Exercise 1.6.18 (Vitali-type covering lemma)

Let B1,…,Bn be a finite collection of open balls in ℝd. Then there exists a subcollection B1′,…,Bm′ of disjoint balls in this collection such that

⋃ i=1nB i ⊆⋃ j=1m3B j′

Answers

Denote B = {B1,…,Bn}. At each step 1 ≤ t ≤ m we will fill out the collection Ft with the balls Bt′ (obviously Ft−1 ⊆Ft), so that the resulting collection Fm = {B1′,…,Bm′} satisfies the end-condition. Thus, consider the following induction.

  • F0 := ∅

  • Define for 1 ≤ t ≤ n:

RESTt := B∕Ft−1 DISJOINT_RESTt := {B ∈RESTt : B ∩ A = ∅ for all A ∈Ft−1} Bt′ ∈DISJOINT_REST  choose such that for all other Bl′∈DISJOINT_REST : m(Bt′) ≥ m(Bl′) Ft := Ft−1 ∪{Bt′}

Let m := #Fn, and enumerate its elements by B1′,…,Bm′. The sets Bi′∩ Bj′ = ∅ are obviously disjoint (otherwise we would come to a contradiction at construction step t = j > i). We now demonstrate that

⋃ i=1nB i ⊆⋃ j=1m3B j′

PIC
Figure 1: On the left: a collection of balls; the green balls are the disjoint subcollection. On the right: the subcollection with three times the radius covers all the balls.

Let Bi from the left-hand side be arbitrary. By construction, there must be a Bt′ which intersects Bi, i.e, Bt′∩ Bi≠∅ (if this would not be the case, then Bi would be disjoint from all B1′,…,Bm′ and would be added to the collection as (m + 1)st element at least at step n). Let Bt′ be the first ball in the collection B1′,…,Bm′ with this property (i.e., all other balls B1′,…,Bt−1′ before Bt′ are disjoint from Bi). Because Bt′ was chosen to be the largest amongst all balls that did not intersect B1′,…,Bj−1′, we conclude that the radius of Bi cannot exceed that of Bj′. If Bi = B(x,r) and Bt′ = B(y,R), set z ∈ Bi ∩ Bt′≠∅. Then from the triangle inequality, we see that

∀ ⁡p ∈ Bi : d(p,y) ≤ d(p,z)+d(z,y) ≤ d(p,x)+d(x,z)+d(z,y) ≤ r+r+R ≤ 3R.

This implies that Bi ⊆ 3Bj.

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2021-02-01 00:00
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