33.7. EXERCISES 669

13. To integrate complex valued functions f : Ω → C, first note that these are defined tobe measurable if the real and imaginary parts are measurable. Then∫

f dµ =∫

(Re f )dµ + i∫

(Im f )dµ

In the context of probability distribution measures described above, explain whyeverything makes sense and for t ∈ Rp

∫Ω

eiX(ω)·tdP =∫Rp

eix·tdλX ≡ φX (t)

This is called the characteristic function. It turns out that these completely character-ize the probability distribution measures but this is a topic for a more advanced bookwhich has important representation theorems not discussed here.

14. Show that there exists a subset of R consisting of everything off a set of measure nomore than ε which contains no intervals.

15. This problem outlines an approach to Stirling’s formula following [26] and [8]. Fromthe above problems, Γ(n+1) = n! for n ≥ 0. Consider more generally Γ(x+1) forx > 0. Actually, we will always assume x > 1 since it is the limit as x → ∞ which isof interest. Γ(x+1) =

∫∞

0 e−ttxdt. Change variables letting t = x(1+u) to obtain

Γ(x+1) = xx+1e−x∫

∞

−1

((1+u)e−u)x du

Next let h(u) be such that h(0) = 1 and (1+u)e−u = exp(− u2

2 h(u))

. Show that

the thing which works is h(u) = 2u2 (u− ln(1+u)). Use L’Hospital’s rule to verify

that the limit of h(u) as u → 0 is 1. The graph of h is illustrated in the followingpicture. Verify that its graph is like this, with an asymptote at u =−1 decreasing andequal to 1 at 0 and converging to 0 as u → ∞.

−1

1

Next change the variables again letting u = s√

2x . This yields, from the original

description of h

Γ(x+1) = xxe−x√

2x∫

∞

−√

x/2exp

(−s2h

(s

√2x

))ds

For s < 1,h(

s√

2x

)> 2−2ln2 = 0.61371 so the above integrand is dominated by