344 CHAPTER 13. LEBESGUE MEASURE

are elements of B and B satisfies 13.2.1 and 13.2.2. If ∪C is not maximal with respectto these two properties, then C was not a maximal chain because then there would existB ⊋ ∪C , that is, B contains C as a proper subset and {C ,B} would be a strictly largerchain in H . Let G = ∪C .

Theorem 13.2.3 (Vitali) Let F be a collection of balls and let

A≡ ∪{B : B ∈F}.

Suppose∞ > M ≡ sup{r : B(p,r) ∈F}> 0.

Then there exists G ⊆F such that G consists of disjoint balls and

A⊆ ∪{B̂ : B ∈ G }.

Proof: Using Lemma 13.2.2, there exists G1 ⊆F ≡F0 which satisfies

B(p,r) ∈ G1 implies r >M2, (13.2.3)

B1,B2 ∈ G1 implies B1∩B2 = /0, (13.2.4)

G1 is maximal with respect to 13.2.3, and 13.2.4.

Suppose G1, · · · ,Gm have been chosen, m≥ 1. Let

Fm ≡ {B ∈F : B⊆ Rn \∪{G1∪·· ·∪Gm}}.

Using Lemma 13.2.2, there exists Gm+1 ⊆Fm such that

B(p,r) ∈ Gm+1 implies r >M

2m+1 , (13.2.5)

B1,B2 ∈ Gm+1 implies B1∩B2 = /0, (13.2.6)

Gm+1 is a maximal subset of Fm with respect to 13.2.5 and 13.2.6.

Note it might be the case that Gm+1 = /0 which happens if Fm = /0. Define

G ≡ ∪∞k=1Gk.

Thus G is a collection of disjoint balls in F . I must show {B̂ : B ∈ G } covers A.Let x ∈ B(p,r) ∈F and let

M2m < r ≤ M

2m−1 .

Then B(p,r) must intersect some set, B(p0,r0) ∈ G1∪·· ·∪Gm since otherwise, Gm wouldfail to be maximal. Then r0 >

M2m because all balls in G1∪·· ·∪Gm satisfy this inequality.