352 CHAPTER 14. THE LEBESGUE INTEGRAL

Modify the above problem slightly to show that it can be assumed that h vanishesoutside of some finite interval.

25. A Lebesgue measurable function f is said to be in L1loc if f X[a,b] has a finite inte-

gral. Show that there is a Borel set of measure zero off which x→∫ x

0 f (t)dm has aderivative a.e.

26. This is on the Lebesgue fundamental theorem of calculus. Let f ∈ L1 (R,m) . ByProblem 24 for each ε, there is g continuous which vanishes off a finite interval suchthat

∫| f −g|dm < ε . Such a g is uniformly continuous by Theorem 6.7.2. Fill in

details to the following argument. For g as just described,

limr→0

1m(B(0,r))

∫B(x,r)

|g(x)−g(y)|dm(y) = 0

Here x is fixed and the function integrated is y→ |g(x)−g(y)|. Now let rn→ 0+ bean arbitrary sequence each less than 1. Then

lim supn→∞

(1

m(B(0,rn))

∫B(x,rn)

| f (y)− f (x)|dm(y))≤

lim supn→∞

1m(B(0,rn))

∫B(x,rn)

| f (y)−g(y)|dm(y)

+ lim supn→∞

1m(B(0,rn))

∫B(x,rn)

|g(y)−g(x)|dm(y)

+ lim supn→∞

1m(B(0,rn))

∫B(x,rn)

|g(x)− f (x)|dm(y)

≤M ( f −g)(x)+ |g(x)− f (x)|

Then

m̄[

x : lim supn→∞

(1

m(B(0,rn))

∫B(x,rn)

| f (y)− f (x)|dm(y))> λ

]≤

m̄ [M ( f −g)> λ/2]+ m̄ [|g− f |> λ/2]

By Theorem 14.12.7 and Problem 11

≤ 10λ

∫| f −g|dm+

2λ

∫|g− f |dm

We are free to choose how close g is to f and so

m̄([

x : lim supn→∞

(1

m(B(0,rn))

∫B(x,rn)

| f (y)− f (x)|dm(y))> λ

])= 0

It follows that

m̄([

x : lim supn→∞

(1

m(B(0,rn))

∫B(x,rn)

| f (y)− f (x)|dm(y))> 0])

= 0