48.4. A MEASURABLE KAKUTANI THEOREM 1565

48.4 A Measurable Kakutani TheoremRecall the Kakutani theorem, Theorem 25.4.4.

Theorem 48.4.1 Let K be a compact convex subset of Rn and let A : K→P (K) such thatAx is a closed convex subset of K and A is upper semicontinuous. Then there exists x suchthat x ∈ Ax. This is the “fixed point”.

Here is a measurable version of this theorem. It is just like the proof of the aboveBrowder lemma.

Theorem 48.4.2 Let K (ω) be compact, convex, and ω → K (ω) a measurable multifunc-tion. Let A(·,ω) : K (ω)→ K (ω) be upper semicontinuous, and let ω → A(x,ω) have ameasurable selection for each x ∈Rn. Then there exists x(ω) ∈ K (ω)∩A(K (ω) ,ω) suchthat ω → x(ω) is measurable.

Proof: Tile Rn with n simplices such that the collection is locally finite and each sim-plex has diameter less than ε < 1. This collection of simplices is determined by a countablecollection of vertices so there exists a one to one and onto map from N to the collectionof vertices. By assumption, for each vertex x, there exists Aε (x,ω) ∈ A

(PK(ω)x,ω

). By

Lemma 48.3.2, ω → PK(ω)x is measurable and by Theorem 48.3.1, there is a measurableselection for ω → A

(PK(ω)x,ω

)which is denoted as Aε (x,ω). By local finiteness, this

function is continuous in x on the set of vertices. Define Aε on all of Rn by the followingrule. If

x ∈ [x0, · · · ,xn],

so x =∑ni=0 tixi, then

Aε (x,ω)≡n

∑k=0

tkAε (xk,ω) .

By local finiteness, this function satisfies ω→Aε (x,ω) is measurable and also x→Aε (x,ε)is continuous. It also maps Rn to K (ω). By Corollary 48.2.14 there is a measurable fixedpoint xε (ω) satisfying xε (ω) ∈ K (ω) and Aε (xε (ω) ,ω) = xε (ω) .

Suppose xε (ω) ∈[xε

0 (ω) , · · · ,xεn (ω)

]so xε (ω) = ∑

nk=0 tε

k (ω)xεk (ω).

claim: The vertices xεk (ω) can be considered measurable also as is tε

k (ω).Proof of claim: Let the simplices in the tiling be {σ k}∞

k=1 and let the vertices of sim-plices in the tiling be

{z j}∞

j=1. Let

Fk := x−1ε (σ k) ,E1 := F1, · · · ,Ek := Fk \∪k

i=1Fi

Then ω is in exactly one of these measurable sets Ek. These measurable sets partition Ω.Let σ k (ω) be the unique simplex for ω ∈ Ek. Thus xε (ω) ∈ σ k (ω) on the measurableset Ek. Its vertices, are zi0 (ω) ,zi1 (ω) , · · · ,zin (ω) . These are xε

0 (ω) , · · · ,xεn (ω) in order.

They are determined in this way on a measurable set so they are measurable Rn valuedfunctions. Then ω → tε

k (ω) is also measurable because there is a continuous mapping tothese scalars from xε (ω) .