1530 CHAPTER 47. THE YANKOV VON NEUMANN AUMANN THEOREM

The set inclusion follows from the observation that if A ∈ B(E) and B ∈ B(X) then A×Bis in B(E×X) and the collection of sets in B(E)×B(X) which are in B(E×X) is a σ

algebra.Therefore, there exists D, a Borel set in E ×X such that f × idX (G) = D∩ (E×X) .

Now from this it follows from Lemma 47.0.7 that D is a Suslin space. Letting Y be {0,1}N,it follows that projY (D) is a Suslin space in Y. By Corollary 47.0.11, we see that projY (D)∈B̂(Y ). Now

projΩ (G) = {ω ∈Ω : there exists x ∈ X with (ω,x) ∈ G}

= {ω ∈Ω : there exists x ∈ X with ( f (ω) ,x) ∈ f × idX (G)}

= f−1 ({y ∈ Y : there exists x ∈ X with (y,x) ∈ D})

= f−1 (projY (D)) .

Now projY (D) ∈ B̂(Y ) and so Lemma 47.0.15 shows f−1 (projY (D)) ∈ Σ̂. This proves thelemma.

Now we are ready to prove the Yankov von Neumann Aumann projection theorem.First we must present another technical lemma.

Lemma 47.0.18 Let X be a Hausdorff space and let G ∈ Σ×B(X) where Σ is a σ algebraof sets of Ω. Then there exists Σ0 ⊆ Σ a countably generated σ algebra such that G ∈Σ0×B(X) .

Proof: First suppose G is a measurable rectangle, G = A×B where A ∈ Σ and B ∈B(X) . Letting Σ0 be the finite σ algebra,

{/0,A,AC,Ω

}, we see that G ∈ Σ0×B(X) . Simi-

larly, if G equals an elementary set, then the conclusion of the lemma holds for G. Let

F ≡ {H ∈ Σ×B(X) : H ∈ Σ0×B(X)}

for some countably generated σ algebra, Σ0. We just saw that F contains the elementarysets. If H ∈F , then HC ∈ Σ0×B(X) for the same Σ0 and so F is closed with respect tocomplements. Now suppose Hn ∈F . Then for each n, there exists a countably generatedσ algebra, Σ0n such that Hn ∈ Σ0n×B(X) . Then ∪∞

n=1Hn ∈ σ ({Σ0n×B(X)}) . We will bedone when we show

σ ({Σ0n×B(X)}∞

n=1)⊆ σ ({Σ0n}∞

n=1)×B(X)

because it is clear that σ ({Σ0n}∞

n=1) is countably generated. We see that

σ ({Σ0n×B(X)}∞

n=1)

is generated by sets of the form A×B where A ∈ Σ0n and B ∈ B(X) . But each such set isalso contained in σ ({Σ0n}∞

n=1)×B(X) and so the desired inclusion is obtained. Therefore,F is a σ algebra and so since F was shown to contain the measurable rectangles, thisverifies F = Σ×B(X) and this proves the lemma.