2026 CHAPTER 61. PROBABILITY IN INFINITE DIMENSIONS

Thus also ∫U

eit(x−m,h)dµ (x) = e−12 t2(Qh,h)

and letting t = 1 this yields ∫U

ei(x−m,h)dµ (x) = e−12 (Qh,h)

From this it follows ∫U

(1− ei(x−m,h)

)dµ = 1− e−

12 (Qh,h)

and since the right side is real, this implies∫U(1− cos(x−m,h))dµ (x) = 1− e−

12 (Qh,h)

Thus1− e−

12 (Qh,h) ≤

∫[||x−m||≤c]

(1− cos(x−m,h))dµ (x)

+2∫[||x−m||>c]

dµ (x)

Now it is routine to show

1− cos t ≤ 12

t2

and so

1− e−12 (Qh,h) ≤ 1

2

∫[||x−m||≤c]

|(x−m,h)|2 dµ (x)

+2µ ([||x−m||> c])

Pick c large enough that the last term is smaller than 1/8. This can be done because thesets decrease to /0 as c→ ∞ and µ is given to be a finite measure. Then with this choice ofc,

78− 1

2

∫[||x−m||≤c]

|(x−m,h)|2 dµ (x)≤ e−12 (Qh,h) (61.8.38)

For each h the integral in the above is finite. In fact∫[||x−m||≤c]

|(x−m,h)|2 dµ (x)≤ c2 ||h||2

Let(Qch,h1)≡

∫[||x−m||≤c]

(x−m,h)(x−m,h1)dµ (x)

and let A denote those h ∈U such that

(Qch,h)< 1.