26.7. DIFFERENTIATION OF MEASURES 965

Only consider those r which are small enough to be contained in B(0,M) so that the collec-tion of such balls has bounded radii. This is a Vitali cover of Npq (M) and so by Corollary26.7.1 there exists a sequence of disjoint balls of this sort, {Bi}∞

i=1 such that

µ (Bi)< pm(Bi) , m(Npq (M)\∪∞i=1Bi) = 0. (26.7.22)

Now for x ∈ Npq (M) ∩ (∪∞i=1Bi) (most of Npq (M)), there exist arbitrarily small balls,

B(x,r) , such that B(x,r) is contained in some set of {Bi}∞

i=1 and

µ (B(x,r))m(B(x,r))

> q.

This is a Vitali cover of Npq (M)∩ (∪∞i=1Bi) and so there exists a sequence of disjoint balls

of this sort,{

B′j}∞

j=1such that

m((Npq (M)∩ (∪∞

i=1Bi))\∪∞j=1B′j

)= 0, µ

(B′j)> qm

(B′j). (26.7.23)

It follows from 26.7.22 and 26.7.23 that

m(Npq (M))≤ m((Npq (M)∩ (∪∞i=1Bi)))≤ m

(∪∞

j=1B′j)

(26.7.24)

Therefore,

∑j

µ(B′j)

> q∑j

m(B′j)≥ qm(Npq (M)∩ (∪iBi)) = qm(Npq (M))

≥ pm(Npq (M))≥ p(m(V )− ε)≥ p∑i

m(Bi)− pε

≥ ∑i

µ (Bi)− pε ≥∑j

µ(B′j)− pε.

It followspε ≥ (q− p)m(Npq (M))

Since ε is arbitrary, m(Npq (M)) = 0. Now Npq ⊆ ∪∞M=1Npq (M) and so m(Npq) = 0. Now

N = ∪p.q∈QNpq

and since this is a countable union of sets of measure zero, m(N) = 0 also. This proves thetheorem.

From Theorem 20.2.5 on Page 605 it follows that if µ is a complex measure then |µ| isa finite measure. This makes possible the following definition.

Definition 26.7.5 Let µ be a real measure. Define the following measures. For E a mea-surable set,

µ+ (E) ≡ 1

2(|µ|+µ)(E) ,

µ− (E) ≡ 1

2(|µ|−µ)(E) .