There is another covering theorem which may also be referred to as the Besicovitch covering theorem. As
before, the balls can be taken with respect to any norm on ℝ^{p}. At first, the balls will be closed but this
assumption will be removed.
Definition 9.3.1A collection of balls, ℱ covers a set E in the sense of Vitali if whenever x ∈ Eand ε > 0, there exists a ball B ∈ℱ whose center is x having diameter less than ε.
I will give a proof of the following theorem.
Theorem 9.3.2Let μ be a Radon measure on ℝ^{p}and let E be a set with μ
(E)
< ∞. Where μis theouter measure determined by μ. Suppose ℱ is a collection of closed balls which cover E in the sense ofVitali. Then there exists a sequence of disjoint balls,
{Bi}
⊆ℱ such that
-( ∞ )
μ E ∖∪ j=1Bj = 0.
Proof:Let N_{p} be the constant of the Besicovitch covering theorem. Choose r > 0 such
that
−1 ( --1---)
(1 − r) 1− 2Np + 2 ≡ λ < 1.
If μ
(E )
= 0, there is nothing to prove so assume μ
(E)
> 0. Let U_{1} be an open set containing E with
(1 − r)
μ
(U )
1
< μ
(E )
and 2μ
(E )
> μ
(U )
1
, and let ℱ_{1} be those sets of ℱ which are contained in U_{1} whose
centers are in E. Thus ℱ_{1} is also a Vitali cover of E. Now by the Besicovitch covering theorem proved
earlier, there exist balls B, of ℱ_{1} such that
E ⊆ ∪Ni=p1 {B : B ∈ Gi}
where G_{i} consists of a collection of disjoint balls of ℱ_{1}. Therefore,
-- N∑p ∑
μ(E) ≤ μ(B)
i=1 B∈Gi
and so, for some i ≤ N_{p},
∑ --
(Np + 1) μ(B) > μ(E) .
B∈Gi
It follows there exists a finite set of balls of G_{i},
Since the balls are closed, you can consider the sets of ℱ which have empty intersection with ∪_{j=1}^{m1}B_{j}
and this new collection of sets will be a Vitali cover of E ∖∪_{j=1}^{m1}B_{j}. Letting this collection of balls play
the role of ℱ in the above argument and letting E ∖∪_{j=1}^{m1}B_{j} play the role of E, repeat the above
argument and obtain disjoint sets of ℱ,
Continuing in this way, yields a sequence of disjoint balls
{Bi}
contained in ℱ and
( ) ( )
μ- E ∖∪ ∞j=1Bj ≤ μ- E ∖∪mkj=1Bj < λkμ-(E )
for all k. Therefore, μ
(E ∖∪∞ B )
j=1 j
= 0 and this proves the Theorem. ■
It is not necessary to assume μ
(E )
< ∞.
Corollary 9.3.3Let μ be a Radon measure on ℝ^{p}. Letting μbe the outer measure determined by μ,suppose ℱ is a collection of closed balls which cover E in the sense of Vitali. Then there exists a sequenceof disjoint balls,
{Bi }
⊆ℱ such that
-( )
μ E ∖∪ ∞j=1Bj = 0.
Proof: Since μ is a Radon measure it is finite on compact sets. Therefore, there are at most countably
many numbers,
{bi}
_{i=1}^{∞} such that μ
(∂B (0,bi))
> 0. It follows there exists an increasing
sequence of positive numbers,
You don’t need to assume the balls are closed. In fact, the balls can be open, closed or anything in
between and the same conclusion can be drawn.
Corollary 9.3.4Let μ be a Radonmeasure on ℝ^{p}. Letting μbe the outer measure determined by μ,suppose ℱ is a collection of balls which cover E in the sense of Vitali, open closed or neither. Then thereexists a sequence of disjoint balls,
{Bi}
⊆ℱ such that
( )
μ-E ∖∪ ∞j=1Bj = 0.
Proof: Let x ∈ E. Thus x is the center of arbitrarily small balls from ℱ. Since μ is a Radon measure,
at most countably many radii, r of these balls can have the property that μ
(∂B (0,r))
= 0. Let
ℱ^{′} denote the closures of the balls of ℱ, B
(x,r)
with the property that μ
(∂B (x,r))
= 0.
Since for each x ∈ E there are only countably many exceptions, ℱ^{′} is still a Vitali cover of E.
Therefore, by Corollary 9.3.3 there is a disjoint sequence of these balls of ℱ^{′},
{--}
Bi
_{i=1}^{∞} for
which
-( --)
μ E ∖ ∪∞j=1Bj = 0
However, since their boundaries have μ measure zero, it follows