Chapter 47

The Yankov von Neumann Aumann the-orem

The Yankov von Neumann Aumann theorem deals with the projection of a product measur-able set. It is a very difficult but interesting theorem. The material of this chapter is takenfrom [29], [30], [10], and [70]. We use the standard notation that for S and F σ algebras,S ×F is the σ algebra generated by the measurable rectangles, the product measure σ

algebra. The next result is fairly easy and the proof is left for the reader.

Lemma 47.0.1 Let (X ,d) be a metric space. Then if d1 (x,y) =d(x,y)

1+d(x,y) , it follows thatd1 is a metric on X and the basis of open balls taken with respect to d1 yields the sametopology as the basis of open balls taken with respect to d.

Theorem 47.0.2 Let (Xi,di) denote a complete metric space and let X ≡∏∞i=1 Xi. Then X

is also a complete metric space with the metric

ρ (x,y)≡∞

∑i=1

2−i di (xi,yi)

1+di (xi,yi).

Also, if Xi is separable for each i then so is X .

Proof: It is clear from the above lemma that ρ is a metric on X . We need to verify X iscomplete with this metric. Let {xn} be a Cauchy sequence in X . Then it is clear from thedefinition that {xn

i } is a Cauchy sequence for each i and converges to xi ∈ Xi. Therefore,letting ε > 0 be given, we choose N such that

∞

∑k=N

2−k <ε

2,

we choose M large enough that for n > M,

2−i di (xni ,xi)

1+di (xni ,xi)

<ε

2(N +1)

for all i = 1,2, · · · ,N. Then letting x ={xi} ,

ρ (x,xn)≤ εN2(N +1)

+∞

∑k=N

2−k <ε

2+

ε

2= ε.

We need to verify that X is separable. Let Di denote a countable dense set in Xi,Di ≡{ri

k

}∞

k=1 . Then let

Dk ≡ D1×·· ·×Dk×{

rk+11

}×{

rk+21

}×·· ·

Thus Dk is a countable subset of X . Let D ≡ ∪∞k=1Dk. Then D is countable and we can

see D is dense in X as follows. The projection of Dk onto the first k entries is dense in

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