Kenneth Kuttler

1998 CHAPTER 61. PROBABILITY IN INFINITE DIMENSIONS

Thus m(

Iνi1··· ,im, j

)= ν (Ai1··· ,im, j) and

ν (Ai1··· ,im) =∞

∑k=1

ν(Ai1··· ,im,k

∞

∑k=1

m(Iνi1··· ,im,k

)= β −α,

the intervals, Iνi1··· ,im, j being disjoint and

Iν

i1 ··· ,im= ∪∞

j=1Iνi1··· ,im, j.

These intervals satisfy the same inclusion properties as the sets {Ai1,··· ,im} . They are juston [0,1) rather than on E. The intervals Iν

i1 ··· ,imkcorrespond to the sets Ai1··· ,im,k and in fact

the Lebesgue measure of the interval is the same as ν(Ai1··· ,im,k

Choosing the sequence {rk} in an auspicious manner

There are at most countably many positive numbers, r such that for ν = µn or µ ,

ν (∂B(ai,r))> 0

This is because ν is a finite measure. Taking the countable union of these countable sets,there are only countably many r such that ν (∂B(ai,r))> 0 for some ai. Let the sequenceavoid all these bad values of r. Thus for

F ≡ ∪∞m=1∪∞

k=1 ∂B(ak,rm)

and ν = µ or µn,ν (F) = 0. Here the rm are all good values such that for all k,m, ∂B(ak,rm)has µ measure zero and µn measure zero.

Claim 1: ∂Ai1,··· ,ik ⊆ F. This really follows from the construction. However, the detailsfollow.

Proof of claim: Suppose C is a Borel set for which ∂C ⊆ F. I need to show ∂Crik ∈ F.

First consider k = 1. Then Cri1 ≡ B(a1,ri)∩C. If x ∈ ∂Cri

1 , then B(x,δ ) contains pointsof B(a1,ri)∩C and points of B(a1,ri)

C ∪CC for every δ > 0. First suppose x ∈ B(a1,ri) .

Then a small enough neighborhood of x has no points of B(a1,ri)C and so every B(x,δ )

has points of C and points of CC so that x ∈ ∂C ⊆ F by assumption. If x ∈ ∂Cri1 , then it

can’t happen that ||x−a1|| > ri because then there would be a neighborhood of x havingno points of Cri

1 . The only other case to consider is that ||x−ai|| = ri but this says x ∈ F.Now assume ∂Cri

j ⊆ F for j ≤ k−1 and consider ∂Crik .

Crik ≡ B(ak,ri)∩C \∪k−1

j=1Crij

= B(ak,ri)∩C∩(∩k−1

j=1

(Cri

)C)

(61.4.10)

Consider x ∈ ∂Crik . If x ∈ int(B(ak,ri)∩C)(int≡ interior) then a small enough ball about

x contains no points of (B(ak,ri)∩C)C and so every ball about x must contain points of(∩k−1

j=1

(Cri

)C)C

= ∪k−1j=1Cri

1998 CHAPTER 61. PROBABILITY IN INFINITE DIMENSIONSThus m (Fins) = V(Aj,.--,in,j) andthe intervals, IY - . being disjoint and1 slmsJIY — Up...iim lms J*These intervals satisfy the same inclusion properties as the sets {Aj, ... ;,,}. They are juston (0,1) rather than on E. The intervals /” , correspond to the sets Aj... ;,,,¢ and in factipesimk simsthe Lebesgue measure of the interval is the same as V (Aj... ji,,k) +Choosing the sequence {7;,} in an auspicious mannerThere are at most countably many positive numbers, r such that for v = u,, or LL,V(OB(aj,r)) >0This is because v is a finite measure. Taking the countable union of these countable sets,there are only countably many r such that v (OB (a;,r)) > 0 for some a;. Let the sequenceavoid all these bad values of r. Thus forF = UP _, Uc OB (ag, 1m)and v = pl or Ll,,, V(F’) =0. Here the r,, are all good values such that for all k,m, OB (ax, 1m)has measure zero and 1,, measure zero.Claim 1: dAi, i, GF. This really follows from the construction. However, the detailsfollow.Proof of claim: Suppose C is a Borel set for which OC C F. I need to show 0C;' € F.First consider k = 1. Then Cj! = B(a,,rj) NC. If x € OC)‘, then B(x, 5) contains pointsof B(a,r;) NC and points of B(a,,r;) UCC for every 5 > 0. First suppose x € B(ay,r;) .Then a small enough neighborhood of x has no points of B(ay,r;)~ and so every B(x, 5)has points of C and points of C© so that x € OC C F by assumption. If x € 0C;', then itcan’t happen that ||x —a,|| > r; because then there would be a neighborhood of x havingno points of C;'. The only other case to consider is that ||x —a;|| =r; but this says x € F.Now assume ac} CF for j <k—1 and consider 0C;'.Ci = Blag,ri) AC\ UIC}Cc= Blag,ri)NCN (ni) (ci) ) (61.4.10)Consider x € OC;’. If x € int (B (ay, 7) NC) (int = interior) then a small enough ball aboutx contains no points of (B (ag,rj) AC) and so every ball about x must contain points ofk—-1 Cc © k= Tj _ —1iGe (cj) ) = UjaiC;