Chapter 31

Differentiation, Radon MeasuresThis is a brief chapter on certain important topics on the differentiation theory for generalRadon measures. For different proofs and some results which are not discussed here, agood source is [47] which is where I first read some of these things.

31.1 Fundamental Theorem Of CalculusIn this section the Besicovitch covering theorem will be used to give a generalization of theLebesgue differentiation theorem to general Radon measures. In what follows, µ will be aRadon measure,

Z ≡ {x ∈ Rn : µ (B(x,r)) = 0 for some r > 0},

Lemma 31.1.1 Z is measurable and µ (Z) = 0.

Proof: For each x ∈ Z, there exists a ball B(x,r) with µ (B(x,r)) = 0. Let C be thecollection of these balls. Since Rn has a countable basis, a countable subset, C̃ , of C alsocovers Z. Let

C̃ = {Bi}∞

i=1 .

Then letting µ denote the outer measure determined by µ ,

µ (Z)≤∞

∑i=1

µ (Bi) =∞

∑i=1

µ (Bi) = 0

Therefore, Z is measurable and has measure zero as claimed.Let M f : Rn→ [0,∞] by

M f (x)≡{

supr≤11

µ(B(x,r))∫

B(x,r) | f |dµ if x /∈ Z0 if x ∈ Z

.

Theorem 31.1.2 Let µ be a Radon measure and let f ∈ L1 (Rn,µ). Then for a.e.x,

limr→0

1µ (B(x,r))

∫B(x,r)

| f (y)− f (x)|dµ (y) = 0

Proof: First consider the following claim which is a weak type estimate of the samesort used when differentiating with respect to Lebesgue measure.

Claim 1: The following inequality holds for Nn the constant of the Besicovitch coveringtheorem.

µ ([M f > ε])≤ Nnε−1 || f ||1

Proof: First note [M f > ε]∩ Z = /0 and without loss of generality, you can assumeµ ([M f > ε])> 0. Next, for each x ∈ [M f > ε] there exists a ball Bx = B(x,rx) with rx ≤ 1and

µ (Bx)−1∫

B(x,rx)| f |dµ > ε.

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