14.12. THE VITALI COVERING THEOREMS 345

Then {B(x,rx) : x ∈ [M f > λ ]} is an open cover of [M f > λ ]. By the Vitali coveringtheorem, there are countably many of these balls {Bi}∞

i=1 which are disjoint and ∪iB̂i ⊇[M f > λ ] . Therefore,

m̄([M f > λ ]) ≤ m(∪iB̂i

)≤

∞

∑i=1

m(B̂i)

= 5∞

∑i=1

m(Bi)≤ 5∞

∑i=1

1λ

∫Bi

| f |dm≤ 5λ

∫| f |dm

the last step resulting from the fact that the balls Bi are disjoint.Next is a version of the Vitali covering theorem which involves covering a set with

disjoint closed balls in such a way that what is left over has measure 0. Here is the conceptof a Vitali covering.

Definition 14.12.8 Let S be a set and let C be a covering of S meaning that everypoint of S is contained in a set of C . This covering is said to be a Vitali covering if for eachε > 0 and x ∈ S, there exists a set B ∈ C containing x, the diameter of B is less than ε, andthere exists an upper bound to the set of diameters of sets of C .

In the following, F will just be a bounded set measurable or not. Actually, you don’teven need to consider it to be bounded but this will not be needed here.

Corollary 14.12.9 Let F be a bounded set and let C be a Vitali covering of F con-sisting of balls. Let r (B) denote the radius of one of these balls. Then assume also thatsup{r (B) : B ∈ C }= M < ∞. Then there is a countable subset of C denoted by {Bi} suchthat m̄

(F \∪N

i=1Bi)= 0,N ≤ ∞, and Bi∩B j = /0 whenever i ̸= j.

Proof: If m̄(F) = 0, there is nothing to show. Assume then that m̄(F) ̸= 0. Let U be abounded open set containing F such that U approximates F so well that m(U)≤ 10

9 m̄(F) .This is possible because of regularity of m shown earlier. Since this is a Vitali covering, foreach x ∈ F, there is one of these balls B containing x such that B̂ ⊆U . Recall this meansyou keep the same center but make the radius 5 times as large. Let Ĉ denote those ballsof C such that B̂⊆U also. Thus, this is also a cover of F . By the Vitali covering theoremabove, Proposition 14.12.3, there are disjoint balls from C {Bi} such that

{B̂i}

covers F .Here the B̂i are open balls. Thus

m̄(F \∪∞

j=1B j)≤ m(U)−m

(∪∞

j=1B j)<

109

m̄(F)−∞

∑j=1

m(B j)

=109

m̄(F)− 15

∞

∑j=1

m(

B̂ j

)≤ 10

9m̄(F)− 1

5m̄(F) = m̄(F)θ 1

Here θ 1 =4145 < 1.

Now F \∪nk=1Bk ⊆

(F \∪∞

k=1Bk)∪(∪n

k=1Bk). Indeed, if x is in F but not in any of the

Bk for k≤ n, then if it fails to be in F \∪∞k=1Bk, so it must be in the complement of this set.

Hence it is in(∩∞

k=1

(F ∩BC

k

))C= ∪∞

k=1

(FC ∪Bk

)but it is not in any of the Bk for k ≤ n

while being in F so it must be in ∪∞k=n+1Bk. It follows, since the Bk are disjoint, that

m̄(F \∪n

j=1B j)≤ m̄

(F \∪∞

j=1B j)+

∞

∑k=n

m(Bk)≤ θ 1m̄(F)+∞

∑k=n

m(Bk) ,θ 1 < 1