1892 CHAPTER 59. BASIC PROBABILITY

and now recall that the inverse Fourier transform maps G onto G . Hence∫f (y)ψ (y)dλ Y = 0

for all ψ ∈ G . Thus this is also so for every ψ ∈ C∞0 (Rp) by an obvious application of

the Stone Weierstrass theorem. Let {φ k} be a sequence of functions in C∞c (Rp) which

converges to

sgn( f )≡{

f̄/ | f | if f ̸= 00 if f = 0

pointwise and in L1 (Rp,λ Y) , each |φ k| ≤ 2. Then for any ψ ∈C∞0 (Rp) ,

0 =∫

f (y)φ n (y)ψ (y)dλ Y→∫| f (y)|ψ (y)dλ Y

Also, the above holds for any ψ ∈Cc (Rp) as can be seen by taking such a ψ and convolvingwith a mollifier. By the Riesz representation theorem, f (y) = 0 λ Y a.e. (The measureµ (E)≡

∫E | f (y)|dλ Y equals 0.)

Proof of the proposition: If the X j are independent, the formula follows from Lemma59.9.6 and Lemma 59.9.4.

Now suppose the formula holds. Thus

n

∏j=1

E(eit j ·X j

)=

∫Rpn· · ·∫Rp2

∫Rp1

eit1·x1eit2·x2 · · ·eitn·xndλ X1dλ X2 · · ·dλ Xn = E(eiP)

=∫Rpn· · ·∫Rp2

∫Rp1

eit1·x1 eit2·x2 · · ·eitn·xndλ X1|x2···xndλ X2|x3···xn · · ·dλ Xn . (59.11.19)

Then from the above Lemma 59.11.2, the following equals 0 for λ Xn a.e. xn.∫Rpn−1

· · ·∫Rp2

∫Rp1

eit1·x1eit2·x2 · · ·eitn−1·xn−1dλ X1dλ X2 · · ·dλ Xn−1−∫Rpn−1

· · ·∫Rp2

∫Rp1

eit1·x1eit2·x2 · · ·eitn−1·xn−1dλ X1|x2···xndλ X2|x3···xn · · ·dλ Xn−1|xn

Let ti = 0 for i = 1,2, · · · ,n−2. Then this implies∫Rpn−1

eitn−1·xn−1dλ Xn−1 =∫Rpn−1

eitn−1·xn−1dλ Xn−1|xn

By the fact that the characteristic function determines the distribution measure, Theorem59.8.4, it follows that for these xn off a set of λ Xn measure zero,λ Xn−1 = λ Xn−1|xn . Return-ing to 59.11.19, one can replace λ Xn−1|xn with λ Xn−1 to obtain∫

Rpn· · ·∫Rp2

∫Rp1

eit1·x1eit2·x2 · · ·eitn·xndλ X1dλ X2 · · ·dλ Xn−1dλ Xn