31.1. FUNDAMENTAL THEOREM OF CALCULUS 1083

≤ µ

([M ( f −g)>

ε

2

])+µ

([| f −g|> ε

2

])(31.1.3)

Now ∫[| f−g|> ε

2 ]| f −g|dµ ≥ ε

2µ

([| f −g|> ε

2

])and so from Claim 1 31.1.3 and hence 31.1.2 is dominated by(

2ε+

Nn

ε

)|| f −g||L1(Rn,µ) .

But by regularity of Radon measures, Cc (Rn) is dense in L1 (Rn,µ) , and so since g in theabove is arbitrary, this shows 31.1.2 equals 0. Now

µ

([x /∈ Z : limsup

r→0

1µ (B(x,r))

∫B(x,r)

| f (y)− f (x)|dµ (y)> 0])

≤∞

∑k=1

µ

([x /∈ Z : limsup

r→0

1µ (B(x,r))

∫B(x,r)

| f (y)− f (x)|dµ (y)>1k

])= 0

By completeness of µ this implies[x /∈ Z : limsup

r→0

1µ (B(x,r))

∫B(x,r)

| f (y)− f (x)|dµ (y)> 0]

is a set of µ measure zero.The following corollary is the main result referred to as the Lebesgue Besicovitch Dif-

ferentiation theorem.

Corollary 31.1.3 If f ∈ L1loc (Rn,µ), then for a.e.x /∈ Z,

limr→0

1µ (B(x,r))

∫B(x,r)

| f (y)− f (x)|dµ (y) = 0 . (31.1.4)

Proof: If f is replaced by f XB(0,k) then the conclusion 31.1.4 holds for all x /∈Fk whereFk is a set of µ measure 0. Letting k = 1,2, · · · , and F ≡ ∪∞

k=1Fk, it follows that F is a setof measure zero and for any x /∈ F , and k ∈ {1,2, · · ·}, 31.1.4 holds if f is replaced byf XB(0,k). Picking any such x, and letting k > |x|+1, this shows

limr→0

1µ (B(x,r))

∫B(x,r)

| f (y)− f (x)|dµ (y)

= limr→0

1µ (B(x,r))

∫B(x,r)

∣∣ f XB(0,k) (y)− f XB(0,k) (x)∣∣dµ (y) = 0.