MATH 417
Assignment #6
1. Let (X, S) be a measurable space, and let f : X → IR be a measurable function.
(a) Show that |f |p is a measurable function for all p > 0.
(b) If f (x) 6= 0 for each x ∈ X, then 1/f is a measurable function.
Proof . (a) Note that |f | is measurable. For a < 0 we have {x ∈ X : |f (x)|p > a} = X;
for a ≥ 0 we have {x ∈ X : |f (x)|p > a} = {x ∈ X : |f (x)| > a1/p }. Hence, for
every a ∈ IR, the set {x ∈ X : |f (x)|p > a} is measurable. This shows that |f |p is
measurable.
(b) If a > 0, then {x ∈ X : 1/f (x) > a} = {x ∈ X : 0 < f (x) < 1/a}; if a = 0, then
{x ∈ X : 1/f (x) > a} = {x ∈ X : f (x) > 0}; if a < 0, then
{x ∈ X : 1/f (x) > a} = {x ∈ X : f (x) > 0} ∪ {x ∈ X : f (x) < 1/a}.
Hence, for every a ∈ IR, the set {x ∈ X : 1/f (x) > a} is measurable. This shows that
1/f is measurable.
2. Let (X, S) be a measurable space, and let f : X → IR be a measurable function.
(a) Show that f −1 (G) ∈ S for every open subset G of IR.
(b) Show that f −1 (B) ∈ S for every Borel subset B of IR.
(c) If φ is a continuous function from IR to IR, then φ ◦ f is measurable.
Proof . (a) Suppose that G is an open subset of IR. Then
G = ∪{(p, q) ⊂ G : p, q ∈ Q
Q, p < q}.
It follows that
f −1 (G) = ∪{f −1 ((p, q)) : p, q ∈ Q
Q, p < q, (p, q) ⊂ G}.
For each pair of p, q ∈ Q
Q with p < q, we have f −1 ((p, q)) ∈ S. As a countable union
of measurable sets, f −1 (G) ∈ S.
(b) Let A be the collection of all subsets A of IR such that f −1 (A) ∈ S. Then
A is a σ-algebra. Indeed, ∅ ∈ A. If A ∈ A, then f −1 (A) ∈ S. It follows that
f −1 (IR \ A) = X \ f −1 (A) ∈ S. Moreover, if Ak ∈ A for k ∈ IN and A = ∪∞
k=1 Ak ,
−1
then f −1 (A) = ∪∞
(Ak ). This verifies that A is a σ-algebra. By part (a), A
k=1 f
contains all open subsets of IR. Therefore, A contains all Borel subsets of IR. Thus,
f −1 (B) ∈ S for every Borel subset B of IR.
(c) Let φ be a continuous function from IR to IR. For each a ∈ IR, we observe that
{x ∈ X : φ ◦ f (x) > a} = f −1 (φ−1 ((a, ∞))). Since φ is continuous, φ−1 ((a, ∞)) is
an open subset of IR. By part (a), f −1 (φ−1 ((a, ∞))) is measurable. This shows that
φ ◦ f is measurable.
3. Let (X, S) be a measurable space, and let (fn )n=1,2,... be a sequence of measurable
functions from X to IR.
(a) Show that the set {x ∈ X : limn→∞ fn (x) = ∞} is measurable.
(b) Let A be the set of those elements x ∈ X such that limn→∞ fn (x) exists as a real
number. Show that A is measurable.
Proof . Let g := lim inf n→∞ fn and h := lim supn→∞ fn . Then g and h are measurable.
(a) We have
{x ∈ X : lim fn (x) = ∞} = {x ∈ X : g(x) = h(x)} ∩ {x ∈ X : h(x) = ∞}.
n→∞
Hence, the set {x ∈ X : limn→∞ fn (x) = ∞} is measurable.
(b) We have
A = {x ∈ X : g(x) = h(x)} ∩ {x ∈ X : −∞ < h(x) < ∞}.
Hence, A is measurable.
4. Let X be a nonempty set and f a function from X to IR. Suppose that (fn )n=1,2,...
is a sequence of functions from X to IR. Let φ be a function from IR to IR.
(a) If the sequence (fn )n=1,2,... converges to f uniformly on X, and if φ is uniformly
continuous, then the sequence (φ ◦ fn )n=1,2,... converges to φ ◦ f uniformly on X.
(b) Suppose in addition that (X, S, µ) is a complete measure space. Moreover, assume
that fn (n = 1, 2, . . .) and f are measurable functions. If the sequence (fn )n=1,2,...
converges to f almost everywhere, and if φ is continuous, then the sequence
(φ ◦ fn )n=1,2,... converges to φ ◦ f almost everywhere.
Proof . (a) Let ε > 0 be given. Since φ is uniformly continuous, there exists δ > 0 such
that |φ(y) − φ(z)| < ε whenever |y − z| < δ. Since the sequence (fn )n=1,2,... converges
to f uniformly on X, there exists a positive integer N such that |fn (x) − f (x)| < δ
whenever n > N and x ∈ X. Consequently, for n > N , |φ ◦ fn (x) − φ ◦ f (x)| < ε for
all x ∈ X. This shows that (φ ◦ fn )n=1,2,... converges to φ ◦ f uniformly on X.
(b) Since (fn )n=1,2,... converges to f almost everywhere, there exists a µ-null set E such
that limn→∞ fn (x) = f (x) ∈ IR for all x ∈ X \ E. Fix an element x ∈ X \ E. For any
ε > 0, there exists some δ > 0 such that |φ(y) − φ(f (x))| < ε whenever |y − f (x)| < δ.
For this δ, there exists a positive integer N such that |fn (x) − f (x)| < δ for all n > N .
Thus, |φ◦fn (x)−φ◦f (x)| < ε for all n > N . This shows that (φ◦fn )n=1,2,... converges
to φ ◦ f on X \ E.
5. Let (X, S, µ) be a complete measure space with µ(X) < ∞. Suppose that f is a measurable function from X to IR and (fn )n=1,2,... is a sequence of measurable functions
from X to IR. Let ε > 0 be fixed. For n ∈ IN, let
En := ∪∞
m=n {x ∈ X : |fm (x) − f (x)| ≥ ε}.
(a) if (fn )n=1,2,... converges to f almost everywhere, then µ(∩∞
n=1 En ) = 0.
(b) Under the above conditions, show that limn→∞ µ(En ) = 0.
Proof . (a) Since (fn )n=1,2,... converges to f almost everywhere, there exists a µnull set K such that limn→∞ fn (x) = f (x) ∈ IR for all x ∈ X \ K. We claim that
∩∞
n=1 En ⊆ K. Indeed, if x ∈ X \ K, then limn→∞ fn (x) = f (x) ∈ IR. For given
ε > 0, there exists n0 ∈ IN such that |fm (x) − f (x)| < ε for all m ≥ n0 . Consequently,
x∈
/ En0 . This justifies our claim that ∩∞
n=1 En ⊆ K. Since µ(K) = 0, it follows that
∞
µ(∩n=1 En ) = 0.
(b) We have En ⊇ En+1 for every n ∈ IN. Since µ(x) < ∞, we obtain
lim µ(En ) = µ(∩∞
n=1 En ) = 0.
n→∞