2000 CHAPTER 61. PROBABILITY IN INFINITE DIMENSIONS

Proof of the claim: From Claim 3 and letting ε > 0 be given, there exists m largeenough that for all n,

supω

d(Zµn

m (ω) ,X µn (ω))< ε/3, sup

ω

d (Zµm (ω) ,X µ (ω))< ε/3.

for ω off a set of measure zero. Now pick ω ∈ [0,1) such that ω is not equal to anyof the end points of any of the intervals,

{Iνi1,··· ,im

}, this countable set of endpoints, a

set of Lebesgue measure zero. Then by Claim 2, there exists N such that if n ≥ N, thend(Zµn

m (ω) ,Zµm (ω)

)< ε/3. Therefore, for such n and this ω,

d (X µn (ω) ,X µ (ω)) ≤ d(X µn (ω) ,Zµn

m (ω))+d(Zµn

m (ω) ,Zµm (ω)

)+d (Zµ

m (ω) ,X µ (ω))

< ε/3+ ε/3+ ε/3 = ε.

This proves the claim.

Showing L (Xν) = ν .

This has mostly proved the theorem except for the claim that L (Xν) = ν for ν =

µn and µ. To do this, I will first show m((Xν)−1 (∂Ai1,··· ,im)

)= 0. By the construction,

ν (∂Ai1,··· ,im) = 0. Let ε > 0 be given and let δ > 0 be small enough that

Hδ ≡ {x ∈ E : dist(x,∂Ai1,··· ,im)≤ δ}

is a set of measure less than ε/2. Denote by Gk the sets of the form Ai1,··· ,ik where(i1, · · · , ik) ∈ Nk. Recall also that corresponding to Ai1,··· ,ik is an interval, Iν

i1,··· ,ik havinglength equal to ν

(Ai1,··· ,ik

). Denote by Bk those sets of Gk which have nonempty inter-

section with Hδ and let the corresponding intervals be denoted by I νk . If ω /∈ ∪I ν

k , thenfrom the construction, Zν

p (ω) is at a distance of at least δ from ∂Ai1,··· ,im for all p≥ k. (IfZν

k (ω) were in some set of Bk, this would require ω to be in the corresponding Iνk and it is

assumed this does not happen. Then for any p > k,Zνp (ω) cannot be in any set of Gp which

intersects Hδ either. If it did, you would need to have ω /∈ ∪I νp but all of these intervals

are inside the intervals I νk .) Passing to the limit as p→ ∞, it follows Xν (ω) /∈ ∂Ai1,··· ,im .

Therefore,(Xν)

−1(∂Ai1,··· ,im)⊆ ∪I ν

k

Recall that Ai1,··· ,ik ⊆ B(aik ,rk

)and the rk→ 0. Therefore, if k is large enough,

ν (∪Bk)< ε

because ∪Bk approximates Hδ closely (In fact, ∩∞k=1 (∪Bk) = Hδ .). Therefore,

m((Xν)

−1(∂Ai1,··· ,im)

)≤ m(∪I ν

k )

= ∑Iνi1 ,··· ,ik

∈I νk

m(Iνi1,··· ,ik

)= ∑

Ai1 ,··· ,ik∈Bk

ν(Ai1,··· ,ik

)= ν (∪Bk)< ε.