14.12. THE VITALI COVERING THEOREMS 343

Suppose G1, · · · ,Gm−1 have been chosen for m ≥ 2. Let Gi denote the collection ofclosures of the balls of Gi. Then let Fm be those balls of F , such that if B is one of theseballs, B has empty intersection with every closed ball of Gi for each i≤ m−1. Then usingLemma 14.12.1, let Gm be a maximal collection of balls from Fm with the property thateach ball has radius larger than

( 23

)mM and their closures are disjoint. Let G ≡ ∪∞

k=1Gk.Thus the closures of balls in G are disjoint. Let x ∈ B(p,r) ∈F \G . Choose m such that(

23

)m

M < r ≤(

23

)m−1

M

Then B(p,r) must have nonempty intersection with the closure of some ball from G1∪·· ·∪Gm because if it didn’t, then Gm would fail to be maximal. Denote by B(p0,r0) a ball inG1∪ ·· ·∪Gm whose closure has nonempty intersection with B(p,r). Thus r0,r >

( 23

)mM.

Consider the picture, in which w ∈ B(p0,r0)∩B(p,r).

p− r p+ r

p0− r0 p0 + r0

x x0

Then for x ∈ B(p,r), |x− p0| ≤ |x−p|+ |p−w|+

≤r0︷ ︸︸ ︷|w−p0|

≤ r+ r+ r0 ≤ 2(

23

)m−1

M+ r0 ≤ 2(

32

) <r0︷ ︸︸ ︷(23

)m

M+ r0 ≤ 4r0

Thus B(p,r) is contained in B(p0,4r0). This shows that A⊆{

B̃ : B ∈ G}⊆∪{B̂ : B∈ G }.

You can easily generalize this proposition to include the case where the balls are notnecessarily open. These balls could be either open, closed, or neither open nor closed.

Proposition 14.12.3 Let F be a collection of balls, open, closed, or neither open norclosed, and let

A≡ ∪{B : B ∈F} .

Let B̂ 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⊆ ∪{B̂ : B ∈ G }

Proof: Let F ′ consist of the open balls obtained from F by keeping the center the sameand multiplying the radius by 20

19 . Denote these open balls by {B′ : B ∈F} . Thus B′ is anopen ball in which the radius is slightly expanded but the center is the same as B∈F . Thenfrom Lemma 14.12.2, there is G ′ ⊆F ′ such that the closures of the balls in G ′ are disjointand

{B̃′ : B′ ∈ G ′

}covers A. Here B̃′ is the closed ball obtained by multiplying the radius

of B′ by four. If B = B(x,r) , then B′ = B(x, 20

19 r)

and so B̃′ = B(x, 80

19 r)⊆ B(x,5r) since

5> 8019 . Thus, letting G be the balls B such that B′ ∈ G ′, it follows the closures of these balls

in G are disjoint since these balls are smaller than the balls of G ′ and A⊆∪{B̂ : B ∈ G }The notion of an outer measure allows the consideration of arbitrary sets, measurable

or not.