14.14. EXERCISES 353

and so, there is a Borel set of measure zero N such that if x /∈ N,

limn→∞

1m(B(0,rn))

∫B(x,rn)

| f (y)− f (x)|dm(y) = 0

and solimr→∞

1m(B(0,r))

∫B(x,r)

| f (y)− f (x)|dm(y) = 0

which implies for x /∈ N,

limr→0

1m(B(0,r))

∫B(x,r)

f (y)dm(y) = f (x)

27. If you only know that f X[a,b] ∈ L1 (R,m) for all finite intervals [a,b] , show exactlythe same conclusion holds.

28. Use the above result to show that if f X[a,b] ∈ L1 (R,m) for all finite intervals [a,b] ,then for a.e. x,

limh→0

1h

∫ x+h

0f (t)dm = f (x)

which can also be termed the Lebesgue fundamental theorem of calculus.

29. This problem outlines an approach to Stirling’s formula following [24] and [9]. Aneasier approach comes from the traditional Stirling’s formula and Problem 16 onPage 247 From the above problems, Γ(n+1) = n! for n ≥ 0. Consider more gen-erally Γ(x+1) for x > 0. Actually, we will always assume x > 1 since it is thelimit as x→ ∞ which is of interest. Γ(x+1) =

∫∞

0 e−ttxdt. Change variables let-ting 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 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 following picture. Verify that itsgraph is like this, with an asymptote at u = −1 decreasing and equal to 1 at 0 andconverging 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 e−(2−2ln2)s2. Consider the integrand in the

above for s ≥ 1. The exponent part is −(√

2√

xs− x ln(

1+ s√

2x

)). Now for