228 CHAPTER 11. ABSTRACT MEASURE AND INTEGRATION

Thus, by continuity in the first entry,

supx∈E

ψ (x,ω) ≥ ψ ( f (ω) ,ω)≥ limn→∞

ψ (sn (ω) ,ω)

≥ supn

supx∈Cn

ψ (x,ω) = supx∈∪nCn

ψ (x,ω) = supx∈E

ψ (x,ω)

Theorem 11.1.11 Let E be a compact metric space and let (Ω,F ) be a measure space.Suppose ψ : E×Ω→R has the property that x→ψ (x,ω) is continuous and ω→ψ (x,ω)is measurable. Then there exists a measurable function, f having values in E such that

ψ ( f (ω) ,ω) = supx∈E

ψ (x,ω) .

Furthermore, ω → ψ ( f (ω) ,ω) is measurable.

Proof: Let C1 be a 2−1 net of E. Suppose C1, · · · ,Cm have been chosen such that Ck isa 2−k net and Ci+1 ⊇Ci for all i. Then consider E \∪

{B(

x,2−(m+1))

: x ∈Cm

}. If this set

is empty, let Cm+1 =Cm. If it is nonempty, let {yi}ri=1 be a 2−(m+1) net for this compact set.

Then let Cm+1 =Cm∪{yi}ri=1 . It follows {Cm}∞

m=1 satisfies Cm is a 2−m net and Cm ⊆Cm+1.

Let{

x1k

}m(1)k=1 equal C1. Let

A11 ≡

{ω : ψ

(x1

1,ω)= max

kψ(x1

k ,ω)}

For ω ∈ A11, define s1 (ω)≡ x1

1. Next let

A12 ≡

{ω /∈ A1

1 : ψ(x1

2,ω)= max

kψ(x1

k ,ω)}

and let s1 (ω)≡ x12 on A1

2. Continue in this way to obtain a simple function, s1 such that

ψ (s1 (ω) ,ω) = max{ψ (x,ω) : x ∈C1}

and s1 has values in C1.Suppose s1 (ω) ,s2 (ω) , · · · ,sm (ω) are simple functions with the property that if m > 1,

d (sk (ω) ,sk+1 (ω)) < 2−k,

ψ (sk (ω) ,ω) = max{ψ (x,ω) : x ∈Ck}sk has values in Ck

for each k+ 1 ≤ m, only the second and third assertions holding if m = 1. Letting Cm ={xk}N

k=1 , it follows sm (ω) is of the form

sm (ω) =N

∑k=1

xkXAk (ω) , Ai∩A j = /0. (11.1.6)