Kenneth Kuttler

380 CHAPTER 13. LEBESGUE MEASURE

Lemma 13.13.3 Let Γ > 1 and B(a,Γr) be a ball and suppose {B(xi,ri)}mi=1 are balls

contained in B(a,Γr) such that r≤ ri and none of these balls contains the center of anotherball. Then there is a constant M (p,Γ) such that m≤M (p,Γ).

Proof: Let zi = xi − a. Then B(zi,ri) are balls contained in B(0,Γr) with no ballcontaining a center of another. Then B

( ziΓr ,

riΓr

)are balls in B(0,1) with no ball containing

the center of another. By compactness, there is a 18Γ

net for B(0,1), {yi}M(p,Γ)i=1 . Thus the

balls B(yi,

18Γ

)cover B(0,1). If m ≥ M (p,Γ) , then by the pigeon hole principle, one of

these B(yi,

18Γ

)would contain some zi

Γr and z jΓr which requires

∥∥ ziΓr −

z jΓr

∥∥ ≤ 14Γ

<r j

4Γr soziΓr ∈ B

( z jΓr ,

r jΓr

). Thus m≤M (p,γ,Γ).

Intersections with small balls

Lemma 13.13.4 Let B be a ball having radius r and suppose B has nonempty intersectionwith the balls B1, · · · ,Bm having radii r1, · · · ,rm respectively, and as before, no Bi containsthe center of any other and the centers of the Bi are not contained in B. Suppose α > 1and r ≤ min(r1, · · · ,rm), each ri < αr. Then there exists a constant M (p,α) such thatm≤M (p,α).

Proof: Let B = B(a,r). Then each Bi is contained in B(a,2r+αr+αr) . This isbecause if y ∈ Bi ≡ B(xi,ri) ,

∥y−a∥ ≤ ∥y−xi∥+∥xi−a∥ ≤ ri + r+ ri < 2r+αr+αr

Thus Bi does not contain the center of any other B j. Then these balls are contained inB(a,r (2α +2)) , and each radius is at least as large as r. By Lemma 13.13.3 there is aconstant M (p,α) such that m≤M (p,α).

Now here is the Besicovitch covering theorem. In the proof, we are considering thesequence of balls described above.

Theorem 13.13.5 There exists a constant Np, depending only on p with the following prop-erty. If F is any collection of nonempty balls in X with

sup{diam(B) : B ∈F}< D < ∞

and if A is the set of centers of the balls in F , then there exist subsets of F , H1, · · · , HNp ,such that each Hi is a countable collection of disjoint balls from F (possibly empty) and

A⊆ ∪Npi=1∪{B : B ∈Hi}.

Proof: To begin with, suppose A is bounded. Let L(p,α) be the constant of Lemma13.13.2 and let Mp = L(p,α) +M (p,α) + 1. Define the following sequence of subsetsof F , G1,G2, · · · ,GMp . Referring to the sequence {Bk} considered in Lemma 13.13.1, letB1 ∈ G1 and if B1, · · · ,Bm have been assigned, each to a Gi, place Bm+1 in the first G j suchthat Bm+1 intersects no set already in G j. The existence of such a j follows from Lemmas13.13.2 and 13.13.4 and the pigeon hole principle. Here is why. Bm+1 can intersect atmost L(p,α) sets of {B1, · · · ,Bm} which have radii at least as large as αr (Bm+1) thanks to

380 CHAPTER 13. LEBESGUE MEASURELemma 13.13.3 Let T > 1 and B(a,Tr) be a ball and suppose {B(x;,rj)}j_, are ballscontained in B(a,Tr) such that r < r; and none of these balls contains the center of anotherball. Then there is a constant M (p,T) such thatm < M(p,T).Proof: Let z; = x;—a. Then B(z;,r;) are balls contained in B(0,Ir) with no ballcontaining a center of another. Then B (4, i) are balls in B(0,1) with no ball containingthe center of another. By compactness, there is a g- net for B(0, 1), fy}? ") Thus theballs B (y;, xr) cover B(0,1). If m > M(p,T), then by the pigeon hole principle, one ofthese B (yi, ar) would contain some i and iz which requires |a eB(#,#). Thusm<M(p,y7,T).Ze By 1 Thfi - rl|| < a < a 80Intersections with small ballsLemma 13.13.4 Let B be a ball having radius r and suppose B has nonempty intersectionwith the balls B,,-++ , Bm having radii r),--- ,’m respectively, and as before, no B; containsthe center of any other and the centers of the B; are not contained in B. Suppose 4 > 1and r <min(r},--- fm), each rj < ar. Then there exists a constant M(p,0) such thatm<M(p,@).Proof: Let B = B(a,r). Then each B; is contained in B(a,2r+ar+ar). This isbecause if y € B; = B(x;, ri),lly —all < lly —x;|| + ||xi—al] Sri t+) <2rt+artarThus B; does not contain the center of any other B;. Then these balls are contained inB(a,r(2@+2)), and each radius is at least as large as r. By Lemma 13.13.3 there is aconstant M (p,q) such thatm <M (p,q). #Now here is the Besicovitch covering theorem. In the proof, we are considering thesequence of balls described above.Theorem 13.13.5 There exists a constant Np, depending only on p with the following prop-erty. If F is any collection of nonempty balls in X withsup {diam (B): BEF} <D<coand if A is the set of centers of the balls in F, then there exist subsets of F, F4,-+-, HN,such that each FG is a countable collection of disjoint balls from F (possibly empty) andA CUM U{B: Be 4G}.Proof: To begin with, suppose A is bounded. Let L(p,a) be the constant of Lemma13.13.2 and let M, = L(p,a)+M(p,a@) +1. Define the following sequence of subsetsof F,4,,%,::- ,Gu,- Referring to the sequence {B;} considered in Lemma 13.13.1, letB, €Y, and if B),--- ,B», have been assigned, each to a Y, place By,+1 in the first Y suchthat B,,41 intersects no set already in Y;. The existence of such a j follows from Lemmas13.13.2 and 13.13.4 and the pigeon hole principle. Here is why. By») can intersect atmost L(p, a) sets of {B,,--- , Bm} which have radii at least as large as @r (Bm+1) thanks to