1534 CHAPTER 48. MULTIFUNCTIONS AND THEIR MEASURABILITY

Then Ωn ∈ C and Ωn∩Ωm = /0 for n ̸= m and ∪∞n=1Ωn = Ω. Let

Dn ≡ {xk : xk ∈ B(xn,1)} .

Now for each n, and ω ∈Ωn, let ψ2 (ω) = xk where k is the smallest index such that xk ∈Dnand B

(xk,

12

)∩Γ(ω) ̸= /0. Thus dist(ψ2 (ω) ,Γ(ω))< 1

2 and

d (ψ2 (ω) ,ψ1 (ω))< 1.

Continue this way obtaining ψk a measurable function such that

dist(ψk (ω) ,Γ(ω))<1

2k−1 , d(ψk (ω) ,ψk+1 (ω)

)<

12k−2 .

Then for each ω,{ψk (ω)} is a Cauchy sequence converging to a point, σ (ω) ∈ Γ(ω).

This has shown that if Γ is measurable, there exists a measurable selection, σ (ω) ∈ Γ(ω).Of course, if Γ(ω) is closed, then σ (ω) ∈ Γ(ω). Note that this had nothing to do with themeasure. It remains to show there exists a sequence of these measurable selections σn suchthat the conclusion of 4.) holds. To do this we define for Γ(ω) closed and measurable,

Γni (ω)≡{

Γ(ω)∩B(xn,2−i

)if Γ(ω)∩B

(xn,2−i

)̸= /0

Γ(ω) otherwise. .

Thus in the case of nonempty intersecton in the above cases,

Γ(ω)∩B(xn,2−(i+1)

)⊆ Γni (ω)⊆ Γ(ω)∩B(xn,2−i).

First we show that Γni is measurable. Let U be open. Then

{ω : Γni (ω)∩U ̸= /0}={

ω : Γ(ω)∩B(xn,2−i)∩U ̸= /0

}∪[{

ω : Γ(ω)∩B(xn,2−i)= /0

}∩{ω : Γ(ω)∩U ̸= /0}

]={

ω : Γ(ω)∩B(xn,2−i)∩U ̸= /0

}∪[(

Ω\{

ω : Γ(ω)∩B(xn,2−i) ̸= /0

})∩{ω : Γ(ω)∩U ̸= /0}

],

a measurable set. By what was just shown, there exists σni, a measurable function suchthat σni (ω) ∈ Γni (ω)⊆ Γ(ω) for all ω ∈Ω. If x ∈ Γ(ω) , then

x ∈ B(xn,2−(i+2)

)whenever xn is close enough to x. Thus both x,σn(i+2) (ω) are in B

(xn,2−(i+2)

)and so∣∣σn(i+1) (ω)− x

∣∣< 2−i. It follows that condition 4.) holds. Note that this had nothing to dowith the measure.

Now we verify that 4.) ⇒ 3.). Suppose there exist measurable selections σn (ω) ∈Γ(ω) satisfying condition 4.). Let U be open. Then

{ω : Γ(ω)∩U ̸= /0}= ∪∞n=1σ

−1n (U) ∈ C .