9.15. MULTIFUNCTIONS AND THEIR MEASURABILITY 271

9.15 Multifunctions and Their MeasurabilityThis is an introduction to the idea of measurable multifunctions. This is a very importanttopic which has surprising usefulness in nonlinear analysis and other areas and not enoughattention is payed to it. As an application, I will give a proof of Kuratowski’s theorem andalso an interesting fixed point result in which the fixed point is a measurable function of ω

in a measure space. One of the main references for this material is the book Papageorgiuand Hu [31] where you can find more of this kind of thing.

9.15.1 The General CaseLet X be a separable complete metric space and let (Ω,F ) be a set and a σ algebra ofsubsets of Ω. A multifunction, is a map from Ω to the nonempty subsets of X . Thus Γ is amultifunction if for each ω, Γ(ω) ̸= /0. For more on the theorems presented in this section,see [31].

Definition 9.15.1 Define Γ− (S)≡ {ω ∈Ω : Γ(ω)∩S ̸= /0} . When

Γ− (U) ∈F

for all U open, we say that Γ is measurable.

More can be said than what follows, but the following is the essential idea for a mea-surable multifunction.

Theorem 9.15.2 The following are equivalent for any measurable space consistingonly of a set Ω and a σ algebra F . Here nothing is known about Γ(ω) other than that isa nonempty set.

1. For all U open in X ,Γ− (U) ∈F where Γ− (U)≡ {ω : Γ(ω)∩U ̸= /0}

2. There exists a sequence, {σn} of measurable functions satisfying σn (ω) ∈ Γ(ω)such that for all ω ∈ Ω,Γ(ω) = {σn (ω) : n ∈ N}. These functions are called mea-surable selections.

Proof: First 1.) ⇒ 2.). A measurable selection will be obtained in Γ(ω). Let D ≡{xn}∞

n=1 be a countable dense subset of X . For ω ∈ Ω, let ψ1 (ω) = xn where n is thesmallest integer such that Γ(ω)∩ B(xn,1) ̸= /0. Therefore, ψ1 (ω) has countably manyvalues, xn1 ,xn2 , · · · where n1 < n2 < · · · . Now the set on which ψ1 has the value xn is asfollows: {ω : ψ1 = xn}=

{ω : Γ(ω)∩B(xn,1) ̸= /0}∩ [Ω\∪k<n {ω : Γ(ω)∩B(xk,1) ̸= /0}] ∈F .

Thus ψ1 is measurable and dist(ψ1 (ω) ,Γ(ω)) < 1. Let Ωn ≡ {ω ∈Ω : ψ1 (ω) = xn} .Then Ωn ∈F and Ωn ∩Ωm = /0 for n ̸= m and ∪∞

n=1Ωn = Ω because if ω is given, Γ(ω)does intersect some B(xn,1) . Let

Dn ≡ {xk ∈ D : 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))<12, d (ψ2 (ω) ,ψ1 (ω))< 1. (9.25)