Definition 9.1.1A Radon measure is a complete Borel measure which is inner and outer regularand is finite on compact sets. Note that m_{p}is an example of a Radon measure.
There are other covering theorems such as the Vitali covering theorem, but the Vitali theorem depends on
adjusting the size of the balls. The Besicovitch covering theorem doesn’t do this. Instead, it estimates the
number of intersections. This is why it has no problem in dealing with arbitrary Radon measures. It is an
amazing result but even though this is the case, it really only depends on simple properties of normed
linear spaces.
The first fundamental observation is found in the following lemma which holds for the context
illustrated by the following picture. This picture is drawn such that the balls come from the usual
Euclidean norm, but the norm could be any norm on ℝ^{p}. Also, it is not particularly important whether the
balls are open or closed. They are just balls which may or may not contain points on their
boundary.
You have a set of balls and A is the union of their centers. The idea is to estimate the number of
possible intersections. Then one uses the pigeon hole principle to obtain a number N such that there are N
sets consisting of balls in original collection of balls and the balls in each of these sets are disjoint and the
union of all balls in all these sets gives A.
PICT
The idea is to consider balls B_{i} which intersect a given ball B such that B contains no center of any B_{i}
and no B_{i} contains the center of another B_{j}. There are two cases to consider, the case where the balls have
large radii and the case where the balls have small radii.
Intersections with big balls
Lemma 9.1.2Let the balls B_{a},B_{x},B_{y}be as shown, having radii r,r_{x},r_{y}respectively. Suppose thecenters of B_{x}and B_{y}are not both in any of the balls shown, and suppose r_{y}≥ r_{x}≥ αr where α isa number larger than 1. Also let P_{x}≡ a + r
-x−a-
||x−a||
with P_{y}being defined similarly. Then it followsthat
||Px − Py||
≥
α−-1
α+1
r. There exists a constant L
(p,α)
depending on α and the dimension, suchthat if B_{1},
⋅⋅⋅
,B_{m}are all balls such that any pair are in the same situation relative to B_{a}as B_{x},and B_{y}, then m ≤ L
(p,α )
.
Proof:From the definition,
∥ ∥
||P − P || = r∥∥-x-−-a − -y−-a--∥∥
x y ∥||x − a|| ||y− a||∥
( )
r ∥x−-y-∥− ---1---|∥y− a∥ − ∥x− a∥|
∥x− a∥ ∥x − a∥
( ∥x− y ∥ 1 )
= r ∥x−-a∥-− ∥x-−-a∥ (∥x − a∥ − ∥y− a∥)
Then since
∥x − a∥
≤ r + r_{x},
∥x − y∥
≥ r_{y},
∥y− a∥
≥ r_{y}, and recalling that r_{y}≥ r_{x}≥ αr
≥ --r--(ry − (r+ rx)+ ry) ≥-r---(ry − (r + ry) +ry)
rx + r rx + r ( )
≥ --r--(r − r) ≥---r--(r − r) ≥----r--- r − 1-r
rx + r y rx + r x rx + 1αrx x α x
r α− 1
= 1+-(1∕α) (1− 1∕α) = α+-1r
This proves the estimate between P_{x} and P_{y}.
Finally, in the case of the balls B_{i} having centers at x_{i}, then as above, let P_{xi} = a + r
-xi−a-
∥xi−a∥
. Then
(Pxi − a )
r^{−1} is on the unit sphere having center 0. Furthermore,
|||| −1 −1|||| −1 −1 α-−-1 α-− 1
(Pxi − a)r − (Pyi − a )r = r ||Pxi − Pyi|| ≥ r rα + 1 = α +1 .
How many points on the unit sphere can be pairwise this far apart? This set is compact and so there exists
a
1
4
( )
α−1
α+1
net having L
(p,α)
points. Thus m cannot be any larger than L
(p,α)
because if it were, then by
the pigeon hole principal, two of the points
(Pxi − a)
r^{−1} would lie in a single ball B
( ( ))
p, 14 αα−+11
so they
could not be
α−1
α+1
apart. ■
The above lemma has to do with balls which are relatively large intersecting a given ball. Next is a
lemma which has to do with relatively small balls intersecting a given ball. Note that in the statement of
this lemma, the radii are smaller than αr in contrast to the above lemma in which the radii of the balls are
larger than αr. In the application of this lemma, we will have γ = 4∕3 and β = 1∕3. These constants will
come from a construction, while α is just something larger than 1 which we will take here to equal
10.
Intersections with small but comparable balls
Lemma 9.1.3Let B be a ball having radius r and suppose B has nonempty intersection with the ballsB_{1},
⋅⋅⋅
,B_{m}having radii r_{1},
⋅⋅⋅
,r_{m}respectively, and as before, no B_{i}has the center of any other and thecenters of the B_{i}are not contained in B. Suppose α,γ > 1 and the r_{i}are comparable with r in the sensethat
1r ≤ ri ≤ αr.
γ
Let B_{i}^{′}have the same center as B_{i}with radius equal to r_{i}^{′} = βr_{i}for some β < 1. If the B_{i}^{′}are disjoint,then there exists a constant M
(p,α, β,γ)
such that m ≤ M
(p,α,β,γ )
. Letting α = 10,β = 1∕3,γ = 4∕3, itfollows that m ≤ 60^{p}.
Proof: Let the volume of a ball of radius r be given by α
(p)
r^{p} where α
(p)
depends on the norm used
and on the dimension p as indicated. The idea is to enlarge B, till it swallows all the B_{i}^{′}.
Then, since they are disjoint and their radii are not too small, there can’t be too many of
them.
This can be done for a single B_{i}^{′} by enlarging the radius of B to r + r_{i} + r_{i}^{′}.
PICT
Then to get all the B_{i}, you would just enlarge the radius of B to r + αr + βαr =
(1+ α + αβ)
r. Then,
using the inequality which makes r_{i} comparable to r, it follows that
( )
m∑ β- p m∑ p p p
α(p) γr ≤ α(p)(βri) ≤ α(p)(1+ α +α β) r
i=1 i=1
Therefore,
( β)p p
m γ- ≤ (1+ α + αβ)
and so m ≤
(1+ α+ α β)
^{p}
(β )
γ
^{−p}≡ M
(p,α,β,γ)
.
From now on, let α = 10 and let β = 1∕3 and γ = 4∕3. Then
(172)p
M (p,α,β,γ) ≤ -3- ≤ 60p
Thus m ≤ 60^{p}. ■
The next lemma gives a construction which yields balls which are comparable as described in the above
lemma. r
(B)
will denote the radius of the ball B.
A construction of a sequence of balls
Lemma 9.1.4Let ℱ be a nonempty set of nonempty balls in ℝ^{p}with
sup{diam (B ) : B ∈ ℱ } ≤ D < ∞
and let A denote the set of centers of these balls. Suppose A is bounded. Define a sequence of balls from ℱ,
{Bj}
_{j=1}^{J}where J ≤∞ such that
3
r(B1) ≥ 4 sup {r(B) : B ∈ ℱ } (9.2)
(9.2)
and if
Am ≡ A ∖ (∪mi=1Bi) ⁄= ∅, (9.3)
(9.3)
then B_{m+1}∈ℱ is chosen with center in A_{m}such that
3
rm+1 ≡ r(Bm+1) ≥ -sup {r : B (a,r) ∈ ℱ, a ∈ Am }. (9.4)
4
(9.4)
Then letting B_{j} = B
(aj,rj)
, this sequence satisfies
r (Bk ) ≤ 4r (Bj ) for j < k, (9.5)
3
(9.5)
{B (aj,rj∕3)}Jj=1 are disjoint, (9.6)
(9.6)
A ⊆ ∪Ji=1Bi. (9.7)
(9.7)
Proof: Consider 9.5. First note the sets A_{m} form a decreasing sequence. Thus from the definition of
B_{j}, for j < k,
where these balls are two which are chosen by the
above scheme such that j > i, then from what was just shown
r r ( 4 1 ) 7
||aj − ai|| ≤ ||aj − x||+ ||x − ai|| ≤-j +-i≤ -+ - ri = -ri < ri
3 3 9 3 9
and this contradicts the construction because a_{j} is not covered by B
(ai,ri)
for any i < j.
Finally consider the claim that A ⊆∪_{i=1}^{J}B_{i}. Pick B_{1} satisfying 9.2. If B_{1},
⋅⋅⋅
,B_{m} have been chosen,
and A_{m} is given in 9.3, then if it equals ∅, it follows A ⊆∪_{i=1}^{m}B_{i}. Set J = m. Now let a be the center of
B_{a}∈ℱ. If a ∈ A_{m} for all m,(That is a does not get covered by the B_{i}.) then r_{m+1}≥
3
4
r
(Ba)
for all m, a
contradiction since the balls B
( rj)
aj,3
are disjoint and A is bounded, implying that r_{j}→ 0.
Thus a must fail to be in some A_{m} which means it got covered by some ball in the sequence.
■
As explained above, in this sequence of balls from the above lemma, if j < k
3
4r (Bk ) ≤ r(Bj)
Then there are two cases to consider,
r(Bj ) ≥ 10r(Bk), r(Bj) ≤ 10r(Bk )
In the first case, we use Lemma 9.1.2 to estimate the number of intersections of B_{k} with B_{j} for j < k. In
the second case, we use Lemma 9.1.3 to estimate the number of intersections of B_{k} with B_{j} for
j < k.
Now here is the Besicovitch covering theorem.
Theorem 9.1.5There exists a constant N_{p}, depending only on p with the following property. If ℱ is anycollection of nonempty balls in ℝ^{p}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, ℋ_{1},
⋅⋅⋅
, ℋ_{Np}, such that eachℋ_{i}is a countable collection of disjoint balls from ℱ (possibly empty) and
+ 60^{p} + 1. Define the following sequence of subsets of ℱ, G_{1},G_{2},
⋅⋅⋅
,G_{Mp}.
Referring to the sequence
{Bk }
just considered, let B_{1}∈G_{1} and if B_{1},
⋅⋅⋅
,B_{m} have been assigned,
each to a G_{i}, place B_{m+1} in the first G_{j} such that B_{m+1} intersects no set already in G_{j}. The
existence of such a j follows from Lemmas 9.1.2 and 9.1.3. Here is why. B_{m+1} can intersect at
most L
(p,10)
sets of
{B1,⋅⋅⋅,Bm }
which have radii at least as large as 10B_{m+1} thanks to
Lemma 9.1.2. It can intersect at most 60^{p} sets of
{B1,⋅⋅⋅,Bm }
which have radius smaller than
10B_{m+1} thanks to Lemma 9.1.3. Thus each G_{j} consists of disjoint sets of ℱ and the set of
centers is covered by the union of these G_{j}. This proves the theorem in case the set of centers is
bounded.
Now let R_{1} = B
(0,5D)
and if R_{m} has been chosen, let
R = B (0,(m + 1)5D )∖ R
m+1 m
Thus, if
|k − m|
≥ 2, no ball from ℱ having nonempty intersection with R_{m} can intersect any ball from ℱ
which has nonempty intersection with R_{k}. This is because all these balls have radius less than
D. Now let A_{m}≡ A ∩ R_{m} and apply the above result for a bounded set of centers to those
balls of ℱ which intersect R_{m} to obtain sets of disjoint balls G_{1}
(Rm )
,G_{2}
(Rm)
,
⋅⋅⋅
,G_{Mp}
(Rm)
covering A_{m}. Then simply define G_{j}^{′}≡∪_{k=1}^{∞}G_{j}
(R2k )
,G_{j}≡∪_{k=1}^{∞}G_{j}
(R2k−1)
. Let N_{p} = 2M_{p}
and
{ } { ′ ′ }
ℋ1,⋅⋅⋅,ℋNp ≡ G1,⋅⋅⋅,GMp, G1,⋅⋅⋅,GMp
Note that the balls in G_{j}^{′} are disjoint. This is because those in G_{j}
(R2k)
are disjoint and if you consider any
ball in G_{j}
(R2k)
, it cannot intersect a ball of G_{j}
(R2m )
for m≠k because
|2k− 2m |
≥ 2. Similar
considerations apply to the balls of G_{j}. ■