31.3. DIFFERENTIATION OF RADON MEASURES 1091

Nowµ (A\∪∞

i=1Bi)+µ (∪∞i=1Bi)≥ µ (A)

and so by 31.3.11 and the fact the Bi are disjoint, it follows from 31.3.12,

aµ (A) ≤ aµ (∪∞i=1Bi) = a

∞

∑i=1

µ (Bi)

≤ ε + ε (µ (A)+ ε)+λ (A) . (31.3.13)

Hence aµ (A)≤ λ (A) since ε > 0 was arbitrary.Now suppose A is a bounded subset of

{x /∈ Z : Dµ λ (x)≤ a

}and let V be a bounded

open set containing A with µ (V )− ε < µ (A) . Then if x ∈ A,

λ (B(x,r))µ (B(x,r))

< a+ ε, B(x,r)⊆V

for values of r which are arbitrarily small. Therefore, by Corollary 13.14.3 again, thereexists a disjoint sequence of these balls, {Bi} satisfying this time,

λ (A\∪∞i=1Bi) = 0.

Then by arguments similar to the above,

λ (A)≤∞

∑i=1

λ (Bi)< (a+ ε)µ (V )< (a+ ε)(µ (A)+ ε) .

Since ε was arbitrary, this proves the lemma in case A is bounded. In general, for

A ∈{

x /∈ Z : Dµ λ (x)≥ a}

One obtains from the first part

λ (A∩B(0,n))≥ aµ (A∩B(0,n))

Then, passing to a limit as n→ ∞ gives the desired result. The case where

A⊆{

x /∈ Z : Dµ λ (x)≤ a}

is similar.

Theorem 31.3.3 There exists a set of measure zero N containing Z such that for x /∈N, Dµ λ (x) exists and also XNC (·)Dµ λ (·) is a µ measurable function. Furthermore,Dµ λ (x)< ∞ µ a.e.

Proof: First I show Dµ λ (x) exists a.e. Let 0 ≤ a < b < ∞ and let A be any boundedsubset of

N (a,b)≡{

x /∈ Z : Dµ λ (x)> b > a > Dµ λ (x)}.

By Lemma 31.3.2,aµ (A)≥ λ (A)≥ bµ (A)