59.11. CHARACTERISTIC FUNCTIONS, INDEPENDENCE 1891

By the upcrossing lemma,

E(

Un[a,b]

)≤ E (|Xn|)+ |a|

b−a≤ K + |a|

b−a

and so by the monotone convergence theorem,

E(U[a,b]

)≤ K + |a|

b−a< ∞

which shows U[a,b] (ω) is finite a.e., for all ω /∈ S[a,b] where P(S[a,b]

)= 0. Define

S≡ ∪{

S[a,b] : a,b ∈Q, a < b}.

Then P(S) = 0 and if ω /∈ S, {Xk}∞

k=1 has only finitely many upcrossings of every inter-val having rational endpoints. Thus, for ω /∈ S, limsupk→∞ Xk (ω) = liminfk→∞ Xk (ω) =limk→∞ Xk (ω)≡ X∞ (ω). Letting X∞ (ω) = 0 for ω ∈ S, Fatou’s lemma implies∫

Ω

|X∞|dP =∫

Ω

lim infn→∞|Xn|dP≤ lim inf

n→∞

∫Ω

|Xn|dP≤ K

and so this proves the theorem.

59.11 Characteristic Functions, IndependenceThere is a way to tell if random vectors are independent by using their characteristic func-tions.

Proposition 59.11.1 If Xi is a random vector having values in Rpi , then the random vec-tors are independent if and only if

E(eiP)= n

∏j=1

E(eit j ·X j

)where P≡ ∑

nj=1 t j ·X j for t j ∈ Rp j .

The proof of this proposition will depend on the following lemma.

Lemma 59.11.2 Let Y be a random vector with values in Rp and let f be bounded andmeasurable with respect to the Radon measure λ Y, and satisfy∫

f (y)eit·ydλ Y = 0

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

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

∫f (y)eit·ydλ Ydt = 0 =

∫f (y)

(∫φ (t)eit·ydt

)dλ Y