1920 CHAPTER 59. BASIC PROBABILITY

≥ 2∫[|x j|≥ 2

u ]

(1− 1|u|(2/u)

)dλ Xn (x)

=∫[|x j|≥ 2

u ]1dλ Xn (x)

= λ Xn

([x :∣∣x j∣∣≥ 2

u

]).

If ε > 0 is given, there exists r > 0 such that if u≤ r,

1u

∫ u

−u(1−ψ (te j))dt < ε/p

for all j = 1, · · · , p and so, by the dominated convergence theorem, the same is true withφ Xn

in place of ψ provided n is large enough, say n≥ N (u). Thus, if u≤ r, and n≥ N (u),

λ Xn

([x :∣∣x j∣∣≥ 2

u

])< ε/p

for all j ∈ {1, · · · , p}. It follows that for u≤ r and n≥ N (u) ,

λ Xn

([x : ||x||

∞≥ 2

u

])< ε.

because [x : ||x||

∞≥ 2

u

]⊆ ∪p

j=1

[x :∣∣x j∣∣≥ 2

u

]This proves the lemma because there are only finitely many measures, λ Xn for n < N (u)and the compact set can be enlarged finitely many times to obtain a single compact set, Kε

such that for all n,λ Xn ([x /∈ Kε ])< ε. This proves the lemma.

Lemma 59.18.3 If φ Xn(t)→ φ X (t) for all t, then whenever ψ ∈S,

λ Xn (ψ)≡∫Rp

ψ (y)dλ Xn (y)→∫Rp

ψ (y)dλ X (y)≡ λ X (ψ)

as n→ ∞.

Proof: Recall that if X is any random vector, its characteristic function is given by

φ X (y)≡∫Rp

eiy·xdλ X (x) .

Also remember the inverse Fourier transform. Letting ψ ∈S, the Schwartz class,

F−1 (λ X)(ψ) ≡ λ X(F−1

ψ)≡∫Rp

F−1ψdλ X

=1

(2π)p/2

∫Rp

∫Rp

eiy·xψ (x)dxdλ X (y)

=1

(2π)p/2

∫Rp

ψ (x)∫Rp

eiy·xdλ X (y)dx

=1

(2π)p/2

∫Rp

ψ (x)φ X (x)dx

1920 CHAPTER 59. BASIC PROBABILITY22f (Frag) dhldAx. (xrao Xe)ax, (x:bsl2 21).If € > 0 is given, there exists r > 0 such that if u < 7,~ [= wee))) dt <e/pu J—ufor all j = 1,--- ,p and so, by the dominated convergence theorem, the same is true with$x, in place of y provided n is large enough, say n > N (u). Thus, if wu <r, andn > N(u),Ax, (|x: > -|) <e/pfor all j € {1,--- ,p}. It follows that for u<randn>N(u),2Ax, (a }) <€.u2 2x: ls. > | CUL_y [0 >This proves the lemma because there are only finitely many measures, Ax, for n < N (u)and the compact set can be enlarged finitely many times to obtain a single compact set, Kesuch that for all n,Ax, ([x € Ke]) < €. This proves the lemma.Lemma 59.18.3 /f @y, (t) + x (t) for all t, then whenever y € G,ax, (W)= | wiy)dax,0) + [ wiy)dax 0) =x (y)becauseas Nn — °°,Proof: Recall that if X is any random vector, its characteristic function is given bydx(y) = [eV aax (x),Also remember the inverse Fourier transform. Letting y € G, the Schwartz class,TAx)(y) = Ax(F ly) = [, F-!wdax1 yxy7 (2m)P/? Lobe mY (x) dxdAx (y)1 of x~ Ga Evol dxGar a V(x) Ox (x) dx