28.5. THE CENTRAL LIMIT THEOREM 777
Lemma 28.5.1 If {Xn}is a sequence of random vectors with values in Rpsuch that
limn→∞
φXn(t)≡ lim
n→∞φ λXn
(t) = ψ (t)
for all t, where ψ (0) = 1 and ψ is continuous at 0, then {λXn}∞
n=1 is tight.
In proving the central limit theorem, one considers the pointwise convergence of char-acteristic functions and then seeks to obtain information about the distribution of the limitfunction. In fact, one is in the situation of the following lemma which is Lemma 28.4.4.
Lemma 28.5.2 If φXn(t)→ φX (t) for all t, then whenever ψ ∈S,
λXn (ψ)≡∫Rp
ψ (y)dλXn (y)→∫Rp
ψ (y)dλX (y)≡ λX (ψ)
as n→ ∞.
The above gives what I want for ψ ∈ S but this needs to be generalized to ψ anybounded uniformly continuous function. The following is Lemma 28.4.5.
Lemma 28.5.3 If φXn(t)→ φX (t) , then if ψ is any bounded uniformly continuous
function,
limn→∞
∫Rp
ψdλXn =∫Rp
ψdλX .
Definition 28.5.4 Let µ be a Radon measure onRp. A Borel set A, is a µ continuityset if 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 [6].
Theorem 28.5.5 If φXn(t)→ φX (t) then λXn (A)→ λX (A) whenever A is a λX
continuity 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
ψkdλX ≤ λX (K)+ ε.
Then by Lemma 28.5.3, applied to the bounded uniformly continuous function ψk,
lim supn→∞
λXn (K)≤ lim supn→∞
∫ψkdλXn =
∫ψkdλX ≤ λX (K)+ ε.
Since ε is arbitrary, this shows limsupn→∞ λXn (K)≤ λX (K) for all K closed.Next suppose V is open and let
ψk (x) = 1−(1− k dist
(x,VC))+.