Math 318 HW #8 Solutions
1. (a) Prove Chebyshev’s inequality, which says that if f is nonnegative and measurable on the
bounded, measurable set A, then
Z
1
f dm.
m{x ∈ A : f (x) ≥ c} ≤
c A
Proof. Let c ∈ R and define
Ac := {x ∈ A : f (x) ≥ c},
which is measurable since f is a measurable function. Then f (x) ≥ c on Ac , so, by
Theorem 25.4(2),
Z
f dm ≥ cm{x ∈ A : f (x) ≥ c)}.
Ac
Therefore,
1
m{x ∈ A : f (x) ≥ c} ≤
c
Z
1
f dm ≤
c
Ac
Z
f dm,
A
where the last inequality follows from Theorem 25.3 and the fact that f is nonnegative.
(b) Show that if
R
A |f | dm
= 0, then f = 0 a.e. on A.
Proof. For any n ∈ N, Chebyshev’s inequality implies that
Z
m{x ∈ A : |f (x)| > 1/n} ≤ n
|f | dm = 0.
A
Letting Bn = {x ∈ A : |f (x)| > 1/n} and B = {x ∈ A : |f (x)| > 0}, we have that
B=
∞
[
Bn ,
n=1
and B1 ⊂ B2 ⊂ · · · . Therefore, Corollary 10.9 implies that
m(B) = lim m(Bn ) = lim 0 = 0.
n→∞
n→∞
But of course B = {x ∈ A : f (x) 6= 0}, so f = 0 a.e.
2. Exercise 26.24. If f is a measurable function such that
B ⊂ A, show that f = 0 a.e. on A.
R
B
Proof. Let
B+ := {x ∈ A : f (x) > 0}
B− := {x ∈ A : f (x) < 0}.
1
f dm = 0 for all measurable sets
Since f is measurable, both B+ and B− are measurable sets. Note that B+ ∪ B− = B, where
B = {x ∈ A : f (x) 6= 0}. The goal is to show that B has measure zero, so it suffices to show
that m(B+ ) = m(B− ) = 0.
Now, for each n ∈ N, define
B+n := {x ∈ A : f (x) > 1/n}
B−n := {x ∈ A : f (x) < −1/n},
which again are measurable sets since f is measurable. Hence,
Z
1
0=
f dm ≥ m(B+n )
n
B
Z +n
1
0=
f dm ≤ − m(B−n ),
n
B−n
which can only be true if m(B+n ) = m(B−n ) = 0.
Moreover,
B+ =
∞
[
B+n
and B− =
n=1
∞
[
B−n
n=1
and we have B+1 ⊂ B+2 ⊂ · · · and B−1 ⊂ B−2 ⊂ · · · . Therefore, by Corollary 10.9,
m(B+ ) = lim m(B+n ) = lim 0 = 0
n→∞
n→∞
m(B− ) = lim m(B−n ) = lim 0 = 0.
n→∞
n→∞
Thus, we conclude that
m(B) = m(B+ ) + m(B− ) = 0 + 0 = 0,
so f = 0 a.e.
3. Exercise 26.36. Let g ∈ L[a, b], h(x) =
[a, b].
R
[a,x] g dm
for x ∈ [a, b]. Show that h is continuous on
Proof. Let An := {x ∈ [a, b] : n − 1 ≤ |g(x)| < n}. Then the An are pairwise disjoint, so the
additivity of the Lebesgue integral implies
Z
∞ Z
X
|g| dm =
|g| dm.
[a,b]
n=1 An
Let > 0. Then the definition of convergence implies there exists N ∈ N such that
Z
∞
X
|g| dm < /2.
n=N +1 An
For
R notational convenience, define BN =
CN |g| dm < /2.
SN
n=1 An
2
and CN = [a, b]\BN ; the above says that
2N
< x ≤ c. Then
Z
Z
g dm
g dm −
|h(c) − h(x)| = [a,c]
[a,x]
Z
g dm
=
[x,c]
Z
|g| dm
≤
[x,c]
Z
Z
=
|g| dm +
|g| dm
[x,c]∩BN
[x,c]∩CN
Z
Z
≤
N dm +
|g| dm
Let x, c ∈ [a, b] such that c −
[x,c]∩BN
CN
<N
+
2N
2
= .
Similar reasoning shows that |h(x) − h(c)| < when c ≤ x < 2N
. Hence, we see that for any
x, c ∈ [a, b] such that |x − c| < δ = 2N , we have |h(x) − h(c)| < , so h is uniformly continuous
on [a, b].
3