1942 CHAPTER 59. BASIC PROBABILITY

That is, for all f ∈S,∫Rn

f (y)dµ (y) =1

(2π)n/2

∫Rn

F (ψ)(y) f (y)dy

=1

(2π)n

∫Rn

f (y)(∫

Rne−iy·x

ψ (x)dx)

dy. (59.21.56)

I will show

f → 1(2π)n

∫Rn

f (y)(∫

Rne−iy·x

ψ (x)dx)

dy

is a positive linear functional and then it will follow from 59.21.56 that µ is unique. Thus itis needed to show the inside integral in 59.21.56 is nonnegative. First note that the integrandis a positive definite function of x for each fixed y. This follows from

∑k, j

e−iy·(xk−x j)ψ (xk−x j)αkα j

= ∑k, j

ψ (xk−x j)(

e−iy·(xk)αk

)e−iy·(x j)α j ≥ 0.

Let t > 0 and

h2t (x)≡1

(4πt)1/2 e−14t |x|

2.

Then by dominated convergence theorem,∫Rn

e−iy·xψ (x)dx = lim

t→∞

∫Rn

e−iy·xψ (x)h2t (x)dx

Letting dη2t = h2t (x)dx, it follows from Lemma 59.21.5 η2t = η t ∗η t and since these aresymmetric measures, it follows from Lemma 59.21.4 the above equals

limt→∞

∫Rn

e−iy·xψ (x)d (η t ∗η t)≥ 0

Thus the above functional is a positive linear functional and so there exists a unique Radonmeasure, µ satisfying∫

Rnf (y)dµ (y) =

1

(2π)n/2

∫Rn

F (ψ)(y) f (y)dy

=1

(2π)n

∫Rn

f (y)(∫

Rne−iy·x

ψ (x)dx)

dy

=1

(2π)n/2

∫Rn

ψ (x)

(1

(2π)n/2

∫Rn

f (y)e−iy·xdy

)dx