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 ∈ QQ, p < q}.

It follows that

f −1(G) = ∪{f −1((p, q)) : p, q ∈ QQ, p < q, (p, q) ⊂ G}.

For each pair of p, q ∈ QQwith 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 −1 −1 ∞ f (IR \ A) = X \ f (A) ∈ S. Moreover, if Ak ∈ A for k ∈ IN and A = ∪k=1Ak, −1 ∞ −1 then f (A) = ∪k=1f (Ak). This verifies that A is a σ-algebra. By part (a), A 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 infn→∞ 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 mea-

surable 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=1En) = 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 µ-

null set K such that limn→∞ fn(x) = f(x) ∈ IR for all x ∈ X \ K. We claim that ∞ ∩n=1En ⊆ 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=1En ⊆ K. Since µ(K) = 0, it follows that ∞ µ(∩n=1En) = 0.

(b) We have En ⊇ En+1 for every n ∈ IN. Since µ(x) < ∞, we obtain

∞ lim µ(En) = µ(∩n=1En) = 0. n→∞