61.7. GAUSSIAN MEASURES 2019

Consequently, αn (r)≤ α0 (r)2n

and also

P(||X ||> tn) = αn (r)P(||X || ≤ r)

≤ P(||X || ≤ r)α0 (r)2n

= P(||X || ≤ r)eln(α0(r))2n. (61.7.27)

Now using the distribution function technique and letting λ > 0,∫E

eλ ||x||2dµ =∫

∞

0µ

([eλ ||x||2 > t

])dt

= 1+∫

∞

1µ

([eλ ||x||2 > t

])dt

= 1+∫

∞

1P([

eλ ||X ||2 > t])

dt. (61.7.28)

From 61.7.27,

P([

exp(

λ ||X ||2)> exp

(λ t2

n)])≤ P([||X || ≤ r])eln(α0(r))2n

.

Now split the above improper integral into intervals,(exp(λ t2

n),exp

(λ t2

n+1))

for n =

0,1, · · · and note that P([

eλ ||X ||2 > t])

is decreasing in t. Then from 61.7.28,

∫E

eλ ||x||2 dµ ≤ exp(λ r2)+ ∞

∑n=0

∫ exp(λ t2n+1)

exp(λ t2n)

P([

eλ ||X ||2 > t])

dt

≤ exp(λ r2)+ ∞

∑n=0

P([

eλ ||X ||2 > exp(λ t2

n)])(

exp(λ t2

n+1)− exp

(λ t2

n))

≤ exp(λ r2)+ ∞

∑n=0

P([||X || ≤ r])eln(α0(r))2nexp(λ t2

n+1)

≤ exp(λ r2)+ ∞

∑n=0

eln(α0(r))2nexp(λ t2

n+1).

It remains to estimate tn+1. From the description of the tn,

tn =

(n

∑k=0

(√2)k)

r = r

(√2)n+1

−1√

2−1≤√

2√2−1

r(√

2)n

and sotn+1 ≤ 5r

(√2)n

Therefore, ∫E

eλ ||x||2 dµ ≤ exp(λ r2)+ ∞

∑n=0

eln(α0(r))2n+λ25r22n.