Math 317 HW #12 Solutions
1. Exercise 7.2.6. Let f : [a, b] → R be increasing on the set [a, b] (i.e., f (x) ≤ f (y) whenever
x < y). Show that f is integrable on [a, b].
Proof. First, note that, for all x ∈ [a, b],
f (a) ≤ f (x) ≤ f (b)
since f is increasing, so, if M = max{|f (a)|, |f (b)|}, we have that
|f (x)| ≤ M for all x ∈ [a, b],
so f is bounded.
Now, let > 0 and choose N > (f (b)−f (a))(b−a)
. Then, if I let xk = a + k(b−a)
for all
N
k = 0, 1, . . . , N , the set P = {a = x0 < x1 < x2 < · · · < xN −1 < xN = b} is a partition of
[a, b]. Now,
U (f, P ) − L(f, P ) =
N
X
(Mk − mk )(xk − xk−1 ) =
k=1
N
X
(Mk − mk )
k=1
b−a
.
N
(1)
Since mk ≥ Mk−1 for all k, meaning that Mk−1 − mk ≤ 0, we have
N
X
(Mk − mk ) = (M1 − m1 ) + (M2 − m2 ) + . . . + (MN − mN )
k=1
= −m1 + (M1 − m2 ) + (M2 − m3 ) + . . . + (MN −1 − mN ) + MN
≤ MN − m1
= f (b) − f (a),
since MN = f (b) and m1 = f (a). Combining this with (1) yields
N
b−a
b−aX
U (f, P ) − L(f, P ) =
(Mk − mk ) ≤
(f (b) − f (a)) < .
N
N
k=1
Since the choice of > 0 was arbitrary, we see that there is such a partition for every > 0,
so Theorem 7.2.8 implies that f is integrable on [a, b].
2. Exercise 7.3.5. Let
(
1
f (x) =
0
if x = 1/n for some n ∈ N
otherwise.
R1
Show that f is integrable on [0, 1] and compute 0 f .
1
Proof. Let > 0. Choose N ∈ N such that N > 2/. Then, for all n ≥ N , we have that
1
1/n < /2. Now, let δ = min{ 4N
, 4N
} and define the partition
1
1
1
1
1
1
P = 0, /2,
− δ,
+ δ,
− δ,
+ δ, . . . , − δ, + δ, 1 − δ, 1
N −1
N −1
N −2
N −2
2
2
1
− δ, so strictly speaking we
(note: for small k, it may be the case that N 1−k + δ < N −k+1
either need to pick δ more carefully to prevent this or re-arrange the elements of P so that
they’re correctly ordered; this can be done—e.g. by making sure δ is less than both N (N1−1)
and N 1−1 − 2 —but doing it carefully seems to me to be more distracting than worthwhile).
Now, sup(f ) = 1 on [0, /2] and on [ N 1−k − δ, N 1−k + δ] for all k ≤ N − 2, and sup(f ) = 0 on
all other subintervals determined by P , so
U (f, P ) = 1 ·
+ 0 + 1 · 2δ + 0 + 1 · 2δ + . . . + 0 + 1 · 2δ + 0 + 1 · δ
2
(the last term coming from the subinterval [1 − δ, 1], which has length δ). Thus
+
+ ... +
+
2 2N
2N
4N
N
−2
X
= +
+
2
2N
4N
U (f, P ) <
k=1
N −1
< +
2
N 2
< .
On the other hand, L(f, P ) = 0, since inf(f ) = 0 on any sub-interval of [0, 1]. Therefore,
U (f, P ) − L(f, P ) = U (f, P ) − 0 = U (f, P ) < .
Since the choice of > 0 was arbitrary, we see that the function f is integrable.
Since L(f, P ) = 0 for all partitions P , it must be the case that
Z 1
0 = L(f ) =
f.
0
3. Exercise 7.3.6. A set A ⊆ [a, b] has content zero if for every > 0 there exists a finite collection
of open intervals {O1 , O2 , . . . , ON } that contain A in their union and whose lengths sum to or less. Using |On | to refer to the length of each interval, we have
A⊆
N
[
On
and
n=1
N
X
|On | ≤ .
n=1
(a) Let f be bounded on [a, b]. Show that if the set of discontinuous points of f has content
zero, then f is integrable.
2
Proof. Since f is bounded on [a, b], there exists M > 0 such that |f (x)| ≤ M for all
x ∈ [a, b].
Let > 0. Let {O1 , O2 , . . . , ON } be open intervals covering Df such that
N
X
|On | ≤ /4M.
n=1
SN
The complement of n=1 On is a finite collection of closed intervals (and possibly singletons) on which f is continuous and, hence, integrable. Thus, there exists a partition
P1 of this complement such that
U (f, P1 ) − L(F, P1 ) < /2.
Construct the partition P from P1 by adding the (finitely many) endpoints of the On .
Then
U (f, P ) − L(f, P ) = (U (f, P1 ) − L(f, P1 )) +
≤ (U (f, P1 ) − L(f, P1 )) +
N
X
n=1
N
X
(Mn − mn )|On |
2M |On |
n=1
< /2 + 2M
N
X
|On |
n=1
≤ /2 + /2
= .
Hence, since the choice of > 0 was arbitrary, we see that such a partition exists for all
, so f is integrable.
(b) Show that any finite set has content zero.
Proof. If A = {x1 , . . . , xN } is finite, let > 0. Then, if Ok = (xk − /2N, xk + /2N ) for
each k = 1, 2, 3, . . ., we have that
A⊆
N
[
On
and
n=1
N
X
|On | =
n=1
N
X
= ,
N
n=1
so the finite set has content zero.
(c) Content zero sets do not have to be finite. They do not have to be countable. Show that
the Cantor set C defined in Section 3.1 has content zero.
Proof. Let > 0. Pick N ∈ N such that
2N −1
<
3N
3
(i.e. N >
ln(3)
ln(2/3)
+ 1) and then define δ such that
1
1
0<δ<
.
−
2 2N −1 3N
Then the set CN consists of a union of 2N −1 closed intervals [xk , yk ] such that yk − xk =
1
. For each k, let
3N
Uk = (xk − δ, yk + δ).
Then notice that Uk ⊇ [xk , yk ], so
C ⊆ CN ⊆
N −1
2[
Uk .
k=1
Also, the length of each Uk is
1
1
|Uk | = (yk + δ) − (xk − δ) = (yk − xk ) + 2δ = N + 2δ < N +
3
3
2N −1
1
− N
3
=
2N −1
.
Therefore,
N −1
2X
|Uk | <
N −1
2X
k=1
k=1
2N −1
= .
Since our choice of > 0 was arbitrary, we see that there exists such a collection of open
sets for any > 0, so C has content zero.
(d) Prove that
(
1
h(x) =
0
if x ∈ C
if x ∈
/C
is integrable, and find the value of the integral.
Proof. By HW #9, Problem 3, the function h is discontinuous precisely on the Cantor set
C. By part (c), the Cantor set has content zero, so part (a) implies that h is integrable.
In the proof of part (a), we saw that we can pick a partition so that the value of the
function near its discontinuities doesn’t matter, so we can ignore the value of h on C for
the purposes of
/ C, so it must be
R 1computing the integral. However, h(x) = 0 for all x ∈
the case that 0 h = 0.
4. Exercise 7.4.4. Decide which of the following conjectures is true and supply a short proof.
For those that are not true, give a counterexample.
(a) If |f | is integrable on [a, b] then f is also integrable on this set.
Answer. False. Define the modified Dirichlet function
(
1
if x ∈ Q
h(x) =
−1 otherwise
on [0, 1]. Then essentially the same argument as for the usual Dirichlet function implies
that h is not integrable, yet |h(x)| = 1 for all x, so |h| is definitely integrable.
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(b) Assume g is integrable and g ≥ 0 on [a, b]. If g(x) > 0 for an infinite number of points
Rb
x ∈ [a, b], then a g > 0.
Answer. False. The function f studied in Problem 2 above is a counterexample.
(c) If g is continuous on [a, b] and g ≥ 0 with g(x0 ) > 0 for at least one point x0 ∈ [a, b],
Rb
then a g > 0.
Answer. True. Since g is continuous and g(x0 ) > 0, there exists δ > 0 such that
|y − x0 | < δ implies that
g(x0 )
|g(y) − g(x0 )| <
.
2
Equivalently, when |y − x0 | < δ, we have that g(y) > g(x2 0 ) .
Now, define the function
(
g(x0 )
if |x − x0 | < δ
2
h(x) =
0
otherwise.
Rb
Then, a h = 2δ g(x2 0 ) = δg(x0 ) > 0, but h(x) ≤ g(x) for all x ∈ [a, b] so, by Theorem
7.4.2(iv),
Z b
Z b
g≥
h = δg(x0 ) > 0.
a
a
Rb
(d) If a f > 0, there is an interval [c, d] ⊆ [a, b] and a δ > 0 such that f (x) ≥ δ for all
x ∈ [c, d].
Answer. True. I’ll prove the contrapositive. Suppose that, for any [c, d] ⊆ [a, b], there
exists x ∈ [c, d] such that f (x) ≤ 0 (this is indeed the negation of the consequent since
[c, d] is a compact set). Then, for any partition
P = {a = x0 < x1 < x2 < · · · < xn−1 < xn = b}
we have that, for all k = 1, 2, . . . , n, there exists ck ∈ [xk−1 , xk ] such that f (ck ) ≤ 0.
Hence, the infimum mk of f over [xk−1 , xk ] is ≤ 0 for all k, so
L(f, P ) =
n
X
mk (xk − xk−1 ) ≤ 0.
k=1
Since the choice of P was arbitrary, this is true for all P , so
Z b
f = L(f ) = sup{L(f, P ) : P ∈ P} ≤ 0.
a
This proves the contrapositive of the desired statement.
5. Exercise 7.4.5. Let f and g be integrable functions on [a, b].
(a) Show that if P is any partition of [a, b], then
U (f + g, P ) ≤ U (f, P ) + U (g, P ).
Provide a specific example where the inequality is strict. What does the corresponding
inequality for lower sums look like?
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Proof. Let P = {a = x0 < x1 < x2 < · · · < xn−1 < xn = b} be a partition of [a, b]. For
each k = 1, 2, . . . , n, let Mk = sup{f (x) + g(x) : x ∈ [xk−1 , xk ]}, let Mk,f = sup{f (x) :
x ∈ [xk−1 , xk ]}, and let Mk,g = sup{g(x) : x ∈ [xk−1 , xk ]}. Then for any x ∈ [xk−1 , xk ],
f (x) + g(x) ≤ Mk,f + Mk,g ,
so Mk,f + Mk,g is an upper bound for the set {f (x) + g(x) : x ∈ [xk−1 , xk ]}. Therefore,
by definition of the supremum, Mk ≤ Mk,f + Mk,g . Hence,
U (f + g, P ) =
≤
=
n
X
k=1
n
X
k=1
n
X
Mk (xk − xk−1 )
(Mk,f + Mk,g )(xk − xk−1 )
Mk,f (xk − xk−1 ) +
k=1
n
X
Mk,g (xk − xk−1 )
k=1
= U (f, P ) + U (g, P ),
as desired.
For an example, let
(
0
f (x) =
1
if x 6= 1/2
if x = 1/2
and let g(x) = −f (x). Then f and g are both integrable on [0, 1] (by, for example,
Problem 3(a) above). Let P = {0, 1} be the trivial partition. Then, since f (x)+g(x) = 0
for all x ∈ [0, 1], U (f + g, P ) = 0. However,
U (f, P ) = 1
and U (g, P ) = 0,
so
U (f, P ) + U (g, P ) = 1
and the above inequality is strict.
The corresponding inequality for lower sums is
L(f + g, P ) ≥ L(f, P ) + L(g, P ).
(b) Review the proof of Theorem 7.4.2 (ii), and provide an argument for part (i) of this
theorem.
Proof. Note, first of all, that the two inequalities in (a) imply that, for any partition P ,
U (f + g, P ) − L(f + g, P ) ≤ (U (f, P ) + U (g, P )) − (L(f, P ) + L(g, P ))
= (U (f, P ) − L(f, P )) + (U (g, P ) − L(g, P )).
Since f is integrable on [a, b], there exist partitions Pn,f such that
lim [U (f, Pn,f ) − L(f, Pn,f )] = 0.
n→∞
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Likewise, since g is integrable, there exist partitions Pn,g such that
lim [U (g, Pn,g ) − L(g, Pn,g )] = 0.
n→∞
Now, for each n, define Pn = Pn,f ∪ Pn,g . Then, since Pn is a common refinement of Pn,f
and Pn,g , Lemma 7.2.3 implies that
L(f, Pn,f ) ≤ L(f, Pn ) ≤ U (f, Pn ) ≤ U (f, Pn,f )
and
L(g, Pn,g ) ≤ L(g, Pn ) ≤ U (g, Pn ) ≤ U (g, Pn,g ).
This combined with the reasoning at the start of the proof implies that
U (f + g, Pn ) − L(f + g, Pn ) ≤ (U (f, P ) − L(f, P )) + (U (g, P ) − L(g, P ))
≤ (U (f, Pn,f ) − L(f, Pn,f )) + (U (g, Pn,g ) − L(g, Pn,g )).
Since 0 ≤ U (f + g, Pn ) − L(f + g, Pn ), the Squeeze Theorem then implies that
lim [U (f + g, Pn ) − L(f + g, Pn )] = 0.
n→∞
Hence,
Rb
a
f + g = lim U (f + g, Pn ) = lim L(f + g, Pn ). But then
Z
L(f, Pn,f )+L(g, Pn,g ) ≤ L(f +g, Pn ) ≤
b
f +g ≤ U (f +g, Pn ) ≤ U (f, Pn,f )+U (g, Pn,g ).
a
Then the Order Limit Theorem tells us that
Z b
Z b
Z b
Z b
Z b
f+
g≤
f +g ≤
f+
g;
a
a
a
a
a
since the terms on the left and right are equal, the above inequalities are actually equalities, as desired.
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