1922 CHAPTER 59. BASIC PROBABILITY

Definition 59.18.5 Let µ be a Radon measure on Rp. A Borel set, A, is a µ continuity setif µ (∂A) = 0 where ∂A≡ A\ int(A) and int denotes the interior.

The main result is the following continuity theorem. More can be said about the equiv-alence of various criteria [19].

Theorem 59.18.6 If φ Xn(t)→ φ X (t) then λ Xn (A)→ λ X (A) whenever A is a λ X continu-

ity set.

Proof: First suppose K is a closed set and let

ψk (x)≡ (1− k dist(x,K))+.

Thus, since K is closed limk→∞ψk (x) = XK (x). Choose k large enough that∫Rp

ψkdλ X ≤ λ X (K)+ ε.

Then by Lemma 59.18.4, applied to the bounded uniformly continuous function ψk,

lim supn→∞

λ Xn (K)≤ lim supn→∞

∫ψkdλ Xn =

∫ψkdλ X ≤ λ X (K)+ ε.

Since ε is arbitrary, this shows

lim supn→∞

λ Xn (K)≤ λ X (K)

for all K closed.Next suppose V is open and let

ψk (x) = 1−(1− k dist

(x,VC))+.

Thus ψk (x) ∈ [0,1] ,ψk = 1 if dist(x,VC

)≥ 1/k, and ψk = 0 on VC. Since V is open, it

followslimk→∞

ψk (x) = XV (x).

Choose k large enough that ∫ψkdλ X ≥ λ X (V )− ε.

Then by Lemma 59.18.4,

lim infn→∞

λ Xn (V )≥ lim infn→∞

∫ψk (x)dλ Xn =

=∫

ψk (x)dλ X ≥ λ X (V )− ε

and since ε is arbitrary,lim inf

n→∞λ Xn (V )≥ λ X (V ).

1922 CHAPTER 59. BASIC PROBABILITYDefinition 59.18.5 Let 1 be a Radon measure on R?. A Borel set, A, is a M continuity setif u (OA) = 0 where 0A =A \ int (A) and int denotes the interior.The main result is the following continuity theorem. More can be said about the equiv-alence of various criteria [19].Theorem 59.18.6 [f x, (t) — x (t) then Ax, (A) > Ax (A) whenever A is a Ax continu-ity set.Proof: First suppose K is a closed set and letVW, (x) = (1—kdist (x,K))*.Thus, since K is closed limy_,.. W;, (x) = 2x (x). Choose k large enough that| wddx <Ax(K) +e.RPThen by Lemma 59.18.4, applied to the bounded uniformly continuous function y;,,lim sup Ax, (K) <lim sup [ y,dax, = / wddx <Ax(K) +e.n—yoo n—yooSince € is arbitrary, this showslim sup Ax, (K) < Ax (K)n—-s00for all K closed.Next suppose V is open and letWw; (x) = 1— (1—kdist (x,V°)) *.Thus y;, (x) € [0,1], yw, = 1 if dist (x,V°) > 1/k, and y, =0 on V©. Since V is open, itfollowslim y;, (x) = 2y (x).k-o0Choose k large enough that| vidax > Ax (V)—e.Then by Lemma 59.18.4,lim inf Ax, (V) > lim inf |v (x)dAx, =noo= [ vy(x)dax > Ax (V)and since € is arbitrary,lim inf Ax, (V) >Ax(V).noo