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^{n}. At first, the balls
will be closed but this assumption will be removed.
Definition 18.3.1A collection of balls, ℱ covers a set, E in the sense of Vitaliif whenever x ∈ E and ε > 0, there exists a ball B ∈ ℱ whose center is x havingdiameter less than ε.
I will give a proof of the following theorem.
Theorem 18.3.2Let μ be a Radon measure on ℝ^{n}and let E be a set withμ
(E )
< ∞. Where μis the outer measure determined by μ. Suppose ℱ is a collection of closedballs which cover E in the sense of Vitali. Then there exists a sequence of disjoint balls,
{Bi}
⊆ℱ such that
( )
μ-E ∖∪∞j=1Bj = 0.
Proof:Let N_{n} be the constant of the Besicovitch covering theorem, Theorem 18.1.4.
Choose r > 0 such that
−1( 1 )
(1 − r) 1− 2N--+-2 ≡ λ < 1.
n
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
(U1)
< μ
(E)
and 2μ
(E)
> μ
(U1)
, 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, Theorem 18.1.4, there exist balls, B, of ℱ_{1} such
that
E ⊆ ∪Nni=1 {B : B ∈ Gi}
where G_{i} consists of a collection of disjoint balls of ℱ_{1}. Therefore,
-- ∑Nn ∑
μ (E) ≤ μ(B)
i=1 B∈Gi
and so, for some i ≤ N_{n},
(Nn + 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
{B }
i
contained in ℱ
and
( ) ( )
μ- E ∖∪∞j=1Bj ≤ μ- E ∖∪mjk=1Bj < λkμ-(E )
for all k. Therefore, μ
(E ∖∪∞ Bj)
j=1
= 0 and this proves the Theorem. ■
It is not necessary to assume μ
(E )
< ∞.
Corollary 18.3.3Let μ be a Radon measure on ℝ^{n}. Letting μbe the outermeasure determined by μ, suppose ℱ is a collection of closed balls which cover E inthe sense of Vitali. Then there exists a sequence of disjoint balls,
{Bi}
⊆ℱ suchthat
-( )
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,
{ri}
_{i=1}^{∞} such that lim_{i→∞}r_{i} = ∞ and
μ
(∂B (0,ri))
= 0. Now let
D1 ≡ {x : ||x|| < r1},D2 ≡ {x : r1 < ||x|| < r2},
⋅⋅⋅,D ≡ {x : r < ||x || < r } ,⋅⋅⋅.
m m −1 m
Let ℱ_{m} denote those closed balls of ℱ which are contained in D_{m}. Then letting E_{m} denote
E ∩ D_{m}, ℱ_{m} is a Vitali cover of E_{m},μ
(Em )
< ∞, and so by Theorem 18.3.2, there exists a
countable sequence of balls from ℱ_{m}
{ m }
Bj
_{j=1}^{∞}, such that μ
( ∞ m)
Em ∖∪j=1B j
= 0. Then
consider the countable collection of balls,
{ m }
Bj
_{j,m=1}^{∞}.
-( ∞ ∞ m ) -( ∞ )
μ E ∖∪ m=1 ∪ j=1 Bj ≤ μ ∪ j=1∂B (0,ri) +
∑∞ -( ∞ m )
+ μ Em ∖ ∪j=1Bj = 0 ■
m=1
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 18.3.4Let μ be a Radonmeasure on ℝ^{n}. Letting μbe the outer measuredetermined by μ, suppose ℱ is a collection of balls which cover E in the sense of Vitali, openclosed or neither. Then there exists a sequence of disjoint balls,
{Bi}
⊆ℱ suchthat
( )
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 18.3.3 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