2004 CHAPTER 61. PROBABILITY IN INFINITE DIMENSIONS

thenν t1···tn (Ft1 ×·· ·×Ftn) = νs1···sp

(Gs1 ×·· ·×Gsp

)(61.5.13)

where if si = t j, then Gsi = Ft j and if si is not equal to any of the indices, tk, then Gsi = Msi .Then for E defined as in Definition 14.4.1, adjusted so that ±∞ never appears as anyendpoint of any interval, there exists a probability measure, P and a σ algebra F = σ (E )such that (

∏t∈I

Mt ,P,F

)is a probability space. Also there exist measurable functions, Xs : ∏t∈I Mt →Ms defined as

Xsx≡ xs

for each s ∈ I such that for each (t1 · · · tn)⊆ I,

ν t1···tn (Ft1 ×·· ·×Ftn) = P([Xt1 ∈ Ft1 ]∩·· ·∩ [Xtn ∈ Ftn ])

= P

((Xt1 , · · · ,Xtn) ∈

n

∏j=1

Ft j

)= P

(∏t∈I

Ft

)(61.5.14)

where Ft = Mt for every t /∈ {t1 · · · tn} and Fti is a Borel set. Also if f is a nonnegative

function of finitely many variables, xt1 , · · · ,xtn , measurable with respect to B(

∏nj=1 Mt j

),

then f is also measurable with respect to F and∫Mt1×···×Mtn

f (xt1 , · · · ,xtn)dν t1···tn

=∫

∏t∈I Mt

f (xt1 , · · · ,xtn)dP (61.5.15)

Theorem 61.5.4 Let X be a real vector space and let X∗ be the space of linear functionalsdefined on X. Also let ψ : X → C. Then ψ is a characteristic function if and only ifψ (0) = 1 and ψ is pseudo continuous at 0.

Proof: Suppose first ψ is a characteristic function as just described. I need to show itis positive definite and pseudo continuous. It is obvious ψ (0) = 1 in this case. Also

ψ

(∑k

akxk

)=∫

X∗exp

(ix∗(

∑k

akxk

))dµ (x∗)

and this is obviously a continuous function of a by the dominated convergence theorem. Itonly remains to verify the function is positive definite. However,

∑k, j

exp(ix∗ (xk− x j))αkα j = ∑k, j

eix∗(xk)αkeix∗(x j)α j ≥ 0

as in the earlier discussion of what it means to be positive definite given on Page 1940.