232 CHAPTER 11. ABSTRACT MEASURE AND INTEGRATION

Definition 11.1.15 Let f : Ω→ [−∞,∞]. Define

[α < f ]≡ {ω ∈Ω : f (ω)> α} ≡ f−1 ((α,∞])

with obvious modifications for the symbols [α ≤ f ] , [α ≥ f ] , [α ≥ f ≥ β ], etc.

Definition 11.1.16 For a set E,

XE(ω) =

{1 if ω ∈ E,0 if ω /∈ E.

This is called the characteristic function of E. Sometimes this is called the indicatorfunction which I think is better terminology since the term characteristic function has an-other meaning. Note that this “indicates” whether a point, ω is contained in E. It is exactlywhen the function has the value 1.

Theorem 11.1.17 (Egoroff) Let (Ω,F ,µ) be a finite measure space,

(µ(Ω)< ∞)

and let 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, By Theorem 11.1.13, [| 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.