E∩[−n,n]
155
Z
E
|f |p
Since
R
E
|f |p < ∞, we can pick n so that
Z
Z
p
|f |p < |f | −
E
E∩[−n,n]
We then have
Z
f − f · χ[−n,n]
p
Z
|f |p =
=
E
Z
E∼[−n,n]
|f |p −
E
Z
|f |p < E∩[−n,n]
Since f · χ[−n,n] ∈ F p , F p is dense in Lp (E).
Now suppose p = ∞. Let = 1/2 and g = 1. Pick f ∈ F ∞ . Then there exists M ≥ 0 such that f = 0
outside [−M, M ], so
||f − g||∞ ≥ 1 > Therefore F ∞ is not dense in L∞ (R).
6. The Riesz Representation Theorem follows from Theorem 5. To show that Theorem 5 holds for
p = 1, let T denote a bounded linear functional on L1 ([a, b]). Define Φ(x) = T (χ[a,x) ). Then for any
[c, d] ⊆ [a, b], we have
|Φ(d) − Φ(c)| = |T (χ[c,d) )| ≤ ||T ||∗ · ||χ[c,d) ||1 = ||T ||∗ · |d − c|
Therefore Φ(·) is Lipschitz on [a, b], so Φ(·) is absolutely continuous on [a, b] by Proposition 7 of Chapter
6. The rest of the proof of Theroem 5 can then be followed.
7. We first prove a few preliminary results. Let
`¯ = {a = (a1 , a2 , · · · ) ∈ `∞ | there exists an index n such that ak = 0 for all k ≥ n.}
Claim 1: `¯ is dense in `p for 1 ≤ p < ∞.
Proof : Suppose b ∈ `p for 1 ≤ p < ∞. Then for any > 0, there exists an index n such that
∞
X
|bk |p < p
k=n
Define
ak =

b
k
0
if k < n
if k ≥ n
and let a = (a1 , a2 , · · · ). Then a ∈ `¯ and
||b − a||p =
∞
X
!1/p
|bk |p
<
k=n
Claim 2: Fix 1 ≤ p < ∞ and let q denote the conjugate of p. Let b = (b1 , b2 , · · · ) denote a sequence.
156
Suppose there exists an M ≥ 0 for which
∞
X
ak · bk ≤ M · ||a||p
k=1
¯ Then b ∈ `q and ||b||q ≤ M .
for all sequences a ∈ `.
Proof : Let ak = sgn(bk ) · |bk |q−1 and define

a
k
āk =
0
Then
n
X
q
|bk | =
k=1
n
X
if k ≤ n
if k > n
ak · bk ≤ M · ||ā||p
(1)
k=1
But since p and q are conjugates, we know that
||ā||p =
n
X
!1/p
|bk |
p(q−1)
k=1
=
n
X
!1/p
|bk |
q
k=1
Rearranging (1), we obtain
n
X
!1/q
q
|bk |
≤M
k=1
Since the expression holds for all n, we can conclude
||b||q ≤ M
Claim 3: Let 1 ≤ p < ∞ and q denote the conjugate of p. Fix c ∈ `q and define the functional
T (a) =
∞
X
ak · ck for all a ∈ `p
k=1
Then T is a bounded, linear functional on `p with ||T ||∗ = ||c||q .
Proof : If c = 0, the results holds trivially. So assume c 6= 0 throughout. By Problem 12 of Chapter
7, we have
|T (a)| ≤
∞
X
|ak · ck | ≤ ||a||p · ||c||q
k=1
157
(2)
Thus T is bounded. Since T is bounded, for all a and b in `p and α and β in R we have
T (α · a + β · b) =
=
∞
X
k=1
∞
X
(α · ak + β · bk ) · ck
(α · ak · ck + β · bk · ck )
k=1
=α·
∞
X
ak · ck + β ·
k=1
∞
X
bk · ck
k=1
= α · T (a) + β · T (b)
Therefore T is linear. By (2), ||T ||∗ ≤ ||c||q . Define
c∗k = ||c||1−q
· sgn(ck ) · |ck |q−1
q
Then ||c∗ ||p = 1 and
T (c∗ ) =
∞
X
c∗k · ck = ||c||1−q
·
q
k=1
∞
X
|ck |q = ||c||q
k=1
Thus ||T ||∗ ≥ ||c||q .
Riesz Representation Theorem for the Dual of `p : Let 1 ≤ p < ∞ and q the conjugate of p. For
each b ∈ `q , define the bounded linear functional Rb on `p as
Rb (a) =
∞
X
ak · bk for all a ∈ `p
k=1
For each bounded linear functional T on `p , there is a unique sequence b ∈ `q for which
Rb = T and ||T ||∗ = ||b||q
Proof : Let T denote a bounded linear functional on `p . Define

1 if k = n
ek (n) =
0 otherwise
and let e(n) = (e1 (n), e2 (n), · · · ). Define
bn = T (e(n))
¯ Then there exists an index n such that ak = 0 for
and let b = (b1 , b2 , · · · ). Pick a = (a1 , a2 , · · · ) ∈ `.
all k ≥ n. We therefore have
∞
X
k=1
a k · bk =
n
X
k=1
ak · bk =
n
X
ak · T (e(k)) = T
k=1
n
X
k=1
158
!
ak · e(k)
= T (a)
(3)
by linearity of T . This expression implies
∞
X
ak · bk = |T (a)| ≤ ||T ||∗ · ||a||p
k=1
¯ Rb = T on `p by Claim 1 and Proposition 3. By
Therefore b ∈ `q by Claim 2. Since Rb = T on `,
Claim 3, ||T ||∗ = ||Rb ||∗ = ||b||q . If Rb0 = T on `p , then
b0k = Rb0 (e(k)) = T (e(k)) = Rb (e(k)) = bk
for all k. Thus b0 = b, so b is unique.
8. We first prove some preliminary results (notation introduced in the solution to Problem 7 is re-used).
Claim 1: `¯ is dense in c0 .
Proof : Let b denote a sequence in c0 and fix > 0. Since {bk } → 0, there exists an index k such that
for all k ≥ n
2
|bk | <
Define
ak =

b
if k < n
k
if k ≥ n
0
Then a ∈ `¯ and
||a − b||∞ = sup |bk | ≤
k≥n
<
2
Claim 2: Fix b ∈ `1 and define the functional
T (a) =
∞
X
ak · bk for all a ∈ c0
k=1
Then T is a bounded, linear functional on c0 with ||T ||∗ = ||b||1 .
Proof : To show that T is bounded, observe that for any a ∈ c0 we have
|T (a)| ≤
∞
X
|ak | · |bk | ≤ ||a||∞ ·
k=1
∞
X
|bk | = ||b||1 · ||a||∞
k=1
Linearity of T (·) was shown in Problem 7. Therefore T is a bounded linear functional on c0 with
||T ||∗ ≤ ||b||1 . To show that ||T ||∗ = ||b||1 , suppose we define ak = sgn(bk ) and
ak (n) =
Then
n
X
k=1
|bk | =

a
if k ≤ n
0
if k > n
k
n
X
ak · bk = T (a(n))
k=1
159
Since ||a(n)||∞ = 1, we must have
||b||1 = sup T (a(n)) ≤ sup ||T ||∗ · ||a(n)||∞ = ||T ||∗
n
n
Claim 3: Let
c̄ = {a = (a1 , a2 , · · · ) ∈ `∞ | there exists a number ā such that an = ā for all n}
If T is a bounded linear functional on c̄, then there exists a real number α such that T (a) = α · ā.
Proof : Pick a ∈ c̄. By linearity, T (a) = ā·T (e) where e = (1, 1, · · · ) denotes the sequence of containing
only ones. Since T is bounded, T (e) ≤ ||T ||∗ · ||e||∞ = ||T ||∗ < ∞. Therefore setting α = T (e) yields
the desired result.
The Dual of c0 and c: For each b ∈ `1 , define the bounded linear functional Rb on c0 as
Rb (a) =
∞
X
ak · bk for all a ∈ c0
k=1
For each bounded linear functional T0 on c0 , there is a unique sequence b ∈ `1 for which
Rb = T0 and ||T0 ||∗ = ||b||1
For each bounded linear functional T on c, there is a unique sequence b ∈ `1 and a real number α for
which
T (a) = α · ā + Rb (a − ā) for all a ∈ c
(4)
where ā = limn→∞ an .
Proof : Let T0 denote a bounded, linear operator on c0 . In Problem 7, we showed how to define a
sequence b = (b1 , b2 , · · · ) such that
T0 (a) =
∞
X
ak · bk for all a ∈ `¯
(5)
k=1
Since `¯ is dense in c0 by Claim 1, (5) must hold on c0 by Proposition 3. By Claim 2, ||T0 ||∗ = ||b||1 .
Uniqueness follow by the same argument used in the solution to Problem 7.
Now let T denote a bounded, linear operator on c. For any a ∈ c, we have
T (a) = T (a − ā · e) + T (ā · e)
by linearity. Since a − ā · e ∈ c0 and ā · e ∈ c̄, we can write this expression as
T (a) = T0 (a − ā · e) + T̄ (ā · e)
where T0 denotes the restriction of T to c0 and T̄ denotes the restriction of T to c̄. The representation
160
in (4) then follows from the result for c0 and Claim 3.
9. For any f ∈ C[a, b], we have
|T (f )| = |f (x0 )| ≤ max |f (x)| = ||f ||max
x∈[a,b]
Therefore T is bounded.
Let g denote a function of bounded variation on [a, b] and let P = {y0 , y1 , · · · , yk } denote a partition
of [a, b]. Let
S(P, f, g) =
k
X
f (ci ) · |g(yi ) − g(yi−1 )|
i=1
where yi ≤ ci ≤ yi−1 . We seek a function g such that for any > 0, there exists a δ > 0 such that
|S(P, f, g) − f (x0 )| < whenever yi − yi−1 < δ for i = 1, · · · , k.
Fix > 0. By Theorem 23 of Chapter 1, there exists a δ > 0 such that |f (x) − f (x0 )| < if |x0 − x| < δ.
Suppose x0 ∈ (a, b] and let g = χ[x0 ,b] . There exists a unique index i∗ such that
yi∗ −1 < x0 ≤ yi∗
This implies
|g(yi ) − g(yi−1 )| =

1
if i = i∗
0
otherwise
so that
|S(P, f, g) − f (x0 )| = |f (ci∗ ) − f (x0 )|
Thus if yi − yi−1 < δ for i = 1, · · · , k, then |S(P, f, g) − f (x0 )| < . Moreover T V (g) = 1 < ∞, so g is
of bounded variation. A similar argument can be used for g = χ(a,b] when x0 = a.
10. By the Extreme Value Theorem, there exists an index x0 such that f (x0 ) = ||f ||max . By Problem 9,
Rb
there exists a function g with T V (g) = 1 such that a f dg = f (x0 ) = ||f ||max .
11. Let x, y belong to [a, b]. Then
|Φ(x) − Φ(y)| = |T (gx ) − T (gy ))|
= |T (gx − gy ))|
≤ ||T ||∗ · ||gx − gy ||max
Let z = min{x, y} and z = max{x, y} Then



0


gx (z) − gy (z) = z − z



z − z
161
for a ≤ z ≤ z
for z < z ≤ z
for z < z ≤ b
But since z − z = |x − y|, we can conclude
|Φ(x) − Φ(y)| ≤ ||T ||∗ · |x − y|
Therefore Φ is Lipschitz.
8.2
Weak Sequential Convergence in Lp
12. For all 1 ≤ p < ∞, we have
1
Z
|fn |p = 1
0
Therefore ||fn ||p = 1 for all n, so ||fn ||p 6→ 0. Thus ||fn || does not converge strongly to 0 in Lp [0, 1]
13. Fix 1 < p < ∞. We can write fn as
fn (x) =
Let f =
α+β
2
α−β
α+β
−
· (−1)k for k/2n ≤ x < (k + 1)/2n , 0 ≤ k < 2n − 1
2
2
on [0, 1]. For any x ∈ [0, 1], we have
x
Z
(fn − f ) ≤
|α − β|
2n+1
x
x
0
which implies
Z
lim
n→∞
We also have
Z
1
Z
fn =
0
f
0
p
|fn |p ≤ |α| + |β|
0
Thus ||fn ||p ≤ |α| + |β| for all n, so {fn } is bounded in Lp [0, 1]. Since ||f ||p =
α+β
2
< ∞, we can apply
Theorem 11 to conclude that fn * f .
If α 6= β, |fn − fm | takes the value |α − β| on a set of measure 1/2 whenever n 6= m. Therefore
||fn − fm ||p ≥
|α−β|
21/p
for n 6= m, which implies that no subsequence of {fn } is Cauchy in Lp [0, 1]. Hence
no subsequence of {fn } can converge strongly in Lp [0, 1].
14. Since h is continuous, it is integrable over [0, T ]. Define
T
Z
|h|
M=
0
Since h is periodic,
Rd
c
h ≤ M for any real numbers c and d. Therefore for any x ∈ [a, b], we have
Z
x
fn =
a
1
n
which implies
Z
lim
n→∞
Z
nx
h ≤
na
x
fn = 0
a
162
M
n
(1)
Let
h̄ = max |h(x)|
x∈[0,T ]
We then have
b
Z
|fn |p =
a
nb
Z
1
n
|h|p ≤ (b − a) · h̄p
na
We can conclude that fn is bounded in Lp [a, b], so fn * 0 by Theorem 11 if 1 < p < ∞.
It remains to show the result holds for p = 1. If h̄ = 0, then fn = 0 and the result follows trivially. So
assume h̄ > 0. Fix g ∈ L∞ [a, b]. If g is a step function, (1) can be used to conclude
b
Z
g · fn = 0
lim
n→∞
a
Now suppose g is not a step function. Since [a, b] is bounded, g ∈ L1 [a, b]. By Proposition 10 of
Rb
Chapter 7, for any > 0 there exists a step function ḡ ∈ L1 [a, b] such that a |g − ḡ| < /h̄. We then
have
b
Z
b
Z
g · fn ≤
Z
b
ḡ · fn +
a
a
|g − ḡ| · |fn |
a
b
Z
ḡ · fn + <
a
which implies
b
Z
g · fn ≤ lim
n→∞
a
Since can be made arbitrarily small, we can conclude
Z
b
g · fn = 0
lim
n→∞
a
Therefore fn * 0 by Proposition 6.
15. If ||f0 ||p = 0, then fn = 0 for all n and the result holds trivially. So suppose ||f0 ||p > 0 and fix
g ∈ Lq (R). By Theorem 11 of Chapter 7, for any > 0 there exists a step function g0 ∈ Lq (R) that
vanishes outside a bounded interval and that satisfies
||g − g0 ||q <
||f0 ||p
We then have
Z
Z
fn · g ≤
R
Z
|fn · (g − g0 )| +
R
fn · g0
R
Z
≤ ||fn ||p · ||g − g0 ||q +
fn · g0
R
Z ∞
<+
f0 (x) · g0 (x + n)dx
−∞
163
(2)
Let gn (x) = g0 (x + n). Then
lim gn (x) = 0
n→∞
for all x ∈ R and ||gn ||q = ||g0 ||q for all n. Therefore {gn } * 0 in Lq (R) by Theorem 12, so
Z
f0 · gn = 0
lim
n→∞
R
by Proposition 6. Taking limits of (2), we obtain
Z
fn · g ≤ lim
n→∞
R
Since can be made arbitrarily small, we must have
Z
fn · g = 0
lim
n→∞
R
Since g was an arbitrary member of Lq (R), {fn } * 0 in Lp (R).
The bottom of page 167 in the text provides an example of a function f0 ∈ L1 (R) for which {fn } does
not converge weakly to 0 in L1 (R).
16. By linearity of integration, we have
Z
2
Z
fn2
(fn − f ) =
E
Z
−2·
E
Z
f · fn +
E
f2
E
Taking limits, we obtain
Z
lim
n→∞
(fn − f )2 = 0
E
Therefore {fn } → f in L2 (E).
17. Let fn = n · χ[0,1/n2 ) and f = 0. Then {fn } → f pointwise a.e. on [0, 1] and {fn } * f in L2 [0, 1] by
Theorem 12. Thus {fn } satisfies properties (i) and (ii). However ||fn ||2 = 1 for all n, so no subsequence
of {||fn ||2 } converges to ||f ||2 = 0 (property (iii)). By Corollary 13, no subsequence of {fn } converges
strongly to f .
18. Since {||fn ||} is unbounded, we can pick an index n1 such that ||fn1 || ≥ 3. Now choose an index
n2 > n1 such that ||fn2 || ≥ 2 · 32 . Notice that if no such index exists, {||fn ||} would be bounded
by max{||f1 ||, · · · , ||fn1 ||, 2 · 32 }. Continuing in this way, we can construct a subsequence {fnk } that
satisfies ||fnk || ≥ αk for all k.
Now suppose ||fn || ≥ αn for all n. Then
α := lim inf
n→∞
||fn ||
≥1
αn
By Problem 38 of Chapter 1, there exists a subsequence of {||fn ||/αn } that convergences to α.
Lastly, suppose {||fn ||/αn } → α ∈ [1, ∞]. Let gn = αn /||fn || · fn . By the homogeneity property of
164
norms, ||gn || = αn = n · 3n . If α < ∞, let g = f /α and choose T ∈ X ∗ . Since {fn } * f , we have
lim T (gn ) = lim
n→∞
n→∞
αn
T (f )
· T (fn ) =
= T (g)
||fn ||
α
If α = ∞, let g = 0. In this case {αn /||fn ||} → 0, so we have
lim T (gn ) = lim
n→∞
n→∞
αn
· T (fn ) = 0 = T (g)
||fn ||
Therefore {gn } * g in X.
19. Suppose {ζn } * ζ. Let T k denote the bounded linear transformation on `p that returns the k-th
element of the sequence. Then
lim ζnk = lim T k (ζn ) = T k (ζ) = ζ k
n→∞
n→∞
Conversely, suppose {ζn } → ζ component-wise. Let T denote a bounded linear functional on `p . Since
{ζn } is bounded in `p , there exists a real number M ≥ 0 such that
||ζn ||p ≤ M for all n
By Problem 7, there exists a unique sequence β ∈ `q such that
T (α) =
∞
X
αk · β k for all α ∈ `p
k=1
Since β ∈ `q , for any > 0 there exists an index N such that
∞
X
|β k |q <
k=N +1
q
M
Applying Hölder’s Inequality, we can derive
|T (ζn ) − T (ζ)| ≤
∞
X
(ζnk − ζ k ) · β k
k=1
=
≤
N
X
∞
X
(ζnk − ζ k ) · β k +
k=1
k=N +1
N
X
∞
X
(ζnk − ζ k ) · β k +
k=1
≤
N
X
N
X
ζnk − ζ
k=N +1
(ζnk − ζ k ) · β k + ||ζn ||p ·
k=1
≤
(ζnk − ζ k ) · β k
(ζnk − ζ k ) · β k + k=1
165
M
k p
!1/p
·
∞
X
k=N +1
β
k q
!1/q
Taking limits as n → ∞, we obtain
lim |T (ζn ) − T (ζ)| ≤ n→∞
Since can be made arbitrarily small, we must have
lim T (ζn ) = T (ζ)
n→∞
Therefore {ζn } * ζ in `p .
20. For any T ∈ (Lp1 [0, 1])∗ and f ∈ Lp2 [0, 1], we have
|T (f )| ≤ ||T ||∗ · ||f ||p1 ≤ ||T ||∗ · ||f ||p2
by Corollary 3 of Chapter 7. Thus T ∈ (Lp2 [0, 1])∗ , so (Lp1 [0, 1])∗ ⊆ (Lp2 [0, 1])∗ . We conclude that
{fn } * f in Lp2 [0, 1] implies {fn } * f in Lp1 [0, 1].
To see that the converse does not hold, suppose p1 > 1. Define
fn = n1/p1 · χ[0,1/n)
and f = 0. Then
||fn ||p2 = np2 /p1 −1
||fn ||p1 = 1,
so {fn } is in Lp2 [0, 1] and Lp1 [0, 1]. For any measurable subset A of [0, 1], we have
Z
fn ≤ n1/p1 −1
A
which implies
Z
Z
lim
n→∞
fn = 0 =
A
f
A
By Theorem 10, {fn } * f in Lp1 [0, 1]. However, limn→∞ ||fn ||p2 = ∞, so {fn } 6* f in Lp2 [0, 1] by
Theorem 7.
21. Suppose p > 1 and fix T ∈ (`p )∗ . By Problem 7, there exists a unique sequence b ∈ `q for which
T (a) =
∞
X
ak · bk
k=1
for all a ∈ `p . But since
∞
X
|bn |q < ∞
n=1
we must have
0 = lim |bn | = lim |T (en )|
n→∞
n→∞
p
Therefore {en } * 0 in ` . But ||en ||p = 1 for all n, so no subsequence of {en } can converge strongly
to 0 in `p .
166
Now suppose p = 1 and consider the bounded linear functional
∞
X
T (a) =
(−1)k · ak
k=1
Then
T (en ) = (−1)n
which does not converge. Therefore {en } cannot converge weakly in `1 .
22. The Radon-Riesz Theorem for `p : Let 1 < p < ∞. Suppose {an } * a in `p . Then
{an } → a in `p if and only if lim ||an ||p = ||a||p
n→∞
Proof for p = 2: Strong convergence implies convergence of the norms by the Triangle Inequality.
It remains to show that weak convergence and convergence of the norms implies strong convergence in
`2 . Suppose {an } is a sequence in `2 for which
{an } → a in `2 and lim
n→∞
For each n, we have
∞
X
(akn − ak )2 =
(akn )2 =
∞
X
(ak )2
k=1
k=1
(akn )2 − 2 · akn · ak + (ak )2
k=1
k=1
2
∞
X
∞
X
q
But a ∈ ` = ` , so
lim
n→∞
∞
X
akn · ak =
k=1
by weak convergence. Therefore
n→∞
(ak )2
k=1
∞
X
lim
∞
X
(akn − ak )2 = 0
k=1
2
Thus {an } → a in ` .
23. For any x ∈ [a, b], define
gx =
Let
Z

χ
[x,b]
if x > a
χ
(a,b]
if x = a
b
h(u)dgx (u) for all h ∈ C[a, b]
Tx (h) =
a
By the example on page 156, Tx is a bounded linear functional. Since {fn } * f , we must have
lim fn (x) = lim Tx (fn ) = Tx (f ) = f (x)
n→∞
n→∞
167
24. For any x ∈ [a, b], let gx = χ[a,x] , Define
b
Z
h · gx for all h ∈ L∞ [a, b]
Tx (h) =
a
Following the remark on page 161, Tx is a bounded, linear functional on L∞ [a, b]. Since {fn } * f , we
have
Z
n→∞
25.
x
Z
x
fn
fn = lim Tx (fn ) = Tx (f ) =
lim
n→∞
a
a
(i) Suppose {fn } * f and {fn } * g in X. Pick T ∈ X ∗ such that T (f − g) = ||f − g||. Then
||f − g|| = T (f − g) = T (f ) − T (g) = lim T (fn ) − lim T (fn ) = 0
n→∞
n→∞
Since ||f − g|| = 0, f − g = 0. Thus f = g.
(ii) Suppose {fn } * f in X. Pick T ∈ X ∗ such that T (f ) = ||f || and ||T ||∗ = 1. Since
T (fn ) ≤ ||T ||∗ · ||fn || = ||fn ||
for all n, we have
||f || = |T (f )| = lim |T (fn )| ≤ lim inf ||fn ||
n→∞
n→∞
26. Suppose by way of contradiction that {||fn ||p }n is unbounded. By the argument in Problem 18, by
taking a subsequence we can assume without loss that {||fn ||p /αn } → α ∈ [1, ∞] where αn = n · 3n .
Now define gn = αn /||fn ||p · fn , and pick g ∈ Lq (E). Then
Z
gn · g =
E
Since {αn /||fn ||p } and {
R
E
αn
||fn ||p
Z
fn · g
E
R
fn · g} are bounded, { E gn · g} is bounded. We also have ||gn ||p = αn . It
was shown in the proof of Theorem 7 that if ||gn ||p = αn , we can construct a function g ∈ Lq that
satisfies
Z
g · gn ≥ n/2
E
R
for all n. But this contradicts the boundedness of { E g · gn }.
8.3
Weak Sequential Compactness
27. Without loss of generality, assume b = 1 and a = 0. Define



0









fn (x) = n · (n + 1) · x − n










1
168
if x ∈ 0,
1
n+1
1
1
if x ∈
,
n+1 n
1
if x ∈
,1
n
Then ||fn ||max = 1 for all n, so {fn } is bounded in C[a, b]. For a fixed n, we have
fn
and
fm
1
n+1
1
n+1
=0
=1
for all m > n. Therefore ||fn − fm ||max = 1 for all m > n, so no subsequence of {fn } can be Cauchy.
28. The example sequence in Problem 21 is bounded in `p for all p but fails to have any strongly convergent
subsequence.
29. If E contains a non-degenerate interval, we can reuse the example on page 173 to construct a sequence
bounded in L1 (E) that does not contain a weakly convergent subsequence.
Now suppose E0 ⊆ E has measure zero. Then for any sequence {fn } in L1 (E0 ),
Z
Z
g · fn = 0 =
g
E0
E0
for all g ∈ L∞ (E0 ). Therefore {fn } * 0 in L∞ (E0 ) by Proposition 6.
30. Suppose {Tn } → T with respect to the || · ||∗ norm. Then for any > 0, there exists N such that
||Tn − T ||∗ < for all n ≥ N . But then for any f ∈ X with ||f || ≤ 1, we have
|Tn (f ) − T (f )| ≤ ||Tn − T ||∗ · ||f || < for all n ≥ N
Therefore limn→∞ Tn (f ) = T (f ) uniformly on {f ∈ X | ||f || ≤ 1}.
Conversely, suppose limn→∞ Tn (f ) = T (f ) uniformly on {f ∈ X | ||f || ≤ 1}. Then for any > 0, there
exists N such that |Tn (f ) − T (f )| < for all n ≥ N and for all f ∈ X with ||f || ≤ 1. But this implies
||Tn − T ||∗ = sup{(Tn − T )(f ) | f ∈ X, ||f || ≤ 1} ≤ for all n ≥ N . We can therefore conclude that {Tn } → T with respect to the || · ||∗ norm.
31. Fix ∈ (0, 1) and δ > 0. Choose n such that 1/n < δ and let A = [0, 1/n]. Then m(A) < δ but
R
f = 1 > . Therefore {fn } is not uniformly integrable.
A n
32. L∞ (E) may not be separable, so Helley’s Theorem need not apply.
33. Let q denote the conjugate of p. Let {an } be a bounded sequence in `p . For each n, define the functional
Tn on `q by
Tn (b) =
∞
X
akn · bk for b ∈ `q
k=1
In the solution to Problem 7, we showed that Tn is a bounded, linear functional on `q with ||Tn ||∗ =
||an ||p . Since {an } is bounded in `p , {Tn } is bounded in (`q )∗ . By Helley’s Theorem, there is a
subsequence {Tnj } and T ∈ (`q )∗ such that
lim Tnj (b) = T (b) for all b ∈ `q
j→∞
169
In the solution to Problem 7, we showed that there exists an a ∈ `p such that
T (b) =
∞
X
ak · bk for all b ∈ `q
k=1
But this implies
lim
∞
X
j→∞
aknj · bk =
k=1
∞
X
ak · bk for all b ∈ `q
k=1
Again applying the Riesz Representation Theorem proven in Problem 7, we can conclude that {anj } *
a in `p .
34. Let X = C[0, 1]. X is a separable space. By the Example on page 156, the functional
Z
1
Tn (g) =
g(x)dfn (x)
0
is in X ∗ for all n. By assumption, there exists a real number M ≥ 0 such that T V (fn ) ≤ M for all n.
We can therefore conclude
|Tn (g)| ≤ ||g||max · T V (fn ) ≤ M · ||g||max for all g ∈ X and all n
But then by Helley’s Theorem, there is a subsequence {Tnk } of {Tn } and T ∈ X ∗ for which
lim Tnk (g) = T (g) for all g ∈ X
k→∞
(1)
By the Remark on page 161, there exists a function f of bounded variation such that
Z
1
g(x) · df (x) for all g ∈ X
T (g) =
0
The expression in (1) therefore implies
Z
lim
k→∞
Thus
35.
nR
1
0
1
Z
0
1
g(x)df (x) for all g ∈ X
g(x)dfnk (x) =
0
o
g(x)dfnk (x) is Cauchy for each g ∈ X.
(i) Assume without loss that M ≥ 1. Fix g ∈ X and > 0. Since S is dense in X, we can pick
g 0 ∈ S such that
||g − g 0 || <
3·M
Since {Tn (g 0 )} is Cauchy, we can pick an index N such that
|Tn (g 0 ) − Tm (g 0 )| <
170
3
for all n, m ≥ N . Then
|Tn (g) − Tm (g)| ≤ |Tn (g) − Tn (g 0 )| + |Tn (g 0 ) − Tm (g 0 )| + |Tm (g) − Tm (g 0 )|
≤ ||Tn ||∗ · ||g − g 0 || + + ||Tm ||∗ · ||g − g 0 ||
3
≤ 2 · M · ||g − g 0 || +
3
<
for all n, m ≥ N . Therefore {Tn (g)} is Cauchy.
(ii) Pick α, β in R and g, g 0 in X. Since Tn is linear for all n, we must have
Tn (α · g + β · g 0 ) = α · Tn (g) + β · Tn (g 0 ) for all n
Taking limits, we conclude that
T (α · g + β · g 0 ) = α · T (g) + β · T (g 0 )
Therefore T is linear. Since each Tn is bounded, we also know that
|Tn (g)| ≤ ||Tn ||∗ · ||g|| ≤ M · ||g|| for all n
Taking limits, we conclude that
|T (g)| ≤ M · ||g||
for all n. Therefore T is bounded.
36. Let fn = 2n+1 · χ(2−(n+1) ,2−n ] and define
Z
Tn (g) =
1
g · fn for all g ∈ L∞ [0, 1]
0
Since
Z
1
|Tn (g)| =
Z
1
g · fn ≤ ||g||∞ ·
0
|fn | = ||g||∞
0
the sequence {Tn } is bounded. Now suppose there exists a subsequence {Tnk } such that {Tnk (g)}
converges for all g ∈ L∞ [0, 1]. Define the function
g0 (x) =

(−1)k
0
if x ∈ (2−(nk +1) , 2−nk ]
otherwise
Then g0 ∈ L∞ [0, 1], so
Tnk (g0 ) = (−1)k for all k
Clearly Tnk (g0 ) does not converge, a contradiction.1
37.
1 This
(i) {fn } * f in Lp (E) because strong convergence implies weak convergence.
solution was found at https://math.stackexchange.com/a/3388752.
171
By the Riesz-Fischer Theorem, a subsequence {fnk } of {fn } converges pointwise a.e. on E to f .
By Proposition 3 of Chapter 5 {fnk } also converges in measure to f on E.
(ii) For p = 1, we can use the example on page 169 to construct a sequence that converges weakly in
Lp (E) but for which no subsequence converges strongly in Lp (E). For 1 < p < ∞, we can use
the example in Problem 13.
Since both sequences in these examples are bounded, no subsequence can converge pointwise
a.e. on Lp (E) by the Bounded Convergence Theorem.
Since no subsequence can converge pointwise, no subsequence can converge in measure to f by
Theorem 4 of Chapter 5.
(iii) Suppose fn = n1/p · χ[0,1/n] . Then fn → 0 pointwise a.e. on [0, 1], but ||fn ||p = 1 for all n.
Therefore no subsequence of {fn } can converge strongly to 0 on [0, 1].
For p = 1, it was shown on page 173 that the above sequence contains no weakly convergent
subsequence. For 1 < p < ∞, weak convergence follows from Theorem 12.
Convergence in measure on E follows by Proposition 3 of Chapter 5.
(iv) Suppose fn = n1/p · χ[0,1/n] and f = 0. By Proposition 3 of Chapter 5, {fn } converges in measure
to f . However ||fn ||p = 1 for all n, so no subsequence of {fn } converges in Lp [0, 1] to 0.
For p = 1, it was shown on page 173 that the above sequence contains no weakly convergent
subsequence. For 1 < p < ∞, weak convergence of a subsequence follows from Theorem 4 of
Chapter 5 and by Theorem 12.
A subsequence converges pointwise a.e. on E to f by Theorem 4 of Chapter 5.
8.4
The Minimization of Convex Functionals
38. In Problem 21, it was shown that {en } * 0 in `p but no subsequence converges strongly to 0 in `p for
1 < p < ∞.
Let 1 < q < ∞ denote the conjugate of p. Then
N
1 X
en
N n=1
=
p
1
→0
N 1/q
Therefore the arithmetic means of {en } converge strongly to 0 in `p .
39. By replacing each an with an − a, we can assume without loss that a = 0. Fix > 0 and pick N such
PN
that |an | < for all n ≥ N . Let A =
n=1 an . Then
n
n
1 X
A
1X
A
an ≤ +
|an | < + n i=1
n
n
n
i=N +1
172
for all n ≥ N . But this expression implies
n
lim sup
n→∞
1X
an ≤ n i=1
Since can be made arbitrarily small, we can conclude
n
1X
an = 0
n i=1
lim
n→∞
40. The Banach-Saks Theorem for `2 : Suppose {an } * a in `2 . Then there is a subsequence {ank }
for which the sequence of arithematic means converges strongly to a in `2 .
Proof : By simply replacing integrals with infinite series, the proof can proceed in the same way as
the proof for the Banach-Saks Theorem given in the text.
41. By the Example on page 176, K is a closed, bounded, convex subset of Lp [a, b]. Linear functionals
are also convex. Therefore by Theorem 17, there exists a minimizing function f0 ∈ K. Since f ∈ K
implies −f ∈ K, we must also have
T (f ) = −T (−f ) ≤ −T (f0 ) for all f ∈ K
Thus |T (f )| ≤ |T (f0 )| for all f ∈ Lp [a, b]. But this implies
|T (f )|
= T
||f ||p
f
||f ||p
≤ |T (f0 )|
for any non-zero f in K. Therefore
|T (f )| ≤ |T (f0 )| · ||f ||p for all f ∈ Lp [0, 1]
We conclude that T is bounded. By the Riesz Representation Theorem, there exists g ∈ Lq [a, b] such
that
Z
T (f ) =
b
f · g for all f ∈ Lp [a, b]
a
By Proposition 2, we know that T (−g ∗ ) = −||T ||∗ But the characterization of f0 from Theorem 17
tells us that
T (f0 ) = inf{T (f ) | f ∈ Lp [a, b] and ||f ||p ≤ 1}
= − sup{T (f ) | f ∈ Lp [a, b] and ||f ||p ≤ 1}
= −||T ||∗
Therefore −g ∗ satisfies
T (−g ∗ ) = T (f0 ) ≤ T (f ) for all f ∈ K
Thus −g ∗ is a minimizer of T on K.
42. We first prove a preliminary uniform integrability result.
173
Claim. Suppose {fn } → f in Lp (E) for some p ∈ [1, ∞). Then {|fn |p } is uniformly integrable.
Proof : Fix > 0. Let A ⊆ E be a measurable set. Since
|fn |p ≤ 2p · (|fn − f |p + |f |p )
we have
Z
p
Z
p
p
|fn | ≤ 2 ·
Z
p
|f |p
|fn − f | + 2 ·
A
A
A
Since |f |p is integrable, there exists δ0 > 0 such that
Z
if m(A) < δ0 , then
|f |p <
A
2p+1
Since {fn } → f in Lp (E), there exists an index N such that
Z
|fn − f |p <
2p+1
E
for all n ≥ N
By Proposition 24 of Chapter 4, there exists δ1 > 0 such that
Z
|fn − f |p <
if m(A) < δ1 , then
A
for 1 ≤ n < N
2p+1
Let δ = min{δ0 , δ1 }. Combining the above results, we can conclude
Z
if m(A) < δ, then
|fn |p < for all n
A
Suppose {fn } → f in Lp1 . Then {||fn ||p1 } → ||f ||p1 < ∞, so there exists M > 0 such that
Z
|fn |p1 ≤ M for all n
E
Since
|ϕ ◦ fn |p2 ≤ |c1 + c2 · |fn |p1 /p2 |p2 ≤ 2p2 · (|c1 |p2 + |c2 |p2 · |fn |p1 )
(1)
we know that
Z
p2
|ϕ ◦ fn |
≤2
p2
p2
· |c1 |
E
p2
· m(E) + |c2 |
Z
·
|fn |
p1
E
≤ 2p2 · (|c1 |p2 · m(E) + |c2 |p2 · M )
<∞
Therefore ϕ ◦ fn ∈ Lp2 (E) for all n. By the same argument, ϕ ◦ f is also in Lp2 (E).
We next show that {|ϕ ◦ fn − ϕ ◦ f |p2 } is uniformly integrable. Fix > 0 and let A denote a measurable
174
subset of E. Since
|ϕ ◦ fn − ϕ ◦ f |p2 ≤ (|ϕ ◦ fn | + |ϕ ◦ f |)p2
≤ 2p2 · (|ϕ ◦ fn |p2 + |ϕ ◦ f |p2 )
≤ 4p2 · (|c1 |p2 + |c2 |p2 · |fn |p1 ) + 2p2 · |ϕ ◦ f |p2
we know that
Z
Z
Z
p2
p2
p2
p2
p1
p2
|fn |
+2 ·
|ϕ ◦ f |p2
|ϕ ◦ fn − ϕ ◦ f | ≤ 4 · |c1 | · m(A) + |c2 | ·
A
A
A
By the above claim, {|fn |p1 } is uniformly integrable. Therefore there exists δ0 > 0 such that
Z
|fn |p1 <
if m(A) < δ0 , then
A
3 · |4 · c2 |p2
Since |ϕ ◦ f |p2 is integrable, there exists δ1 > 0 such that
Z
if m(A) < δ1 , then
|ϕ ◦ f |p2 <
A
Let δ = min
3 · 2 p2
, δ0 , δ1 . Then
3 · |4 · c1 |p2
Z
if m(A) < δ, then
|ϕ ◦ fn − ϕ ◦ f |p2 < A
as desired.
We are now ready to show
Z
|ϕ ◦ fn − ϕ ◦ f |p2 → 0
E
Fix > 0. For any κ > 0, define
En (κ) = {x ∈ E : |fn (x)| < κ and |f (x)| < κ}
Since ϕ is continuous on R, ϕ is uniformly continuous on the closed and bounded interval [−κ, κ].
Therefore there exists δ(κ) > 0 such that for all x ∈ En (κ),
if |fn (x) − f (x)| ≤ δ(κ), then |(ϕ ◦ fn )(x) − (ϕ ◦ f )(x)| <
Now define
An (κ) = {x ∈ En (κ) : |fn (x) − f (x)| < δ(κ)}
175
2 · m(E)
1/p2
Then
Z
Z
p2
|ϕ ◦ fn − ϕ ◦ f |
|ϕ ◦ fn − ϕ ◦ f |
=
E
p2
Z
+
E∼An (κ)
Z
≤
|ϕ ◦ fn − ϕ ◦ f |p2 +
m(An (κ)) ·
m(E)
2
|ϕ ◦ fn − ϕ ◦ f |p2 +
2
E∼An (κ)
Z
≤
|ϕ ◦ fn − ϕ ◦ f |p2
An (κ)
E∼An (κ)
for all n and κ. Since {|ϕ ◦ fn − ϕ ◦ f |p2 } is uniformly integrable, there exists η > 0 such that
Z
if m(E ∼ An (κ)) < η for all n ≥ N , then
|ϕ ◦ fn − ϕ ◦ f |p2 <
E∼An (κ)
for all n ≥ N
2
Now
m(E ∼ An (κ)) ≤
m({x ∈ E : |fn (x)| ≥ κ}) + m({x ∈ E : |f (x)| ≥ κ}) + m({x ∈ E : |fn (x) − f (x)| ≥ δ(κ)})
By Markov’s Inequality,
Z
M
1
·
|fn |p1 ≤ p1 for all n
κp1 E
κ
Z
1
m({x ∈ E : |f | ≥ κ}) ≤ p1 ·
|f |p1
κ
E
Z
1
m({x ∈ E : |fn (x) − f (x)| ≥ δn (κ)}) ≤
·
|fn (x) − f (x)|p1
δ(κ)p1 E
m({x ∈ E : |fn | ≥ κ}) ≤
Choose κ0 such that
κ0 >
max{M 1/p1 , ||f ||p1 }
(η/3)1/p1
Since {fn } → f in Lp1 , there exists an index N such that
Z
|fn (x) − f (x)|p1 <
E
η · δ(κ0 )p1
for all n ≥ N
3
Thus
m(E ∼ An (κ0 )) < η for all n ≥ N
so
Z
|ϕ ◦ fn − ϕ ◦ f |p2 < for all n ≥ N
E
as desired.
43. Define
T (g) = ||g − f0 ||p for all g ∈ Lp (E)
176
Then for any λ in [0, 1] and g, h in Lp (E), we have
T (λ · g + (1 − λ) · h) = ||λ · g + (1 − λ) · h||p
≤ ||λ · g||p + ||(1 − λ) · h||p
≤ λ · ||g||p + (1 − λ) · ||h||p
= λ · T (g) + (1 − λ) · T (h)
Therefore T is convex. Now suppose {gn } → g in Lp (E). Then
|T (gn ) − T (g)| = ||gn − f0 ||p − ||g − f0 ||p
≤ ||gn − g||p
→0
Therefore T is continuous. By Theorem 17, there exists g0 ∈ C such that
||g0 − f0 ||p = T (g0 ) ≤ T (g) = ||g − f0 ||p for all g ∈ C
44.
(i) By direct computation, it is straight-forward to show that

k


 x − n · 1 − 2n+1
Z x

2
fn =

0


x − k + 1
2n
if
k
k
1
≤ x < n + 2n+1
2n
2
2
if
k
1
k+1
+ 2n+1 ≤ x <
2n
2
2n
for 0 ≤ k ≤ 2n − 1. Therefore
Z
x
fn ≤
0
1
1
1
− 2n+1 ≤ n for all x ∈ [0, 1] and all n
n
2
2
2
which implies
Z
lim
n→∞
x
Z
x
f for all x ∈ [0, 1]
fn = 0 =
0
0
(ii) Let En denote the set of values for which fn = 1 on E. Then
n
m(En ) = 2 ·
1
1
− 2n+1
n
2
2
=1−
1
2n+1
From the Fréchet Inequality, we know that
m
N
\
n=1
!
En
≥n−
N
X
n=1
1
2n+1
Therefore
m(E) = lim m
N →∞
177
1
− (n − 1) = +
2
N
\
n=1
!
En
≥
1
2
N +1
1
2
(iii) Direct computation reveals that
||fn ||1 = 2 · 1 −
1
2n+1
≤ 2 for all n
Therefore {||fn ||1 } is a bounded sequence. Since E is measurable set and
Z
lim
n→∞
Z
fn = m(E) > 0 =
E
f
E
we know that {fn } cannot converge weakly to f in L1 [0, 1] by Theorem 10. Theorem 11 is not
violated because it does not apply when p = 1.
(iv) As in part (iii), {fn } cannot converge weakly to f by Theorem 10.
Now by direct computation, we have
||fn ||p = 1 −
1
2n+1
n+1
2
−1
≥ (n+1)/p
2
≥ 2(n+1)/q
+
1
2n+1
n+1
· (2
− 1)
p
1/p
where q denotes the conjugate of p. Since 2(n+1)/q is unbounded, {||fn ||p } is not a bounded
sequence. Therefore Theorem 11 need not apply.
45. Let {gn } denote the sequence of Radamacher functions defined in the example on page 163. It was
shown in that example that {gn } is in L2 [0, 1] but does not contain any Cauchy subsequence. Fix
> 0. By Theorem 12 of Chapter 7, for each n there exists a continuous real-valued function fn on
[0, 1] such that
||fn − gn ||2 ≤ Let {fnk } denote a subsequence of {fn }. Fix an index N . Since {gnk } is not Cauchy, there exists
nk , nj ≥ N such that
||gnj − gnk ||2 > 3 · But by the Triangle Inequality, we have
||fnk − fnj ||2 ≥ ||fnk − gnj ||2 − ||fnj − gnj ||2
≥ ||gnj − gnk ||2 − ||fnk − gnk ||2 − ||fnj − gnj ||2
>
Therefore {fnk } is not Cauchy. We conclude that {fn } does not contain a Cauchy subsequence.
178