936 CHAPTER 26. INTEGRALS AND DERIVATIVES

Now

ε >∫| f −g|dm≥

∫[| f−g|> λ

2 ]| f −g|dm

≥ λ

2m([| f −g|> λ

2

])This along with the weak estimate of Theorem 26.1.2 implies

m([

x : limsupr→0

1m(B(x,r))

∫B(x,r)

| f (y)− f (x)|dm > λ

])<

(2λ

5k +2λ

)|| f −g||L1(Rk)

<

(2λ

5k +2λ

)ε.

Since ε > 0 is arbitrary, it follows

mn

([x : limsup

r→0

1m(B(x,r))

∫B(x,r)

| f (y)− f (x)|dm > λ

])= 0.

Now let

N =

[x : limsup

r→0

1m(B(x,r))

∫B(x,r)

| f (y)− f (x)|dm > 0]

and

Nn =

[x : limsup

r→0

1m(B(x,r))

∫B(x,r)

| f (y)− f (x)|dm >1n

]It was just shown that m(Nn) = 0. Also, N =∪∞

n=1Nn. Therefore, m(N) = 0 also. It followsthat for x /∈ N,

limsupr→0

1m(B(x,r))

∫B(x,r)

| f (y)− f (x)|dm = 0

and this proves a.e. point is a Lebesgue point.Of course it is sufficient to assume f is only in L1

loc

(Rk).

Corollary 26.1.6 (Fundamental Theorem of Calculus) Let f ∈ L1loc(Rk). Then there exists

a set of measure 0,N, such that if x /∈ N, then

limr→0

1m(B(x,r))

∫B(x,r)

| f (y)− f (x)|dy = 0.

Proof: Consider B(0,n) where n is a positive integer. Then fn ≡ f XB(0,n) ∈ L1(Rk)

and so there exists a set of measure 0, Nn such that if x ∈ B(0,n)\Nn, then

limr→0

1m(B(x,r))

∫B(x,r)

| fn(y)− fn(x)|dy

= limr→0

1m(B(x,r))

∫B(x,r)

| f (y)− f (x)|dy = 0.

Let N = ∪∞n=1Nn. Then if x /∈ N, the above equation holds.