Real Analysis Homework: #1
Yingwei Wang
∗
Department of Mathematics, Purdue University, West Lafayette, IN, USA
1
Banach space
P
P∞
Question: Let (xn ) ⊂ X be a Banach space, and ∞
n=1 kxn k is convergent. Proof that
n=1 xn is
convergent in X.
P
P∞
Proof: Suppose ∞
n=1 kxn k = M < ∞, then ∀ ε > 0, ∃N, s.t.
n=N kxn k < ε.
Pn
Let sn = i=1 xi , then ∀n > m > N ,
ksn − sm k = k
n
X
i=m
xi k 6
∞
X
i=m
kxi k <
∞
X
kxi k < ε.
i=N
Hence, (sn ) is a Cauchy sequence and must converge to an element in X.
2
Continuous function
Question: f : (X, τX ) → (Y, τY ) is continuous ⇔ ∀x0 ∈ X and any neighborhood V of f (x0 ), there
is a neighborhood U of x0 such that f (U ) ⊂ V .
Proof: “⇒”: Let x0 ∈ X f (x0 ) ∈ Y . For each open set V containing f (x0 ), since f is
continuous, f −1 (V ) which containing x0 is open. Then, there is a neighborhood U of x0 such that
x0 ∈ U ⊂ f −1 (V ), that is to say f (U ) ⊂ V .
S
“⇐”: Let V ∈ Y . ∀y ∈ V , choose Vy satisfy y ∈ Vy ⊂ V, Vy ∈ τY . Then V = Vy .
Let x = f −1 (y), then ∀Vy , ∃Ux ∈ τX , s.t. x ∈ Ux and f (Ux ) ⊂ Vy .
S
Let U = Ux , then U ∈ τX and f (U ) = V .
Therefore, f : (X, τX ) → (Y, τY ) is continuous.
∗
E-mail address: wywshtj@gmail.com; Tel : 765 337 3504
I
Yingwei Wang
3
Real Analysis
Closure
3.1
T
Question: Topological space (X, τ ), x ∈ X, E ⊂ X, x ∈ Ē ⇔ ∀ neighborhoodV of x, E V 6= ∅.
Proof: Actually, this question is equal to the following one:
T
(*) Topological space (X, τ ), x ∈
/ Ē ⇔ There exist an open set V containing x that E V = ∅.
We just need to prove (*).
T
“⇒ ” If x ∈
/ Ē, then V = X\Ē is an open set containing x and E V = ∅.
T
“⇐” If there exist an open set V containing x that E V = ∅, then X\V is a closed set
containing E.
By the definition of closer Ē, that is the intersection of all closed sets containing E, the set
X\V must contain E.
Thus, x ∈
/ Ē.
3.2
Question: Metric space (X, d), x ∈ X, E ⊂ X, x ∈ Ē ⇔ there is a sequence (en )∞
n=1 in E such
that limn→∞ d(en , x) = 0.
Proof: Similarly as Section 3.1, the above question is equal to the following one:
(**) Metric space (X, d), x ∈
/ Ē ⇔ ∀e ∈ E, ∃ ε0 , s.t. d(e, x) > ε0 .
We just need to prove (**).
“⇒ ” Suppose x ∈
/ Ē. Since X\Ē is an open set, then ∃B(x, δ) ⊂ X\Ē. That is to say
∀e ∈ E, d(e, x) > δ.
“⇐” TIf ∀e ∈ E, ∃ ε0 , s.t. d(e, x) > ε0 , then B(x, ε20 ) is an open set that containing x and
B(x, ε20 ) E = ∅. So X\B(x, ε20 ) is a closed set containing E.
By the definition of Ē, Ē ⊂ X\B(x, ε20 ).
Since x ∈ B(x, ε20 ), x ∈
/ Ē.
4
Distance between point and set
If E is a nonempty subset of a metric space X, define the distance from x ∈ X to E by
d(x, E) = inf d(x, z).
z∈E
4.1
(a) Prove that d(x, E) = 0 ⇔ x ∈ Ē.
Proof: If inf z∈E d(x, z) = 0, then ∀ε, ∃ zε , s.t. d(x, zε ) < ε.
II
(4.1)
Yingwei Wang
Real Analysis
Choosing ε = n1 , then we can get a sequence (zn )∞
n=1 satisfying limn→∞ d(x, zn ) = 0.
According to Section 3.2, x ∈ Ē.
“⇐” If x ∈ Ē, by the previously Section 3.2, there is a sequence (zn )∞
n=1 in E such that
limn→∞ d(zn , x) = 0.
Then d(x, E) = inf z∈E d(x, z) = limn→∞ d(zn , x) = 0.
4.2
(b) Prove that x 7→ d(x, E) is a uniformly continuous function on X, by showing that
|d(x, E) − d(y, E)| ≤ d(x, y).
for all x ∈ X, y ∈ X.
Proof: By the definition of (4.1), we can get:
∀x, y ∈ X, ∀ε > 0, ∃zx ∈ E, s.t. d(x, E) = d(x, zx ) + ε, ∃zy ∈ E, s.t. d(y, E) = d(y, zy ) + ε.
Without loss of generality, we suppose that d(x, E) ≤ d(y, E). Then when ε is sufficient small,
d(y, zy ) ≤ d(y, zx ).
According to triangle inequality,
d(x, y) ≥ d(y, zx ) − d(x, zx ) ≥ d(y, zy ) − d(x, zx ) = d(x, E) − d(y, E).
Besides, in the case that d(x, E) ≥ d(y, E), we can get d(x, y) ≥ d(y, E) − d(x, E). So we can
conclude that |d(x, E) − d(y, E)| ≤ d(x, y).
∀x0 ∈ X, ∀ε > 0, just choose δ = 2ε , then ∀x ∈ B(x0 , δ), we have |d(x, E) − d(x0 , E)| ≤
d(x, x0 ) < δ < ε. That is to say x 7→ d(x, E) is a uniformly continuous function on X.
III
Real Analysis Homework: #2
Yingwei Wang
∗
Department of Mathematics, Purdue University, West Lafayette, IN, USA
1
Banach space
Question: Let C([a, b]) denote the linear space of continuous function f : [a, b] → R. Show that
C[a, b] is a Banach space with respect to the norm
kf k = max{|f (t)| : t ∈ [a, b]}.
Proof: Let (fn )∞
n=1 be a Cauchy sequence in C([a, b]). For ∀t ∈ [a, b], ∀m, n ∈ N, |fm (t)−fn (t)| ≤
kfn − fm k, which means (fn (t))∞
n=1 is a Cauchy sequence in R and must converge to an element
in R. So we can define a function f : [a, b] → R that
f (t) = lim fn (t),
n→∞
∀t ∈ [a, b].
First, we will show that fn → f when n → ∞. ∀ε > 0, there is an N s.t. ∀n, m ≥ N we have
kfn − fm k < ε. For each x ∈ [a, b] we have
|fn (x) − f (x)|
= lim |fn (x) − fm (x)|
m→∞
≤ lim kfn − fm k
m→∞
≤ ε,
for all n ≥ N . Then for all n ≥ N ,
kfn − f k = max {|fn (x) − f (x)|} ≤ ε.
x∈[a,b]
Thus fn → f .
∗
E-mail address: wywshtj@gmail.com; Tel : 765 237 7149
I
Yingwei Wang
Real Analysis
Second, we will show that f ∈ C([a, b]). Choose a point t0 ∈ [a, b]. Since fn ∈ C([a, b]), for the
previous ε, ∃ δn , s.t. ∀t ∈ (t0 − δ, t0 + δ), |fn (t) − fn (t0 )| < ε. For t ∈ (t0 − δ, t0 + δ), n > N we
have
|f (t) − f (t0 )|
≤ |f (t) − fn (t)| + |fn (t) − fn (t0 )| + |fn (t0 ) − f (t0 )|
<ε+ε+ε
< 3ε.
Thus f ∈ C([a, b]) and C([a, b]) is a Banach space.
2
Banach space
1
Question:
of all the functions f : N → R with the property that
P∞Let l (N) be the linear space
|f |1 := n=1 |f (n)| < ∞. Prove that (l1 (N), k · k1 ) is a Banach space.
1
Proof: Let (fi )∞
i=1 be a Cauchy sequence in l (N). For ∀n ∈ N, ∀i, j ∈ N, |fi (t) − fj (t)| ≤
∞
kfi − fj k, which means (fi )i=1 (n) is a Cauchy sequence in R and must converge to an element in
R. So we can define a function f : N → R that
f (n) = lim fi (n),
i→∞
∀n ∈ N.
First, we will show that fi → f when i → ∞. ∀ε > 0, there is an I s.t. ∀i, j ≥ I we have
kfi − fj k < ε. For each n ∈ N we have
|fi (n) − f (n)|
= lim |fi (x) − fj (x)|
j→∞
≤ lim kfi − fj k
j→∞
≤ ε,
for all i ≥ I. Then for all i ≥ N ,
kfi − f k = max{|fi (n) − f (n)|} ≤ ε.
n∈N
Thus fi → f .
Second, we will show that f ∈ l1 (N).
kf k1
=
=
∞
X
n=1
∞
X
|f (n)|
| lim fi (n)|
i→∞
n=1
≤ lim
i→∞
∞
X
n=1
II
|fi (n)| < ∞.
Yingwei Wang
Real Analysis
Then f ∈ l1 (N) and (l1 (N), k · k1 ) is a Banach space.
3
Application of Baire Theorem
Question: Let f : R → R be a smooth function (i.e. C ∞ ). Suppose that for each t ∈ R there is
nt ∈ N such that f (nt ) (t) = 0. Prove that there is an interval I of positive length such that the
restriction of f to I is a polynomial.
Proof: Define Tn := {t ∈ R : f (n) (t) = 0}. It is easy to verify that for each n the restriction
of f to Tn is a polynomial with at most (n + 1) degree. Then we just need to show that ∃n0 , s.t.
(Tn0 )◦ 6= ∅.
First, we claim that ∀n ≥ 1, Tn is a closed set. Consider the set Tnc = R\Tn : ∀t ∈ Tnc , f (n) 6= 0.
Since f ∈ C ∞ , f (n) is continuous, we know that ∃ δ, s.t. ∀x ∈ (t − δ, t + δ), f (n) (x) 6= 0. That is
to say (t − δ, t + δ) ⊂ Tnc . Then R\Tn is an open set while Tn is a closed set.
∞
S
Second, we claim that
Tn = R. ∀t ∈ R, ∃ nt ∈ N, s.t. f (nt ) (t) = 0, which means t ∈ Tnt .
Thus,
∞
S
n=1
Tn = R.
n=1
According to the Baire Category Theorem, ∃n0 , s.t. (Tn0 )◦ 6= ∅. That is to say ∃ interval I ⊂
(Tn )◦ , s.t. the restriction of f to I is a polynomial.
4
Outer measure
Question: Let m∗ (A) denote the outer measure of A ⊂ R. Show that if B ⊂ R and m∗ (B) = 0,
then m∗ (A ∪ B) = m∗ (A).
Proof: On one hand,
A ⊂ (A ∪ B) ⇒ m∗ (A) ≤ m∗ (A ∪ B).
On the other hand,
m∗ (A ∪ B) ≤ m∗ (A) + m∗ (B) = m∗ (A) + 0 = m∗ (A).
Thus,
m∗ (A ∪ B) = m∗ (A).
5
Continuity of function
P 1
Question: Let (xn )∞
n be an enumeration of Q. Define f : R → R by f (x) =
2n . where the
summation is extended over all n such that xn < x. Prove that f is discontinuous at each rational
number and that f is continuous at each irrational number.
III
Yingwei Wang
Real Analysis
Proof: Define Ax = {i ∈ N : xi < x & xi ∈ {xn }∞
n = Q}. Then f (x) =
P
n∈Ax
5.1
1
2n ,
∀x ∈ R.
Rational points ⇒ Discontinuity
Fix xr ∈ Q, r ∈ N. Let xn ∈ Q and xn > xr . It is obviously that r ∈ (Axn \ Axr ). Then
X
X
X
1
1 1 1
−
=
> r.
|f (xn ) − f (xr )| = p
q
i
2 2 2
p∈Axn 2
q∈Axr
i∈(Axn \Axr )
which means
lim
x∈Q, x→x+
r
f (x) 6= f (xr ). That is to say f is discontinuous at each rational number.
5.2
Irrational points ⇒ Continuity
Since
∞
P
n=1
1
2n
= 1, we can know that ∀ε > 0, ∃N ∈ N, s.t.
∞
P
n=N +1
1
2n
< ε. Let the set Sε =
{x1 , x2 , · · · , xN }. We can rearrange the elements of the set Sε such that x1 < x2 < · · · < xN .
∀α ∈ (R \ Q), we can choose δ > 0 as the following way:
 1
if α < x1 ,
 2 (x1 − α)
1
(min{α
−
x
,
x
−
α})
if
xi < α < xi+1 ,
δ=
i
i+1
 21
(α
−
x
)
if
α
> xN .
N
2
Then (Ax+δ \ Ax−δ ) ∩ ASε = ∅. That is to say ∀n ∈ (Ax+δ \ Ax−δ ), n ≥ N + 1.
P
1
So ∀x ∈ (x − δ, x + δ), |f (x) − f (α)| ≤ |f (x + δ) − f (x − δ)| =
2n <
n∈(Ax+δ \Ax−δ )
which means f is continuous at the point α ∈ (R \ Q).
IV
∞
P
n=N +1
1
2n
< ε,
Real Analysis Homework: #3
Yingwei Wang
∗
Department of Mathematics, Purdue University, West Lafayette, IN, USA
1
Measure inequality
Question: Let (X, A, µ) be a measure space. Let (Ak )∞
k=1 be a sequence of sets in A. Prove that
!
∞ \
∞
[
µ
Ak ≤ lim inf µ(Ak ).
k→∞
n=1 k=n
Proof: Let Bn =
∞
T
Ak , then Bn = Bn+1 ∩ An , Bn ⊂ Bn+1 . So
k=n
µ(
⇒
⇒
µ
µ
m
[
Bn ) = µ(Bm ) =
n=1
m
[
n=1
∞
[
n=1
2
∞
\
Ak ⊂ Ak , ∀k ≥ m
k=m
Bn
!
≤ µ(Ak ), ∀k ≥ m
Bn
!
≤ lim inf µ(Ak ).
k→∞
Example of measure
Question: Let µ be a measure on (R, B), where B are the Borel sets, such that µ([0, 1)) = 1 and
µ(x + B) = µ(B) for all x ∈ R and B ∈ B. Prove that
(1) µ([0, 1/n)) = 1/n for all integers n ≥ 1 and
(2) µ([a, b)) = b − a for all real numbers a < b.
∗
E-mail address: wywshtj@gmail.com; Tel : 765 237 7149
I
Yingwei Wang
Real Analysis
Proof: (1) By the assumption that µ(x + B) = µ(B), we can know that µ([0, 1/n)) =
µ([1/n, 2/n)) = · · · = µ([(n − 1)/n, 1)). Then
1 = µ([0, 1]) =
n
X
µ([(k − 1)/n, k/n)) = nµ([0, 1/n))
k=1
⇒
µ([0, 1/n)) = 1/n.
(2) Since for any a < b, µ([a, b)) = µ([0, b − a)), we only need to consider the cases that
a = 0, b > 0.
If b ∈ Q, then b = pq where p, q ∈ N, q 6= 0.
p−1
X
p
k k+1
1
p
µ([0, b)) = µ([0, )) =
µ([ ,
)) = p µ([0, )) = = b.
q
q
q
q
q
k=0
If b ∈ R\Q, then ∃ (bn )∞
n=1 ⊂ Q s.t. lim bn = b. Then
n→∞
µ([0, b)) = lim µ([0, bn )) = lim bn = b.
n→∞
3
n→∞
Lebesgue outer measure
Question: Let m∗ be the Lebegue outer measure on R. For any two sets A, B ⊂ R, prove the
inequality:
m∗ (A) + m∗ (B) ≤ 2m∗ (A△B) + 2m∗ (A ∩ B).
(3.1)
Proof: For the set A, we have
m∗ (A)
≤ m∗ (A\B) + m∗ (A ∩ B)
(since A = (A\B) ∪ (A ∩ B))
≤ m (A△B) + m (A ∩ B). (since (A\B) ⊂ (A△B))
∗
∗
(3.2)
Similarly, for the set B, we have
m∗ (B) ≤ m∗ (A△B) + m∗ (A ∩ B).
(3.3)
By (3.2)-(3.3), we can get (3.1).
4
Zero measure set
Question: Let m denote the Lebesgue measure on R. Let A ⊂ R be a Lebesgue measurable set.
Suppose that m(A ∩ [a, b]) < (b − a)/2 for all a < b real numbers. Show that m(A) = 0.
II
Yingwei Wang
Real Analysis
Proof: On one hand, by the definition of Lebesgue measure,
m(A) = inf{
∞
X
l(In ) : In ⊂ R are open intervals, A ⊂
n=1
∞
[
In },
n=1
we know that ∀ε > 0, ∃ {In }∞
n=1 , s.t.
m(A) ≥
∞
X
l(In ) − ε,
n=1
where A ⊂
∞
S
In and Ii ∩ Ij = ∅, i 6= j.
n=1
We can choose ε = 14 m(A) in the above inequality, then we can get
∞
4X
m(A) ≥
l(In ).
5
(4.1)
n=1
On the other hand, by the assumption, m(A ∩ In ) < 12 l(In ), ∀n ∈ N+ . So we have
m(A) =
∞
X
∞
m(A ∩ In ) ≤
n=1
1X
l(In ).
2
(4.2)
n=1
From (4.1)-(4.2) we can conclude that m(A) = 0.
5
Measure function
Question: Let m denote the Lebesgue measure on R. Let A ⊂ R be a Lebegue measurable
set with m(A) < ∞. Show that the function f : R → [0, ∞), f (x) = m(A ∩ (−∞, x)) is
continuous. Deduce that for every β ∈ [0, m(A)] there is a Lebesgue measurable set B ⊂ A such
that m(B) = β.
Proof:
5.1
The continuity of the auxiliary function g
Since m(A) < ∞, by the Proposition 15 on Page 63 in Royden’s book, given ∀ε > 0, there is a
finite union U of open intervals such that
m∗ (U △A) < ε.
III
(5.1)
Yingwei Wang
Suppose U =
Real Analysis
n
S
(ai , bi ), and a = a1 , b = bn , then U ⊂ [a, b].
i=1
On the interval [a, b], we can define the function g such that
g(x) = m(A ∩ [a, x)),
∀x ∈ [a, b].
Then for ∀∆x > 0,
A ∩ [a, x + ∆x) = (A ∩ [a, x))
⇒
g(x + ∆x) ≤ g(x) + ∆x
⇒
g(x + ∆x) − g(x) ≤ ∆x
[
(A ∩ [x, x + ∆x))
For ∆x < 0, we can get the similar result:
g(x + ∆x) − g(x) ≥ −∆x
Then we have
|g(x + ∆x) − g(x)| ≤ |∆x|
which means g ∈ C([a, b]).
5.2
The continuity of f
For ∀x ∈ [a, b], choose ∆x < ε, then
|g(x) − f (x)| ≤ m(U △A) < ε
⇒
|f (x + ∆x) − f (x)| ≤ |f (x + ∆x) − g(x + ∆x)| + |g(x + ∆x) − g(x)| + |g(x) − f (x)| < 3ε.
which means f ∈ C([a, b]).
For x ≤ a, |f (x)| ≤ m(U △A) < ε, so f ∈ C((−∞, a]). Similarly, we have f ∈ C([b, ∞)).
Thus, f ∈ C(R).
5.3
The Intermediate Value Theorem
Since g(a) = 0, g(b) = m(A), by the Intermediate Value Theorem of continuous functions, for
∀β ∈ [0, m(A)] there is an x0 ∈ [a, b] such that g(x0 ) = β. Then we can choose B = A ∩ [a, x0 ].
IV
Real Analysis Homework: #4
Yingwei Wang
∗
Department of Mathematics, Purdue University, West Lafayette, IN, USA
1
Measurable function
Question: Let (X, A, µ) be a measure space. Let (fn )∞
n=1 be a sequence of measurable
functions: fn : X → R. Show that the set of points x where the limit lim fn exists
n→∞
(finite or infinite) is a measurable set.
Proof: Let Ak = {x ∈ X : |fn (x) − fm (x)| <
1
k,
∀m, n > k}, A = lim Ak , where
k→∞
m, n, k ∈ N. It is easy to know that each Ak is measurable set since fn is a measurable
function. We have this observation: Ak+1 ⊂ Ak ⇒ A ⊂ Ak , ∀k. Thus, A is a measurable
set.
Let B = {x ∈ R : lim fn (x) exists}. I will show that A = B.
n→∞
On one hand, let x ∈ A, then for ∀k ∈ N, |fn (x) − fm (x)| < k1 , ∀m, n > k, which
mean fn (x) is a Cauchy sequence in R. So lim fn (x) exists and then x ∈ B.
n→∞
On the other hand, let x ∈ B, then ∀ε > 0, ∃N ∈ N s.t. |fn (x) − fm (x)| < ε, ∀m, n >
N , which means x ∈ AN . Then if ε → 0, N → ∞, x ∈ A.
Now we can conclude that B is a measurable set.
2
Differential function
Question: Let f : R → R be a differential function. Prove that f ′ is Lebesgue measurable.
∗
E-mail address: wywshtj@gmail.com; Tel : 765 237 7149
I
Yingwei Wang
Real Analysis
Proof: Define a sequence of functions: gn (x) =
f (x+1/n)−f (x)
,
1/n
x ∈ R, n ∈ N. Then
f is measurable
3
⇒
gn (x) is measurable , ∀n,
⇒
f ′ (x) = lim gn is measurable.
n→∞
The sets with measure zero
Question: Let f : R → R be defined by f (x) = x5 + sin x. Suppose that a set A ⊂ R has
Lebesgue measure zero. Show that f (A) has Lebesgue measure zero.
Proof: I want to prove that for any function f ∈ C 1 (R), the image of the set with
Lebesgue measure zero has also Lebesgue measure zero.
∞
S
AN , m(A) = 0 ⇒ m(AN ) = 0, ∀N ∈ N. We
Let AN = [−N, N ] ∩ A, then A =
N =1
will focus the problem on each AN .
By the definition of Lebesgue measure, for ∀ε > 0, there exists a sequence of intervals
(In )∞
n=1 , Ii ∩ Ij = ∅, i 6= j and AN ⊂ ∪In , s.t.
X
l(In ) < ε, since m(AN ) = 0.
(3.1)
n
Let In = (an , bn ), I¯n = [an , bn ], then m(In ) = m(I¯n ). Then we have
X
X
m(f (AN )) <
m(f (In )) =
m(f (I¯n )).
n
(3.2)
n
Since f is continuous, we can know that
m(f (I¯n )) = max f (x) − min f (x).
x∈I¯n
x∈I¯n
Since I¯n is a closed interval and f ∈ C 1 (I¯n ), we can find x1 , x2 ∈ I¯n s.t. f (x1 ) =
max f (x), f (x2 ) = min f (x). Without loss of generality, we can assume that x1 ≤ x2 .
x∈I¯n
x∈I¯n
Then
m(f (I¯n ))
= f (x1 ) − f (x2 )
≤ |f ′ (ξ)|(x2 − x1 ),
′
≤ max f (x) l(In )
x∈I¯n
≤ max f ′ (x) ε.
x∈I¯n
II
ξ ∈ [x1 , x2 ],
Yingwei Wang
Real Analysis
Since maxx∈I¯n f ′ (x) < ∞ and let ε → 0, we can get m(f (In )) = 0.
By (3.2) we have m(f (AN )) = 0. So
X
m(f (A)) = m(f (∪AN )) ≤
m(f (AN )) = 0.
N
4
Measurable function
Question:
P Let ff(x):n R → R be a Lebesgue measurable function. Define g : R → R by
g(x) = ∞
n=1 n!(n+3) . Prove that g is Lebesgue measurable.
Pn
ak
Proof: First, we claim that ∀a ∈ R, the series sn =
k=1 k!(k+3) is convergent.
∀a ∈ R, ∃N ∈ N, s.t. |a| < N . Then for ∀k > 2N , we have
N
|a|N
|a|k−2N
1
ak
≤ |a|
·
·
k!(k + 3) 1 · · · N (N + 1) · · · (2N ) (2N + 1) · · · k · k + 3 .
Let M =
|a|k−2N
(2N +1)···k
<
|a|N
1···N , we can choose
1 k−2N
, we have
2
k > M , s.t.
|a|N
1···N
·
1
k+3
k−2N
ak
1
.
k!(k + 3) < 2
< 1. Since
|a|N
(N +1)···(2N )
P
P
1 k−2N
is convergent, we have for fixed a, ∞
Since for fixed N , ∞
k=1 2
k=1
also convergent.
P
f (x)n
Second, let fn = nk=1 n!(n+3)
which is measurable and
< 1,
ak
k!(k+3)
is
g(x) = lim fn (x), for ∀x ∈ R.
n→∞
Hence, we know that g(x) is also measurable.
5
Measurable function
Question: Let f : R → R be a Lebesgue measurable function. Suppose that A ⊂ R is a
Borel set. Show that the set {x ∈ A : f (x) > x} is Lebesgue measurable.
Proof: Define a function g : R → R by g(x) = f (x) − x, then g(x) is measurable (by
the Theorem 6 in the page 259 of Royden’s book). So the set B = {x ∈ R : g(x) > 0} is
measurable. Then, {x ∈ A : f (x) > x} = A ∩ B is measurable.
III
Real Analysis Homework: #5
Yingwei Wang
∗
Department of Mathematics, Purdue University, West Lafayette, IN, USA
Note: In this paper, {f (x) satisfies some property} = {x : f (x) satisfies some property}
1
Integrable function
1.1
a
Question: Show that if f is integrable then the set {f (x) 6= 0} is of σ-finite measure.
Proof: On one hand,
{f (x) 6= 0} = {|f (x)| > 0} =
∞
[
{|f (x)| ≥
n=1
On the other hand,
Z
Z
∞ > |f |dµ ≥
1
{|f (x)|≥ n
}
|f |dµ ≥
which means
µ{|f (x)| ≥
1
}.
n
1
1
µ{|f (x)| ≥ },
n
n
(1.1)
∀n ∈ N,
1
} < ∞, ∀n ∈ N.
n
(1.2)
(1.3)
From (1.1) and (1.3), we can know that the set {f (x) 6= 0} is of σ-finite measure.
1.2
b
Question: Show that if f is integrable, f ≥ 0, then f = lim ϕn for some increasing sequence
of simple functions each of which vanishes outside a set of finite measure.
Proof: By proposition 7 on Page 260 in Royden’s book, since the set {f (x) 6= 0} is of
σ-finite measure, we can find a sequence (ϕn ) of simple functions defined on {f (x) 6= 0} with
ϕn+1 ≥ ϕn such that f = lim ϕn and each ϕn vanishes outside a set of finite measure.
Then we can define ϕn (x) = 0 on the set {f (x) = 0}, and get the conclusion.
∗
E-mail address: wywshtj@gmail.com; Tel : 765 237 7149
I
Yingwei Wang
1.3
Real Analysis
c
Question: Show that Rif f is integrable with respect to µ, then given ǫ > 0 there is a simple
function ϕ such that |f − ϕ|dµ < ǫ.
Proof: By the assumption, f + and f − are nonnegative integrable functions. By (b), there
are increasing sequence (φn ) and (ϕm ) such that f + = lim φn and f − = lim ϕm . By the
Monotone Convergence Theorem, we have
Z
Z
+
f dµ = lim φn ,
Z
Z
−
f dµ = lim ϕm .
So given ǫ > 0, there are φN and ϕM such that
Z
Z
ǫ
+
f dµ − φN dµ < ,
2
Z
Z
ǫ
f − dµ − ϕM dµ < .
2
Let ϕ = φN − ϕM . Then ϕ is also a simple function and satisfies
Z
|f − ϕ|dµ
Z
Z
≤
|f + − φN |dµ + |f − − ϕM |dµ
Z
Z
Z
Z
+
−
=
f dµ − φN dµ +
f dµ − ϕM dµ
<
2
ǫ.
Measurable set
Question: Let f : R → R be defined by f (x) = x3 − 2x2 + x − 1. Show that if A ⊂ R is
Lebesgue measurable, then f (A) ⊂ R is Lebesgue measurable.
Proof: Since f (x) is a polynomial, then it is measurable and there exists a sequence of
step functions (ϕn ) s.t. ϕn (x) → f (x), ∀x ∈ R.
By assumption, A ⊂ R is Lebesgue measurable, which means ∀E ⊂ R,
m(E) = m(E ∩ A) + m(E ∩ Ac ).
II
Yingwei Wang
Real Analysis
For each step function ϕn (x), if A is measurable, then ϕn (A) is just a set of points so it
is measurable. Thus f (A) = lim ϕn (A) is measurable.
n→∞
3
Limits
Question: Let (X, A, µ) be a measure space with µ(X) < ∞. Let f : X → (−1, 1) be a
measurable function. Consider the sequence
Z
an =
(1 + f + · · · + f n )dµ.
X
Show that either (an ) converges to a finite number or otherwise lim an = ∞. (In other words
n→∞
(an ) cannot have two distinct limit points.)
P
n
for ∀x ∈ X. It
Proof: Since ∀x ∈ X, |f (x)| < 1, theP
series ∞
n=0 |f (x)| is convergent
P∞
∞
n
2n
follows
that
all
of
the
functions
g(x)
=
(f
(x))
and
h
(x)
=
(f
0
n=0
n=0 (x)) , h1 (x) =
P∞
2n+1 are absolutely convergent.
n=0 (−f (x))
Let A = {f ≥ 0}, B = {f < 0}, then X = A ∪ B.
Z X
n
+
an =
(f (x))j dµ
A j=0
( R P
R P
Z X
n
k
f 2j dµ − B kj=0 (−f )2j+1 dµ if n = 2k + 1,
−
j
j=0
B
R Pk
R Pk
an =
(f (x)) dµ =
2j
2j−1 dµ if n = 2k.
B j=0
j=0 f dµ − B
j=0 (−f )
B
Hence,
lim an =
n→∞
lim (a+
n→∞ n
+
a−
n)
=
Z
g(x)dµ +
A
Z
h0 (x)dµ −
A
Z
h1 (x)dµ.
B
which means lim an can not have two distinct limit points.
n→∞
4
Convergence
Question: Show that the following sequence is convergent in R.
Z 1
cos(x + 1/n) − cos x
an = n
dx.
x3/2
1/n
Proof: Consider the function
fn (x) =
cos(x + 1/n) − cos x X[1/n,1] (x)
,
1/n
x3/2
III
x ∈ [0, 1],
Yingwei Wang
Real Analysis
where XA (x) is the characteristic function.
It is obvious that
sin x
lim fn = − 3/2 ,
n→∞
x
x
Let f (x) = − sin
, then |f (x)| ≤
x3/2
x
x3/2
=
1
x1/2
x ∈ [0, 1].
since sin x ≤ x, x ∈ [0, 1].
1
Let g(x) = x1/2 , then ∃N ∈ N, s.t. ∀n > N, |fn (x)| < g(x). Then by Lebesgue dominated
convergence theorem, we have
Z 1
Z 1
Z 1
Z 1
sin x
1
lim an = lim
fn (x)dx =
lim fn (x)dx =
− 3/2 dx <
dx = 2.
1/2
n→∞
n→∞ 0
x
0 n→∞
0
0 x
So an is convergent.
5
Infinitive sum
Question: Let (X, A, µ) be a complete measure space. Let (fn )∞
n=1 be a sequence of Ameasurable functions, fn : X → R. Suppose that
∞ Z
X
|fn |dµ < ∞.
(5.1)
n=1 X
P
Show that the series ∞
n=1 fn (x) is absolutely convergent µ-almost everywhere on X.
Proof: First, claim that
!
Z
∞
∞ Z
X
X
|fn | dµ =
|fn |dµ.
(5.2)
X
Let Fn =
n
P
|fi |, F∞ =
i=1
∞
P
n=1 X
n=1
|fi |, ∀x ∈ X. Then the statement (5.2) becomes
i=1
Z
F∞ dµ = lim
n→∞
X
n Z
X
i=1
|fi |dµ = lim
Z
n→∞ X
X
Fn dµ.
(5.3)
Since Fn (x) ≤ Fn+1 (x) and Fn → F∞ , by the Monotone Convergence Theorem, we can
get (5.3).
Second, by the assumption (5.1) and the claim (5.2), we can get
!
Z
∞
∞ Z
X
X
|fn | dµ =
|fn |dµ < ∞,
(5.4)
X
n=1
n=1 X
P
P∞
which means the measure µ ({ ∞
n=1 |fn | = ∞}) = 0. That is to say the series
n=1 fn (x) is
absolutely convergent µ-almost everywhere on X.
IV
Real Analysis Homework: #6
∗
Yingwei Wang
Department of Mathematics, Purdue University, West Lafayette, IN, USA
1
Integral function
Question: Let f : R → R be a Lebesgue integrable function. Prove that g(x) is also
Lebesgue integrable.
Proof: First, if f (x) is a non-negative simple function, then
f (x) =
g(x) =
n
X
k=1
n
X
ck χEi (x),
∀x ∈ R,
ck χEi +{1} (x),
∀x ∈ R.
k=1
Hence, g(x) is also a non-negative simple function and Lebesgue integrable.
Second, if f (x) is a non-negative measurable function, then there exists a sequence
of non-negative simple monotonic increasing functions {ϕk (x)}∞
k=1 s.t.
lim ϕk (x) = f (x),
k→∞
∀x ∈ R.
It is obvious that {ϕk (x − 1)}∞
k=1 is also a monotonic increasing sequence and
lim ϕk (x − 1) = f (x − 1),
k→∞
∗
E-mail address: wywshtj@gmail.com; Tel : 765 237 7149
I
∀x ∈ R.
Yingwei Wang
Real Analysis
Hence,
Z
g(x) dm
ZR
=
f (x − 1) dm
Z
lim
ϕk (x − 1) dm
k→∞ R
Z
lim
ϕk (x) dm
k→∞ R
Z
f (x)dm < ∞.
R
=
=
=
R
So we have g(x) is Lebesgue integrable on R.
Finally, if f (x) is an arbitrary function on R, we can rewrite f as f = f + − f − ,
where both f + and f − are non-negative integrable function, and then get the same
conclusion about g(x).
2
Zero function
Question: Let f : [0, 1] → R be a Lebesgue measurable Rfunction. Suppose that for
any function g : [0, 1] → R, f g is Lebesgue integrable and [0,1] f g dm = 0. Prove that
f = 0 almost everywhere with respect to the Lebesgue measure.
Proof: Suppose that there exists a set A ⊂ [0, 1], m(A) > 0, s.t. ∀x ∈ A, f (x) 6= 0.
Without loss of generality, we can assume that f (x) > 0, ∀x ∈ A.
∀ε > 0, we can find a sequence of intervals In = (a, b) ⊂ [0, 1] such that A ⊂ ∪In
and m(A\ ∪ In ) < ε. Choose g(x) as follows:
−(x − an )(x − bn ), x ∈ [an , bn ],
g(x) =
0,
x ∈ [0, 1]\ ∪ I¯n .
It is easy to verify that g(x) is continuous and satisfies: g(x) > 0, x ∈ ∪In , and
g(x) = 0, x ∈ [0, 1]\ ∪ I¯n . Hence, f g ≥ 0, x ∈ [0, 1] and f g > 0, x ∈ A. So we have
Z
Z
Z
Z
f g dm =
f gdm =
f gdm +
f gdm,
[0,1]
∪In
A
II
∪In \A
Yingwei Wang
Real Analysis
R
On one hand, A f gdm > 0. On the other hand, since m(A\ ∪ In ) < ε, we can
R
R
R
choose ε such that ∪In \A f gdm < A f gdm. It follows that [0,1] f g dm > 0, which
R
contradicts that [0,1] f g dm = 0, ∀g ∈ C 1 ([0, 1]).
Thus, f = 0, a.e. x ∈ [0, 1].
Note: Another method is by the statement that “Any measurable function can be
approximated by a sequence of continuous functions”.
3
σ-finite
Question: Let (X, A), µ be a measure space and let f : X → R be A-measurable.
Suppose that µ is σ-finite.R Suppose also that there is c ≥ 0 such that for all E ∈ A
with µ(E) < ∞ one has | E f dµ| ≤ c. Show that f is Lebesgue integrable on X.
Proof: Since µ is σ-finite, we can choose a sequence of sets (Xn ) such that X = ∪Xn ,
X1 ⊂ X2 · · · ⊂ Xn ⊂ · · · ⊂ X, lim Xn = X and
n→∞
µ(Xn ) < ∞, ∀n ∈ N.
Define fn (x) = f (x)χXn , ∀x ∈ X, then
lim fn (x) = f (x) and |fn (x)| ≤ |f (x)|, ∀x ∈ X.
n→∞
Since µ(Xn ), by assumption,
Z
Z
fn dµ = X
Xn
f dµ ≤ c,
∀n ∈ N.
By Lebesgue dominated control theorem, we have
Z
Z
Z
f dµ = lim
fn dµ ≤ lim fn dµ ≤ c,
n→∞
n→∞
X
X
X
which implies f is Lebesgue integrable on X.
III
Yingwei Wang
4
Real Analysis
Limits
Question: Let (X, A), µ be a measure space. Let (fn )∞
n=1 be a sequence of A-measurable
functions, fn : X → [0, +∞]. Show that
Z
Z
lim inf fn dµ ≤ lim inf
fn dµ.
X n→∞
n→∞
X
Proof: Let gn (x) = inf fn (x), f (x) = lim inf fn (x), ∀x ∈ X. Then gn are nonnegn→∞
k≥n
ative and gn increases to f , lim gn (x) = f (x). Therefore,
n→∞
Z
gn dµ ≤ inf
X
Z
k≥n X
fk dµ.
R
If we take the limit as n → ∞, on the left side we obtain XR f dµ by the monotone
convergence theorem, while on the right side we obtain lim inf X fn dµ.
n→∞
IV
Real Analysis Homework: #7
Yingwei Wang
∗
Department of Mathematics, Purdue University, West Lafayette, IN, USA
1
Limits
Question: Compute the following limits and justify your calculation.
R∞
−n
(i) lim 0 1 + nx
sin nx dx.
n→∞
−n
Solution: Let fn (x) = 1 + nx
sin nx , then |fn (x)| ≤ e−x , ∀x ∈ [0, ∞). Besides, lim fn (x) =
n→∞
0. Then we have
Z ∞
lim
fn dx = 0.
n→∞ 0
(ii) lim
Rn
n→∞ 0
1−
x n
ex/2 dx.
n
Solution: Let fn (x) = 1 −
x n x/2
e χ[0,n] ,
n
f (x) = e−x/2 then lim fn (x) = f (x).
n→∞
Let gn = 2f − fn , then gn monotonic increasingly converges to f as n → ∞. Besides, when
n is sufficient large, gn ≥ 0. By Monotonic Convergence Theorem, we can see that
lim gn (x) = f (x).
n→∞
So we have
lim
Z
n→∞ 0
(ii) lim
Rn
n→∞ 0
1+
n
x n x/2
e dx = lim
1−
n→∞
n
x n −2x
e dx.
n
Solution: Let fn (x) = 1 +
f (x), ∀x ∈ [0, ∞).
So we have
lim
Z
n→∞ 0
∗
n
x n −2x
e χ[0,n] ,
n
Z
∞
fn dx =
0
Z
∞
f dx = 2.
0
f (x) = e−x then lim fn (x) = f (x) and fn (x) ≤
n→∞
x n −2x
1+
e dx = lim
n→∞
n
Z
E-mail address: wywshtj@gmail.com; Tel : 765 237 7149
I
∞
fn dx =
0
Z
∞
f dx = 1.
0
Yingwei Wang
2
Real Analysis
Limits
Question: LetR (X, A, µ) be measure space and let f : X → [0, +∞) be A-measurable. Suppose
that 0 < c = X f dµ < ∞. Show that

α Z
 +∞, if 0 < α < 1,
f
lim
n ln 1 +
dµ =
c,
if α = 1,
n→∞ X

n
0,
if 1 < α < ∞.
Proof: Let
α f
= n ln 1 +
n
α n
f
= ln 1 +
n
α !n1−α
fα n
= ln
1+ α
n
α!
α n
f
= n1−α · ln
1+ α
n
fn (x)
If 0 < α < 1, then
lim fn (x) = +∞ · f α = +∞.
n→∞
By Fatou’s Lemma, we have
Z
Z
Z
fn (x)dµ ≥ lim
fn (x)dµ ≥
lim
n→∞ X
n→∞ X
lim fn (x)dµ = +∞.
X n→∞
If α = 1, then
lim fn (x) = f (x).
n→∞
Besides, ∀x ∈ X, fn (x) ≤ f (x). By Lebesgue Dominated Control Theorem, we have
Z
Z
lim
fn (x)dµ =
f (x)dµ = c.
n→∞ X
X
If 1 < α < ∞, then
lim fn (x) = 0 · f α = 0.
n→∞
By Monotonic Convergence Theorem, we have
Z
Z
lim
fn (x)dµ =
0 dµ = 0.
n→∞ X
X
II
Yingwei Wang
3
Real Analysis
Limits
Question: Let (X, A, µ) be measure space. Let (fn )∞
n=1 be sequence of A-measurable functions, fnR : X → [0, +∞).
Suppose
that
lim
f
n→∞ n (x) =R f (x) for all x
R
R ∈ X and that
limn→∞ X fn (x)dµ = X f (x)dµ < ∞. Show that limn→∞ A fn (x)dµ = A f (x)dµ for every set A ∈ A.
Proof: For any set A ∈ A, define gn (x) = fn (x)χA , hn (x) = fn (x)χX\A , the we have
lim fn (x) = f (x),
n→∞
lim gn (x) = f (x)χA ,
n→∞
lim hn (x) = f (x)χX\A ,
n→∞
gn (x) + hn (x) = fn (x),
∀n and ∀x.
On one hand, by Fatou’s Lemma, we have
Z
Z
f χA dµ ≤ lim
gn dµ,
n→∞ X
Z
ZX
f χX\A dµ ≤ lim
hn dµ,
ZX
Zn→∞ X
Z
f dµ ≤ lim
gn dµ + lim
hn dµ.
⇒
n→∞ X
X
(3.1)
(3.2)
(3.3)
n→∞ X
On the other hand,
Z
Z
Z
Z
Z
Z
lim
gn dµ + lim
hn dµ = lim
gn dµ +
hn dµ ≤ lim
fn dµ =
f dµ. (3.4)
n→∞ X
n→∞ X
n→∞
X
n→∞ X
X
By (3.3) and (3.4), we can see that
Z
Z
Z
lim
gn dµ + lim
hn dµ =
f dµ.
n→∞ X
n→∞ X
n→∞ X
So we can get
⇒
Z
Z
fn (x)χA dµ =
f (x)χA dµ
n→∞ X
X
Z
Z
lim
fn (x)dµ =
f (x)dµ.
lim
n→∞ A
A
III
(3.5)
X
Then by (3.1), (3.2) and (3.5), we can see that
Z
Z
f χA dµ = lim
gn dµ,
n→∞ X
ZX
Z
f χX\A dµ = lim
hn dµ,
X
X
(3.6)
(3.7)
Yingwei Wang
4
Real Analysis
The absolute continuity of integral
Question: Let f : R → [0, ∞] be a Lebesgue integrable function. Show that for
R each α > 0
there is β > 0 such that if B ⊂ R is Lebesgue measurable and m(B) < β, then B f dm < α.
Proof: Since f is nonnegative and Lebesgue integrable, ∀α > 0, ∃ a nonnegative simple
function ϕ(x), such that
Z
Z
Z
α
f − ϕdm =
f dm −
ϕdm < .
2
R
R
R
α
Let ϕ(x) < M , then we can choose B ⊂ R s.t. m(B) < 2M
, then
Z
Z
Z
f dm
≤
f − ϕdm +
ϕdm
B
B
ZB
f − ϕdm + M · m(B)
≤
R
α
α
< +M ·
= α.
2
2M
IV
Real Analysis Homework: #8
Yingwei Wang
∗
Department of Mathematics, Purdue University, West Lafayette, IN, USA
Note: In this paper, {f (x) satisfies some property} = {x : f (x) satisfies some property}
1
Integral function
Question: Let f : (0, 1) → R be a Lebesgue integrable function. Suppose that
Z
f dm for all x ∈ (0, 1).
(1.1)
(0,x)
Show that f (x) = 0 a.e.
R1
Proof: First I want to claim that 0 f dm = 0. Let fn = f χ(0,1−1/n) , then |f χ(0,1/n) (x)| ≤
|f (x)|, ∀n ∈ N and x ∈ (0, 1) By Lebesgue dominated convergence theorem,
Z 1
Z
Z
f dm = lim
fn dm = lim
f dm = 0.
0
n→∞ (0,1)
n→∞ (0,1−1/n)
Second, let A = {f (x) > 0}. Suppose m(A) > 0, then ∃ closed set B ⊂ A, m(B) > 0.
Let C = [0, 1]\B, then C is an open set, i.e. C = ∪(an , bn ).
Z
Z
Z
f dm +
f dm =
f dm = 0,
B
C
(0,1)
Z
Z
⇒
f dm 6= 0, (since
f dm > 0)
C
B
Z
⇒
f dm 6= 0,
∪(an ,bn )
Z
⇒
∃ (an , bn ), s.t.
f dm 6= 0,
(an ,bn )
Z
Z
⇒
f dm −
f dm 6= 0,
(0,bn )
∗
(0,an )
E-mail address: wywshtj@gmail.com; Tel : 765 237 7149
I
Yingwei Wang
Real Analysis
which contradicts with (1.1).
So m(A) = 0. Similarly, we can prove that m({f (x) < 0}) = 0. That is to say f (x) = 0
a.e.
2
Convergence of L1 norm and pointwise
Question: Let (fn ) be a convergent sequence in L1 (X, A, µ) with limit f . Show that (fn )
has a subsequence (fnk ) such that fnk → f for almost x ∈ X.
Proof: Since (fn ) be a convergent sequence in L1 (X, A, µ), then (fn ) is a Cauchy sequence in L1 (X, A, µ), which means
Z
|fn − fm |dµ → 0, as n, m → ∞.
X
⇒ ∀ε > 0, ∃N , s.t. ∀n, m > N ,
m({|fn − fm | =
6 0}) < ε.
⇒ ∀i, ∃ki ∈ N s.t. ∀n, m > ki ,
m({|fn − fm | ≥
1
1
}) < i .
i
2
2
We can assume that k1 < k2 < · · · < ki < ki+1 < · · · , and define
Ei = {|fki − fki+1 | ≥
then m(Ei ) <
Let
1
},
2i
1
.
2i
E=
∞ [
∞
\
Ei ,
j=1 i=j
then m(E) = 0.
∀x ∈ X\E, ∃j s.t.
x∈X
∞
[
Ei .
i=j
Then ∀i ≥ j,
|fki +1 (x) − fki (x)| <
II
1
,
2i
Yingwei Wang
Real Analysis
P
P
which means
|fki +1 (x) − fki (x)| is convergent, so fk1 + (fki +1 (x) − fki (x)) is also
convergent.
Denote lim fki (x) = g(x), ∀x ∈ X\E, then |fnk − g| → 0, as n → ∞, ∀x ∈ X\E. So
ki →∞
Z
|fnk − g|dµ =
X
⇒
Z
|fnk − g|dµ → 0,
X\E
kfnk − gkL1 → 0.
Since kfnk − f kL1 → 0, by the completeness of L1 (X, A, µ), f = g for almost all x ∈ X.
So we have a subsequence (fki ) which is convergent to f for almost all x ∈ X.
3
Integral functions and continuous functions
Question: Let f : [0, 1] → R be a Lebesgue integrable function. Show that there is a
sequence of continuous functions fn : [0, 1] → R which converges pointwise to f almost
everywhere.
Proof: First, I want to introduce the Lusin’s Theorem (Problem 3.31 on Page 74 in
Royden’s book):
Theorem 3.1 (Lusin). Let f be a measurable real-valued function on an interval [a, b].
Then given δ, there is a continuous function ϕ on [a, b] such that m({f 6= ϕ}) < δ.
By this theorem, for f ∈ L1 ([0, 1]), ∀n ∈ N, ∃ a sequence of continuous functions
ϕn ∈ C([0, 1]) s.t.
1
m({ϕn 6= f }) < n .
2
Then we have
Z
X
|ϕn − f |dm =
Z
|ϕn − f |dm → 0, as n → ∞,
{ϕn 6=f }
since m({δn 6= f }) → 0 as n → ∞.
It implies that ϕn converges to f in the L1 norm. The conclusion of the last problem
tells us that (ϕn ) has a subsequence (ϕnk ) such that ϕnk → f for almost x ∈ X.
III
Real Analysis Homework: #9
Yingwei Wang
∗
Department of Mathematics, Purdue University, West Lafayette, IN, USA
1
Absolutely continuous
Question: Let f, g : [a, b] → R be two absolutely continuous functions. Prove or disprove
that h(x) = ef (x)g(x) is absolutely continuous on [a, b].
Proof: First, I want to show that if both f and g are absolutely continuous, then f g is
also absolutely continuous.
There exists M > 0 such that |f (x)| ≤ M and |g(x)| ≤ M for ∀x ∈ [a, b]. Given ε > 0,
there exits δ > 0 such that
n
X
i=1
n
X
ε
ε
|f (xi ) − f (yi )| <
, and
|g(xi ) − g(yi )| <
,
2M
2M
i=1
for any finite collection {(xi , yi )} of disjoint intervals in [a, b] with
Then
n
X
≤
i=1 |xi
− yi | < δ.
|f (xi )g(xi ) − f (yi )g(yi )|
i=1
n
X
|f (xi )||g(xi ) − g(yi )| +
i=1
<
Pn
n
X
|g(yi )||f (xi ) − f (yi )|
i=1
ε ε
+ = ε.
2 2
Second, I want to show that if q : [a, b] → [a′ , b′ ] is absolutely continuous on [a, b] and
p : [a′ , b′ ] → R is Lipschitz continuous on [a′ , b′ ], then p(q(x)) is absolutely continuous on
[a, b].
∗
E-mail address: wywshtj@gmail.com; Tel : 765 237 7149
I
Yingwei Wang
Real Analysis
There exists N > 0 such that for ∀x, y ∈ [a′ , b′ ],
|p(x) − p(y)| ≤ N |x − y|.
Given ε, there exits δ > 0 such that
n
X
|q(xi ) − q(yi )| <
i=1
ε
N
for any finite collection {(xi , yi )} of disjoint intervals in [a, b] with
Then
n
X
|p(q(xi )) − p(q(yi ))|
Pn
i=1 |xi
− yi | < δ.
i=1
≤N
n
X
|q(xi ) − q(yi )|
i=1
< ε.
Finally, since ex is Lipschitz continuous in any finite interval [a′ , b′ ], so h = ef g is
absolutely continuous if both f and g are absolutely continuous.
2
Bounded variation
Rb
Question: Let f ∈ BV [a, b]. Show that Tab f ≥ a |f ′ (x)|dx.
Proof: By Jordan Theorem, f (x) can be written as the difference of two monotone
increasing functions on [a, b],
f (x) = g(x) − h(x),
where
1
g(x) = (Tax f + f (x)),
2
1
h(x) = (Tax f − f (x)).
2
It is easy to verify that both g and h are increasing, so g′ and h′ are nonnegative.
By Lebesgue Theorem,
Z b
g′ (x)dx ≤ g(b) − g(a),
a
Z b
h′ (x)dx ≤ h(b) − h(a).
a
II
Yingwei Wang
Real Analysis
Since f ′ = g′ − h′ , |f ′ | ≤ g′ + h′ . Then we have
Z
3
b
′
|f (x)|dx ≤
a
Z
a
b
g′ (x) + h(x)dx ≤ g(b) − g(a) + h(b) − h(a) = Tab f.
Bounded variation
Question: Let f ∈ BV [a, b]. Show that f has at most countably many points of discontinuity.
Proof: First, I want to show that ∀x0 ∈ [a, b], both limx→x+ f (x) and limx→x− f (x)
0
0
exist.
By Jordan Theorem, f = g − h where g and h are monotone increasing functions on
[a, b]. Let G = supx∈[a,x0 ) g(x) and H = supx∈[a,x0 ) h(x). It is obvious that A, B < ∞.
Given ε > 0, there exists δ > 0 such that
ε
< g(x0 − δ) ≤ G,
2
ε
H − < h(x0 − δ) ≤ H.
2
G−
Then for ∀x ∈ (x0 − δ, x0 ), we have
ε
ε
< g(x) ≤ G ⇒ 0 ≤ G − g(x) < ,
2
2
ε
ε
H − < h(x) ≤ H ⇒ 0 ≤ H − h(x) < .
2
2
G−
It follows that
|G − H − f (x)| ≤ (G − g(x)) + (H − h(x)) < ε,
for ∀x ∈ (x0 − δ, x0 ). So limx→x+ f (x) = G − H. Similarly we can know that limx→x− f (x)
0
0
also exists.
Second, I want to show that the set of discontinuities of f is at most countably.
Let E = {x ∈ [a, b] : f (x+) 6= f (x−)}, E1 = {x ∈ [a, b] : g(x+) > g(x−)}, E2 = {x ∈
[a, b] : h(x+) > h(x−)}. Since g and h are monotone increasing and f = g − h, we have
E = E1 ∪ E2 .
For ∀x ∈ E1 , ∃rx ∈ Q such that g(x−) < rx < g(x+). Besides, if x1 < x2 , then
g(x1 +) ≤ g(x2 −) so rx1 6= rx2 . Thus x → rx is a bijection between E1 and a subset of Q,
which means E1 is at most countably.
Similarly, E2 is at most countably and further E = E1 ∪ E2 is also at most countably.
III
Yingwei Wang
4
Real Analysis
L1 space
Question: Let (fn ) be a sequence of nonnegative functions in L[0, 1] such that
R1
/ L1 [0, 1].
and 1/n dt ≤ 1/n for all n ≥ 1. If g(x) = supn fn (x) show that g ∈
R1
0
fn (t)dt
Proof: Suppose that g ∈ L1 [0, 1], I want to get a contradiction.
By the absolutely continuity of the integral, for ε = 13 , there exists δ such that for any
measurable subset E ⊂ [0, 1], if m(E) < δ, then
Z
1
gdm < .
(4.1)
3
E
Choosing an integer n satisfying n ≥ 3 and 1/n < δ, then since
1/n, we have
Z 1
Z 1/n
fn dt = 1 −
fn dt ≥ 1 − 1/n,
0
R1
0
fn (t)dt and
R1
1/n
dt ≤
1/n
for all n > 3.
By the definition of g, we have
Z
1/n
0
But by (4.1), we have
gdt ≥
Z
0
1/n
2
fn dt ≥ 1 − 1/n ≥ 1 − 1/3 ≥ .
3
(4.2)
Z
(4.3)
1/n
0
1
gdt < .
3
It is obvious that (4.2) contradicts with (4.3).
IV
Real Analysis Homework: #10
Yingwei Wang
∗
Department of Mathematics, Purdue University, West Lafayette, IN, USA
1
Convergence series
Question: Let (an )∞
n=1 be sequence of non-negative numbers such that
∞
X
n=1
an < ∞.
(1.1)
Let r1 , r2 , . . . be an enumeration of the rational numbers. Show that the series
∞
X
a2n
|x − rn |
n=1
(1.2)
converges for almost all x ∈ R.
Proof: I found two methods to prove this. Hopefully, both of them are correct.
1.1
Method one: integral
Consider the new series:
∞
X
n=1
∗
p
an
|x − rn |
If (1.3) converges, then (1.2) is sure to converge.
If |x − rn | ≥ 1 for some n, then
a
p n
≤ an .
|x − rn |
E-mail address: wywshtj@gmail.com; Tel : 765 237 7149
I
(1.3)
(1.4)
Yingwei Wang
Real Analysis
So the convergence of the series is obvious.
If |x − rn | < 1 for some n, then rn − 1 < x < rn + 1. Consider the integral
Z
rn +1
rn −1
=
n=1
|x − rn |
dx
Z rn +1
1
1
√
√
an
dx
dx +
rn − x
x − rn
rn −1
rn
Z 1
Z 1
1
1
√ dt +
√ dt
an
t
t
0
0
4an ,
=
P∞
rn
Z
=
which means
an
p
√ an
|x−rn |
(1.5)
is convergent almost everywhere for x ∈ (rn − 1, rn + 1).
In sum, (1.3) converges for x ∈ R, a.e. and so does (1.2).
1.2
Method two: measure
Consider the set
Enε = {x : |x − rn | < εan },
(1.6)
m(Enε ) = 2εan .
(1.7)
then the measure of this set is
For ∀x ∈ R\ ∪ Enε , we have
⇒
⇒
As ε → 0, m(∪Enε ) → 0. Hence,
an
≤
|x − rn |
a2n
≤
|x − rn |
∞
X
a2
1
ε
an
ε
∞
1X
≤
an .
|x − rn |
ε
n
n=1
a2n
n=1 |x−rn |
P∞
II
n=1
converges for almost all x ∈ R.
(1.8)
Yingwei Wang
2
Real Analysis
Bounded variation functions
Question: Find all functions f ∈ BV [0, 1] with the property that
f (x) + (T0x f )1/2 = 1, ∀x ∈ [0, 1].
(2.1)
Solution: Let x = 0 in Eqn.(2.1), we can get
f (0) = 1.
Since F (x) = (T0x f )1/2 is an increasing function, f (x) should be a decreasing function.
Hence, F (x) = (T0x f )1/2 = (f (0) − f (x))1/2 . So
⇒
⇒
⇒
3
f (x) + (f (0) − f (x))1/2 = 1,
(f (0) − f (x)) = (1 − f (x))2 ,
(1 − f (x)) = (1 − f (x))2 ,
f (x) = 1,
∀x ∈ [0, 1].
Absolutely continuous function
√
Question: Let f : [0, 1] → [0, 1], f (x) = x. Show that f ∈ AC[0, 1].
R
√
t
Proof: Since x = 12 0 t−1/2 dt, by the Absolutely continuity of the integral (Propo√
sition 4.14 in Page 88 of Royden’s book), it is easy to know that f (x) = x ∈ AC[0, 1].
4
Non-absolutely continuous function
Question: Construct f : [0, 1] → [0, 1], f ∈ AC[0, 1] such that
Solution: Let f (x) be as following
2 2 1
x sin ( x ), 0 < x ≤ 1,
f (x) =
0,
x = 0.
√
f∈
/ AC[0, 1].
(4.1)
It is easy to know that f (x) ∈ AC[0, 1] since f ′ (x) = 2x sin2 (1/x)−2 sin(1/x) cos(1/x)
is bounded in [0, 1].
III
Yingwei Wang
However, for
Real Analysis
√
f,
p
f (x) =
x sin( x1 ), 0 < x ≤ 1,
0,
x = 0,
we choose the partition
0<
1
1
1
1
1
1
<
<
< ··· <
< <
< 1.
nπ + π/2
nπ
(n − 1)π + π/2
π + π/2
π
π/2
Then
n p
n
X
X
p
2
1
1
| f (xk + 1) − f (xk )| = | sin 1 − | +
+
.
π
kπ + π/2 nπ + π/2
k=0
k=0
√
√
√
So T01 f = ∞ and f ∈
/ BV [0, 1] and further f ∈
/ AC[0, 1].
IV
(4.2)
Real Analysis Homework: #11
Yingwei Wang
∗
Department of Mathematics, Purdue University, West Lafayette, IN, USA
1
Open sets
Question: Show that every open subset U of R can be written uniquely as the union of a
countable family of mutually disjoint open intervals.
Proof: Let U ∈ R be open. For x1 ∈ U , consider the interval I1 = (a1 , b1 ) where
a1 = inf{t ∈ R : (t, x) ⊂ U },
b1 = sup{t ∈ R : (x, t) ⊂ U }.
Then for x2 = U \I1 , consider the interval I2 = (a2 , b2 ) where
a2 = inf{t ∈ R : (t, x) ⊂ U \I1 },
b2 = sup{t ∈ R : (x, t) ⊂ U \I1 }.
∞
In this way, we get a sequence of disjoint open intervals {In }∞
n=1 such that ∪n=1 In = U .
2
Measurable sets
Question: Let f : R → R be a continuous functions with the property that m∗ (f (A)) = 0
whenever m∗ (A) = 0, A ⊂ R. Show that if A ⊂ R is Lebesgue measurable, then f (A) is
Lebesgue measurable.
Proof: That A is measurable means ∀ε > 0, ∃ disjoint open intervals {In }∞
n=1 s.t.
A ⊂ ∪In and m(∪In − A) < ε.
P
Since m∗ (A) = inf{ l(In ) : A ⊂ ∪In }, we have m∗ (∪In ) → m∗ (A) and m∗ (∪In −A) →
0 as ε → 0.
By the assumption, m∗ (f (∪In − A)) → 0, so m(∪f (In ) − f (A)) < ε, for ∀ε > 0.
Since ∪f (In ) is an open set and f (A) ⊂ ∪f (In ), we know that f (A) is measurable.
∗
E-mail address: wywshtj@gmail.com; Tel : 765 237 7149
I
Yingwei Wang
3
Real Analysis
Absolutely continuity
Question: Let f : R → R ∈ AC[a, b] for all a < b, a, b ∈ R. Show that if A ⊂ R is Lebesgue
measurable, then f (A) is Lebesgue measurable.
Proof: It is suffices to show that f ∈ AC[a, b] for all a < b, a, b ∈ R implies that f has
the property that m(f (E)) = 0 whenever m(E) = 0, A ⊂ R.
That f is absolutely continuous means ∀ε > 0, ∃ δ such that for disjoint intervals
{Ii = (xi , yi )}ni=1 satisfying
n
X
(yi − xi ) < δ,
i=1
then
n
X
|f (yi ) − f (xi )| < ε.
i=1
Suppose E ⊂ R and m(E) = 0. Let G = ∪(xi , yi ) ⊃ E. Choose ci , di ∈ [xi , yi ] such that
f ([xi , yi ]) = [f (ci ), f (di )],
then
m(f (E)) ≤ m(f (G)) ≤
X
|f (di ) − f (ci )| < ε.
So m(f (E)) = 0.
By the conclusion of the Problem 2, we can know that if E ⊂ R is Lebesgue measurable,
then f (E) is Lebesgue measurable.
4
Lebesgue measure
Question: Let f ∈ AC[a, b] for all a < b where a, b ∈ R. Suppose that f is strictly
increasing. Show that if A is Lebesgue measurable then
Z
m(f (A)) =
f ′ (t)dt.
(4.1)
A
Proof: Case 1: A is an open interval A = (a0 , b0 ).
On one hand, since f is strictly increasing, m(f (A)) = f (b0 ) − f (a0 ); on the other hand,
R ′
R b0 ′
(4.1) holds.
A f (t)dt = a0 f (t)dt = f (b0 ) − f (a0 ). Thus,
F
Case 2: If A is an open set, then A = In where Ii ∩ Ij = ∅, i 6= j, In = (an , bn ).
II
Yingwei Wang
Real Analysis
On one hand,
m(f (A)) =
∞
X
(f (bn ) − f (an )) = f (b∞ ) − f (a1 ).
n=1
On the other hand,
Z
∞ Z
X
′
f (t)dt =
A
bn
′
f (t)dt =
n=1 an
∞
X
(f (bn ) − f (an )) = f (b∞ ) − f (a1 ).
n=1
It implies the (4.1).
Case 3: If A is a Gσ -set, it suffices to show that
Z
f ′ (x)dx ≤ m∗ (f (A)).
(4.2)
A
Choose open intervals In such that f (A) ⊂ ∪In , then E ⊂ ∪Jn , where Jn = f −1 (In ).
(n)
(n)
Choose sequences {αk } and {βk } such that
(n)
lim αk
= inf {x},
(n)
lim β
k→∞ k
= sup {x}.
x∈Jn
k→∞
x∈Jn
Then we get
Z
As a sequence,
Z
′
f (x)dx = lim
f (x)dx ≤
A
(n)
βk
k→∞ α(n)
k
Jn
′
Z
Z
′
f (x)dx ≤
f ′ (x)dx ≤ m(In ).
XZ
f ′ (x)dx ≤ m(In ).
Jn
∪Jn
By the definition of outer measure, we can get (4.2).
5
Signed measure
Question: Let µ = µ+ − µ− be the Jordan decomposition of a signed measure µ defined on
a measurable space (X, A). Define the total variation of µ to be the measure |µ| = µ+ + µ− .
Show that for each A ∈ A,
( n
)
X
|µ|(A) = sup
|µ(Ai )| : (Ai )ni is a partition of A with Ai ∈ A, n ≥ 1 .
(5.1)
i=1
III
Yingwei Wang
Real Analysis
Proof: Let E, F be a Hahn decomposition for µ. That is to say X = E ∪ F , E ∩ F = ∅,
and E is positive while F is negative respect to µ.
On one hand, we know that
µ+ A = µ(A ∩ E),
µ− A = −µ(A ∩ F ),
⇒
|µ|(A) = µ(A ∩ E) − µ(A ∩ F ).
On the other hand,
Z
Z
(χA∩E − χA∩F ) dµ =
A
1dµ −
A∩E
Z
(5.2)
1dµ = µ(A ∩ E) − µ(A ∩ F ).
(5.3)
A∩F
By Eqn.(5.2) and Eqn.(5.3), we know that
Z
|µ|(A)
=
(χA∩E − χA∩F ) dµ
A
=
n Z
X
i=1
(χAi ∩E − χAi ∩F ) dµ,
Ai
where (Ai )ni=1 is a partition of A with Ai ∈ A.
R
Since for each Ai , Ai (χAi ∩E − χAi ∩F ) dµ ≥ |µ(Ai )|, so we have
|µ|(A) =
n Z
X
i=1
(χAi ∩E − χAi ∩F ) dµ ≥ sup
Ai
( n
X
)
|µ(Ai )| ,
i=1
(5.4)
for any partition (Ai )ni=1 .
Besides, if we choose A1 = A ∩ E and A2 = A ∩ F , then
|µ|(A) = µ(A1 ) − µ(A2 ) = |µ(A1 )| + |µ(A2 )|,
which means
sup
( n
X
)
|µ(Ai )|
i=1
≥ |µ|(A).
By Eqn.(5.4) and (5.5), we can get the conclusion (5.1).
IV
(5.5)