26.1. THE FUNDAMENTAL THEOREM OF CALCULUS 935

Definition 26.1.4 Let f ∈ L1(Rk,m

). A point, x ∈ Rk is said to be a Lebesgue point if

limsupr→0

1m(B(x,r))

∫B(x,r)

| f (y)− f (x)|dm = 0.

Note that if x is a Lebesgue point, then

limr→0

1m(B(x,r))

∫B(x,r)

f (y)dm = f (x) .

and so the symmetric derivative exists at all Lebesgue points.

Theorem 26.1.5 (Fundamental Theorem of Calculus) Let f ∈ L1(Rk). Then there exists aset 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: Let λ > 0 and let ε > 0. By density of Cc(Rk)

in L1(Rk,m

)there exists g ∈

Cc(Rk)

such that ||g− f ||L1(Rk) < ε . Now since g is continuous,

limsupr→0

1m(B(x,r))

∫B(x,r)

| f (y)− f (x)|dm

= limsupr→0

1m(B(x,r))

∫B(x,r)

| f (y)− f (x)|dm

− limr→0

1m(B(x,r))

∫B(x,r)

|g(y)−g(x)|dm

= limsupr→0

(1

m(B(x,r))

∫B(x,r)

| f (y)− f (x)|− |g(y)−g(x)|dm)

≤ limsupr→0

(1

m(B(x,r))

∫B(x,r)

|| f (y)− f (x)|− |g(y)−g(x)||dm)

≤ limsupr→0

(1

m(B(x,r))

∫B(x,r)

| f (y)−g(y)− ( f (x)−g(x))|dm)

≤ limsupr→0

(1

m(B(x,r))

∫B(x,r)

| f (y)−g(y)|dm)+ | f (x)−g(x)|

≤ M ([ f −g]) (x)+ | f (x)−g(x)| .

Therefore, [x : limsup

r→0

1m(B(x,r))

∫B(x,r)

| f (y)− f (x)|dm > λ

]⊆

[M ([ f −g])>

λ

2

]∪[| f −g|> λ

2

]