342 CHAPTER 14. THE LEBESGUE INTEGRAL

notation which is appropriate for this generalization. The main ideas are all present in onedimension. A ball will just be an interval. It might have no endpoints, one endpoint, orboth endpoints.

Lemma 14.12.1 Let F be a countable collection of balls satisfying

∞ > M ≡ sup{r : B(p,r) ∈F}> 0

and let k ∈ (0,∞) . Then there exists G ⊆F such that

If B(p,r) ∈ G then r > k, (14.8)

If B1,B2 ∈ G then B1∩B2 = /0, (14.9)

G is maximal with respect to 14.8 and 14.9. (14.10)

By this is meant that if H is a collection of balls satisfying 14.8 and 14.9, then H cannotproperly contain G .

Proof: If no ball of F has radius larger than k, let G = /0. Assume therefore, that someballs have radius larger than k. Let F ≡ {Bi}∞

i=1 ,Bi∩B j = /0 if i ̸= j. Now let Bn1 be thefirst ball in the list which has radius greater than k. If every ball having radius larger than khas closure which intersects Bn1 , then stop. The maximal set is {Bn1} . Otherwise, let Bn2be the next ball having radius larger than k for which Bn2 ∩Bn1 = /0. Continue this wayobtaining {Bni}

Ni=1, a finite or infinite sequence of balls having radius larger than k whose

closures are disjoint. N = ∞ if the process never stops. N is some number if the processdoes stop. Then let G ≡ {Bni}

Ni=1. To see G is maximal with respect to 14.8 and 14.9,

suppose B ∈F , B has radius larger than k, and G ∪{B} satisfies 14.8 and 14.9. Then atsome point in the process, B would have been chosen because it would be the ball of radiuslarger than k which has the smallest index at some point in the construction. Therefore,B ∈ G and this shows G is maximal with respect to 14.8 and 14.9.

Lemma 14.12.2 Let F be a collection of open balls, and let

A⊆ ∪{B : B ∈F} .

Suppose B̃ denotes the closed ball with the same center as B but four times the radius. LetB̂ denote the open ball with same center as B but five times the radius.

∞ > M ≡ sup{r : B(p,r) ∈F}> 0.

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

A⊆ ∪{

B̃ : B ∈ G}⊆ ∪{B̂ : B ∈ G }

Proof: First of all, it follows from Theorem 4.8.20 on Page 71 that there is a countablesubset F̃ of F which also covers A. Thus ∪F̃ ⊇ A⊇ ∪F ⊇ ∪F̃ and so ∪F̃ = A. Thusit can be assumed that F is countable.

By Lemma 14.12.1, there exists G1 ⊆ F which satisfies 14.8, 14.9, and 14.10 withk = 2M

3 . That is, G1 consists of balls having radii larger than 2M3 and their closures are

disjoint and G1 is as large as possible.