When dealing with probability distribution functions or some other Radon measure, it is
necessary to have a better covering theorem than the Vitali covering theorem which works
well for Lebesgue measure. However, for a Radon measure, if you enlarge the ball by making
the radius larger, you don’t know what happens to the measure of the enlarged ball except
that its measure does not get smaller. Thus the thing required is a covering theorem which
does not depend on enlarging balls.
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^{n}.
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 18.1.1Let the balls B_{a},B_{x},B_{y}be as shown, having radii r,r_{x},r_{y}respectively. Suppose the centers of B_{x}and B_{y}are not both in any of the balls shown,and suppose r_{y}≥ r_{x}≥ αr where α is a number larger than 1. Also let P_{x}≡ a+r
x||x−−-aa||
with P_{y}being defined similarly. Then it follows that
||Px − Py||
≥
αα−+11
r. There existsa constant L
(n,α)
depending on α and the dimension, such that if B_{1},
⋅⋅⋅
,B_{m}are allballs such that any pair are in the same situation relative to B_{a}as B_{x}, and B_{y}, thenm ≤ L
(n,α)
.
Proof:From the definition,
∥∥ x − a y− a ∥∥
||Px − Py|| = r∥∥||x-−-a|| − ||y−-a||∥∥
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
(n,α)
points. Thus m cannot be any larger than
L
(n,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
αα−+11
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 18.1.2Let B be a ball having radius r and suppose B has nonempty intersectionwith the balls B_{1},
⋅⋅⋅
,B_{m}having radii r_{1},
⋅⋅⋅
,r_{m}respectively, and as before, no B_{i}has thecenter of any other and the centers of the B_{i}are not contained in B. Suppose α,γ > 1 andthe r_{i}are comparable with r in the sense that
1-r ≤ 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
(n,α,β,γ )
such that m ≤ M
(n,α, β,γ)
. Lettingα = 10,β = 1∕3,γ = 4∕3, it follows that m ≤ 60^{n}.
Proof: Let the volume of a ball of radius r be given by α
(n)
r^{n} where α
(n)
depends on
the norm used and on the dimension n 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- n ∑m n n n
α(n) γr ≤ α (n)(βri) ≤ α(n)(1+ α + αβ) r
i=1 i=1
Therefore,
(β )n
m -- ≤ (1+ α +α β)n
γ
and so m ≤
(1+ α+ α β)
^{n}
( β)
γ
^{−n}≡ M
(n,α,β,γ)
.
From now on, let α = 10 and let β = 1∕3 and γ = 4∕3. Then
( )
172 n n
M (n,α,β,γ) ≤ 3 ≤ 60
Thus m ≤ 60^{n}. ■
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 18.1.3Let ℱ be a nonempty set of nonempty balls in ℝ^{n}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 ofballs from ℱ,
{Bj}
_{j=1}^{J}where J ≤∞ such that
r(B1) ≥ 3sup{r (B ) : B ∈ ℱ } (18.2)
4
(18.2)
and if
Am ≡ A ∖(∪mi=1Bi) ⁄= ∅, (18.3)
(18.3)
then B_{m+1}∈ℱ is chosen with center in A_{m}such that
r ≡ r (B ) ≥ 3sup{r : B (a,r) ∈ ℱ, a ∈ A } . (18.4)
m+1 m+1 4 m
(18.4)
Then letting B_{j} = B
(aj,rj)
, this sequence satisfies
r (Bk ) ≤ 4r (Bj ) for j < k, (18.5)
3
(18.5)
J
{B (aj,rj∕3)}j=1 are disjoint, (18.6)
(18.6)
A ⊆ ∪J B . (18.7)
i=1 i
(18.7)
Proof: Consider 18.5. First note the sets A_{m} form a decreasing sequence. Thus from the
definition of B_{j}, for j < k,
r (Bk ) ≤ sup{r : B (a,r) ∈ ℱ,a ∈ Ak −1}
4
≤ sup{r : B (a,r) ∈ ℱ,a ∈ Aj −1} ≤ 3r (Bj )
where these balls are two
which are chosen by the above scheme such that j > i, then from what was just
shown
( )
||a − a|| ≤ ||a − x||+ ||x− a || ≤ rj+ ri≤ 4 + 1 r = 7r < r
j i j i 3 3 9 3 i 9 i i
and this contradicts the construction because a_{j} is not covered by B
(ai,ri)
.
Finally consider the claim that A ⊆∪_{i=1}^{J}B_{i}. Pick B_{1} satisfying 18.2. If B_{1},
⋅⋅⋅
,B_{m} have
been chosen, and A_{m} is given in 18.3, 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}≥
34
r
(Ba)
for all m, a contradiction since the
balls B
(aj, rj3 )
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 18.1.1 to estimate the number of intersections of B_{k} with B_{j}
for j < k. In the second case, we use Lemma 18.1.2 to estimate the number of intersections of
B_{k} with B_{j} for j < k.
Now here is the Besicovitch covering theorem.
Theorem 18.1.4There exists a constant N_{n}, depending only on n with thefollowing property. If ℱ is any collection of nonempty balls in ℝ^{n}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_{Nn},such that each ℋ_{i}is a countable collection of disjoint balls from ℱ (possibly empty)and
+ 60^{n} + 1. Define the following sequence of subsets of ℱ,
G_{1},G_{2},
⋅⋅⋅
,G_{Mn}. 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 18.1.1 and
18.1.2. Here is why. B_{m+1} can intersect at most L
(n,10)
sets of
{B1,⋅⋅⋅,Bm }
which
have radii at least as large as 10B_{m+1} thanks to Lemma 18.1.1. It can intersect at
most 60^{n} sets of
{B1,⋅⋅⋅,Bm }
which have radius smaller than 10B_{m+1} thanks to
Lemma 18.1.2. 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_{Mn}
(Rm )
covering A_{m}. Then simply define
G_{j}^{′}≡∪_{k=1}^{∞}G_{j}
(R2k)
,G_{j}≡∪_{k=1}^{∞}G_{j}
(R2k−1)
. Let N_{n} = 2M_{n} and
{ℋ1,⋅⋅⋅,ℋN } ≡ {G′,⋅⋅⋅,G′ ,G1,⋅⋅⋅,GM }
n 1 Mn n
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}. ■