27.4. CHARACTERISTIC FUNCTIONS AND INDEPENDENCE 745

Lemma 27.4.2 Let Y be a random vector with values in Rp and let f be bounded andmeasurable with respect to the Radon measure λY , and satisfy∫

f (y)eit·ydλY = 0

for all t ∈ Rp. Then f (y) = 0 for λY a.e. y.

Proof: You could write the following for φ ∈ G∫φ (t)

∫f (y)eit·ydλY dt = 0 =

∫f (y)

(∫φ (t)eit·ydt

)dλY

Recall that the inverse Fourier transform maps G onto G . Hence∫

f (y)ψ (y)dλY = 0 forall ψ ∈ G . Thus this is also so for every ψ ∈C∞

0 (Rp)⊇C∞c (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,λY ) , each |φ k| ≤ 2. Then for any ψ ∈C∞0 (Rp) ,

0 =∫

f (y)φ n (y)ψ (y)dλY →∫| f (y)|ψ (y)dλY

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 λY a.e. (The measureµ (E)≡

∫E | f (y)|dλY equals 0.) ■

Proof of the proposition: If theX j are independent, the formula follows from Lemma27.2.6 and Lemma 27.2.4.

Now suppose the formula holds. Thus ∏nj=1 E

(eit j ·X j

)=∫

Rpn· · ·∫Rp2

∫Rp1

eit1·x1eit2·x2 · · ·eitn·xndλX1dλX2 · · ·dλXn = E(eiP)

=∫Rpn· · ·∫Rp2

∫Rp1

eit1·x1 eit2·x2 · · ·eitn·xndλX1|x2···xndλX2|x3···xn · · ·dλXn . (27.8)

Then from the above Lemma 27.4.2, the following equals 0 for λXn a.e. xn.∫Rpn−1

· · ·∫Rp2

∫Rp1

eit1·x1eit2·x2 · · ·eitn−1·xn−1dλX1dλX2 · · ·dλXn−1−

∫Rpn−1

· · ·∫Rp2

∫Rp1

eit1·x1eit2·x2

· · ·eitn−1·xn−1dλX1|x2···xndλX2|x3···xn · · ·dλXn−1|xn

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

eitn−1·xn−1dλXn−1 =∫Rpn−1

eitn−1·xn−1dλXn−1|xn