382 CHAPTER 13. LEBESGUE MEASURE

If µ (E) = 0, there is nothing to prove so assume µ (E) > 0. Let U1 be an open set con-taining E with (1− r)µ (U1)< µ (E) and 2µ (E)> µ (U1) , and let F1 be those sets of Fwhich are contained in U1 whose centers are in E. Thus F1 is also a Vitali cover of E. Nowby the Besicovitch covering theorem proved earlier, there exist balls, B, of F1 such that

E ⊆ ∪Npi=1 {B : B ∈ Gi}

where Gi consists of a collection of disjoint balls of F1. Therefore,

µ (E)≤Np

∑i=1

∑B∈Gi

µ (B)

and so, for some i≤ Np,

(Np +1) ∑B∈Gi

µ (B)> µ (E) .

It follows there exists a finite set of balls of Gi, {B1, · · · ,Bm1} such that

(Np +1)m1

∑i=1

µ (Bi)> µ (E) (13.14.36)

and so

(2Np +2)m1

∑i=1

µ (Bi)> 2µ (E)> µ (U1) .

Since 2µ (E)≥ µ (U1) , 13.14.36 implies

µ (U1)

2N2 +2≤ 2µ (E)

2N2 +2=

µ (E)N2 +1

<m1

∑i=1

µ (Bi) .

Also U1 was chosen such that (1− r)µ (U1)< µ (E) , and so

λ µ (E)≥ λ (1− r)µ (U1) =

(1− 1

2Np +2

)µ (U1)

≥ µ (U1)−m1

∑i=1

µ (Bi) = µ (U1)−µ

(∪m1

j=1B j

)= µ

(U1 \∪m1

j=1B j

)≥ µ

(E \∪m1

j=1B j

).

Since the balls are closed, you can consider the sets of F which have empty intersectionwith ∪m1

j=1B j and this new collection of sets will be a Vitali cover of E \∪m1j=1B j. Letting

this collection of balls play the role of F in the above argument and letting E \∪m1j=1B j

play the role of E, repeat the above argument and obtain disjoint sets of F ,

{Bm1+1, · · · ,Bm2} ,