2304 CHAPTER 67. THE EASY ITO FORMULA

where yi is an arbitrary vector in Rn. It follows that for all choices of y j ∈ Rn,

∫Ω

g(ω)exp

(m

∑j=0

yTj Wt j (ω)

)dP = 0.

Now the mapping

y = (y0, · · · ,ym)→∫

Ω

g(ω)exp

(m

∑j=0

yTj Wt j (ω)

)dP

is analytic on C(m+1)n and equals zero on R(m+1)n so from standard complex variable the-ory, this analytic function must equal zero on C(m+1)n, not just on R(m+1)n. In particular,for all y = (y0, · · · ,ym) ∈ Rn(m+1),

∫Ω

g(ω)exp

(m

∑j=0

iyTj Wt j (ω)

)dP = 0. (67.9.17)

This left side equals

∫Ω

g+ (ω)exp

(m

∑j=0

iyTj Wt j (ω)

)dP−

∫Ω

g− (ω)exp

(m

∑j=0

iyTj Wt j (ω)

)dP

where g+ and g− are the positive and negative parts of g. By the Lemma 67.9.2 and theobservation at the end, this equals

∫Rnm

exp

(m

∑j=0

iyTj x j

)dν+−

∫Rnm

exp

(m

∑j=0

iyTj x j

)dν−

where ν+ (B)≡∫

Ωg+ (ω)XB (Wt1 (ω) , · · · ,Wtm (ω))dP and ν− is defined similarly. Then

letting ν be the measure ν+−ν−, it follows that

0 =∫Rnm

exp

(m

∑j=0

iyTj x j

)dν (y)

and this just says that the inverse Fourier transform of ν is 0. It follows that ν = 0. Thus∫Ω

g(ω)XB (Wt1 (ω) , · · · ,Wtm (ω))dP

=∫

Ω

g(ω)XW−1m (B) (ω)dP = 0

for every B Borel in Rnm where

Wm (ω)≡ (Wt1 (ω) , · · · ,Wtm (ω))

Let K be the π system defined as W−1m (B) for B of the form ∏

mi=1 Ui where Ui is open

in Rn, this for some m a positive integer. This is indeed a π system because it includes