Kenneth Kuttler

1526 CHAPTER 47. THE YANKOV VON NEUMANN AUMANN THEOREM

Now define a closed set,

C ≡ ∩∞k=1 f (E (m1,m2, · · · ,mk−1,mk)).

The sets f (E (m1,m2, · · · ,mk−1,mk)) are decreasing as k→ ∞ and so

µ∗ (C) = lim

k→∞µ∗(

f (E (m1,m2, · · · ,mk−1,mk)))≥ µ

∗ (A)− ε.

We wish to verify that C ⊆ A. If we can do this we will be done because C, being a closedset, is measurable and so

µ∗ (Ω) = µ

∗ (C)+µ∗ (Ω\C)≥ µ

∗ (A)− ε +µ∗ (Ω\A) .

Since ε is arbitrary, this will conclude the proof. Therefore, we only need to verify thatC ⊆ A.

What we know is that each f (E (m1,m2, · · · ,mk−1,mk)) is contained in A. We do notknow their closures are contained in A. We let m≡{mi}∞

i=1 where the mi are defined above.Then letting

K ≡{

n ∈ NN : ni ≤ mi for all i},

we see that K is a closed, hence complete subset of NN which is also totally bounded due tothe definition of the distance. Therefore, K is compact and so f (K) is also compact, henceclosed due to the assumption that Ω is a Hausdorff space and we know that f (K)⊆ A. Weverify that C = f (K) . We know f (K)⊆C. Suppose therefore, p∈C. From the definition ofC, we know there exists rk ∈ E (m1,m2, · · · ,mk−1,mk) such that d

(f(rk), p)< 1

k . Denote

by r̃k the element of NN which consists of modifying rk by taking all components after thekth equal to one. Thus r̃k ∈ K. Now

{r̃k}

is in a compact set and so taking a subsequence

we can have r̃k → r ∈ K. But from the metric on NN, it follows that ρ

(r̃k,rk

)< 1

2k−2 .

Therefore, rk→ r also and so f(rk)→ f (r) = p. Therefore, p ∈ f (K) and this proves the

theorem.Note we could have proved this under weaker assumptions. If we had assumed only

that every point has a countable basis (first axiom of countability) and Ω is Hausdorff, thesame argument would work. We will need the following definition.

Definition 47.0.10 Let F be a σ algebra of sets from Ω and let µ denote a finite measuredefined on F . We let Fµ denote the completion of F with respect to µ. Thus we let µ∗ bethe outer measure determined by µ and Fµ will be the σ algebra of µ∗ measurable subsetsof Ω. We also define F̂ by

F̂ ≡ ∩{Fµ : µ is a finite measure defined on F

Also, if X is a topological space, we will denote by B(X) the Borel sets of X .

With this notation, we can give the following simple corollary of Theorem 47.0.9. Thisis really quite amazing.

1526 CHAPTER 47. THE YANKOV VON NEUMANN AUMANN THEOREMNow define a closed set,C= Nef (E (mj,m2,+°° ,M—1,M)).The sets f (E (m,mz,--- ,mg_1,m)) are decreasing as k + oo and soue (C) = jim py" (7 (E (m),m2, otk] m4) 2 we (A) —é.We wish to verify that C C A. If we can do this we will be done because C, being a closedset, is measurable and sowu" (Q) =" (C)+M"(Q\C) =u" (A)—€ +" (Q\A).Since € is arbitrary, this will conclude the proof. Therefore, we only need to verify thatCCA.What we know is that each f (E (m,mz,--- ,mg_1,mx)) is contained in A. We do notknow their closures are contained in A. We let m = {m;};~ , where the m; are defined above.Then lettingK = {ne NN : mj <m;for all i},we see that K is a closed, hence complete subset of N‘ which is also totally bounded due tothe definition of the distance. Therefore, K is compact and so f (K) is also compact, henceclosed due to the assumption that Q is a Hausdorff space and we know that f (K) C A. Weverify that C= f (K). We know f (K) CC. Suppose therefore, p € C. From the definition ofC, we know there exists r* € E (m,,m2,-++ ,mg_1,m,) such that d (f (r*) .P) < i: Denoteby ré the element of N which consists of modifying r* by taking all components after thek" equal to one. Thus ré € K. Now {r‘} is in a compact set and so taking a subsequencewe can have r + r € K. But from the metric on NN, it follows that p (rr) < aeTherefore, r“ — r also and so f (r*) — f(r) = p. Therefore, p € f (K) and this proves thetheorem.Note we could have proved this under weaker assumptions. If we had assumed onlythat every point has a countable basis (first axiom of countability) and © is Hausdorff, thesame argument would work. We will need the following definition.Definition 47.0.10 Let F be ao algebra of sets from Q and let u denote a finite measuredefined on F. We let Fy, denote the completion of F with respect to U. Thus we let u* bethe outer measure determined by 1 and F, will be the o algebra of 1* measurable subsetsof Q.. We also define F byF=n {Fy : LL is a finite measure defined on F \ .Also, if X is a topological space, we will denote by B(X) the Borel sets of X.With this notation, we can give the following simple corollary of Theorem 47.0.9. Thisis really quite amazing.