1082 CHAPTER 31. DIFFERENTIATION, RADON MEASURES

Let F be this collection of balls so that [M f > ε] is the set of centers of balls of F . By theBesicovitch covering theorem,

[M f > ε]⊆ ∪Nni=1 {B : B ∈ Gi}

where Gi is a collection of disjoint balls of F . Now for some i,

µ ([M f > ε])/Nn ≤ µ (∪{B : B ∈ Gi})

because if this is not so, then

µ ([M f > ε]) ≤Nn

∑i=1

µ (∪{B : B ∈ Gi})

<Nn

∑i=1

µ ([M f > ε])

Nn= µ ([M f > ε]),

a contradiction. Therefore for this i,

µ ([M f > ε])

Nn≤ µ (∪{B : B ∈ Gi}) = ∑

B∈Gi

µ (B)≤ ∑B∈Gi

ε−1∫

B| f |dµ

≤ ε−1∫Rn| f |dµ = ε

−1 || f ||1 .

This shows Claim 1.Claim 2: If g is any continuous function defined on Rn, then for x /∈ Z,

limr→0

1µ (B(x,r))

∫B(x,r)

|g(y)−g(x)|dµ (y) = 0

andlimr→0

1µ (B(x,r))

∫B(x,r)

g(y)dµ (y) = g(x). (31.1.1)

Proof: Since g is continuous at x, whenever r is small enough,

1µ (B(x,r))

∫B(x,r)

|g(y)−g(x)|dµ (y)≤ 1µ (B(x,r))

∫B(x,r)

ε dµ (y) = ε.

31.1.1 follows from the above and the triangle inequality. This proves the claim.Now let g ∈ Cc (Rn) and x /∈ Z. Then from the above observations about continuous

functions,

µ

([x /∈ Z : limsup

r→0

1µ (B(x,r))

∫B(x,r)

| f (y)− f (x)|dµ (y)> ε

])(31.1.2)

≤ µ

([x /∈ Z : limsup

r→0

1µ (B(x,r))

∫B(x,r)

| f (y)−g(y)|dµ (y)>ε

2

])+µ

([x /∈ Z : |g(x)− f (x)|> ε

2

]).