10.9. SOME IMPORTANT GENERAL THEORY 291

Proof: First consider a measurable set E where µ (E) < ∞. Let K ⊆ E ⊆ V whereµ (V \K)< ε. Now let K ≺ h≺V. Then∫

|h−XE |dµ ≤∫

XV\Kdµ = µ (V \K)< ε. (10.9)

By Corollary 10.7.7, there is a sequence of simple functions converging pointwise to fsuch that for m,n > N, ε

2 >∫(|sn− sm|)dµ. Then let n→ ∞ and apply the dominated

convergence theorem to get∫| f − sm|dµ ≤ ε

2 . However, from 10.9, there is g in Cc (Ω)such that

∫|sm−g|dµ < ε

2 and so∫| f −g|dµ ≤

∫| f − sm|dµ +

∫|sm−g|dµ < ε. ■

10.9 Some Important General Theory10.9.1 Eggoroff’s TheoremYou might show that a sequence of measurable real or complex valued functions convergeson a measurable set. This is Proposition 9.1.8 above. Eggoroff’s theorem says that if theset of points where a sequence of measurable functions converges is all but a set of measurezero, then the sequence almost converges uniformly in a certain sense.

Theorem 10.9.1 (Egoroff) Let (Ω,F ,µ) be a finite measure space, µ (Ω)< ∞ andlet fn, f be complex valued functions such that Re fn, Im fn are all measurable and

limn→∞

fn(ω) = f (ω)

for all ω /∈ E where µ(E) = 0. Then for every ε > 0, there exists a set,

F ⊇ E, µ(F)< ε,

such that fn converges uniformly to f on FC.

Proof: First suppose E = /0 so that convergence is pointwise everywhere. It followsthen that Re f and Im f are pointwise limits of measurable functions and are thereforemeasurable. Let Ekm = {ω ∈Ω : | fn(ω)− f (ω)| ≥ 1/m for some n > k}. Note that

| fn (ω)− f (ω)|=√

(Re fn (ω)−Re f (ω))2 +(Im fn (ω)− Im f (ω))2

and so,[| fn− f | ≥ 1

m

]is measurable. Hence Ekm is measurable because

Ekm = ∪∞n=k+1

[| fn− f | ≥ 1

m

].

For fixed m,∩∞k=1Ekm = /0 because fn converges to f . Therefore, if ω ∈ Ω there exists

k such that if n > k, | fn (ω)− f (ω)| < 1m which means ω /∈ Ekm. Note also that Ekm ⊇

E(k+1)m. Since µ(E1m)< ∞, Theorem 9.2.4 on Page 242 implies

0 = µ(∩∞k=1Ekm) = lim

k→∞µ(Ekm).

Let k(m) be chosen such that µ(Ek(m)m) < ε2−m and let F = ∪∞m=1Ek(m)m. Then µ(F) <

ε because µ (F)≤ ∑∞m=1 µ

(Ek(m)m

)< ∑

∞m=1 ε2−m = ε.