There are other covering theorems such as the Vitali covering theorem, but this 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 and so fits in well with linear algebra.
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}. 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.
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 21.2.1Let the balls B_{a},B_{x},B_{y}be as shown, having radii r,r_{x},r_{y}respectively. Supposethe centers of B_{x}and B_{y}are not both in any of the balls shown, and suppose r_{y}≥ r_{x}≥ αr where αis a 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||
≥
αα−+11
r. There exists a constant L
(n,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
(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
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 21.2.2Let 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 ≤ r ≤ αr.
γ i
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)
. Letting α = 10,β = 1∕3,γ = 4∕3, itfollows 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
where these balls are two which are chosen by the
above scheme such that j > i, then from what was just shown
rj ri ( 4 1 ) 7
||aj − ai|| ≤ ||aj − x||+ ||x − ai|| ≤-3 + 3-≤ 9 + 3 ri = 9ri < ri
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 21.10. If B_{1},
⋅⋅⋅
,B_{m} have been chosen,
and A_{m} is given in 21.11, 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
(aj, rj)
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
3r (Bk ) ≤ r(Bj)
4
Then there are two cases to consider,
r(Bj ) ≥ 10r(Bk), r(Bj) ≤ 10r(Bk )
In the first case, we use Lemma 21.2.1 to estimate the number of intersections of B_{k} with B_{j} for j < k. In
the second case, we use Lemma 21.2.2 to estimate the number of intersections of B_{k} with B_{j} for
j < k.
Now here is the Besicovitch covering theorem.
Theorem 21.2.4There exists a constant N_{n}, depending only on n with the following property. If ℱ isany 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 eachℋ_{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 21.2.1 and 21.2.2. 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 21.2.1. It can intersect at most 60^{n} sets of
{B1,⋅⋅⋅,Bm }
which have radius smaller than
10B_{m+1} thanks to Lemma 21.2.2. 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
Rm+1 = B (0,(m + 1)5D )∖ Rm
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}
(R )
m
,G_{2}
(R )
m
,
⋅⋅⋅
,G_{Mn}
(R )
m
covering A_{m}. Then simply define G_{j}^{′}≡∪_{k=1}^{∞}G_{j}
(R )
2k
,G_{j}≡∪_{k=1}^{∞}G_{j}
(R )
2k−1
. Let N_{n} = 2M_{n}
and
{ℋ1,⋅⋅⋅,ℋNn } ≡ {G′1,⋅⋅⋅,G′M ,G1,⋅⋅⋅,GMn}
n
Note that the balls in G_{j}^{′} are disjoint. This is because those in G_{j}
(R )
2k
are disjoint and if you consider any
ball in G_{j}
(R )
2k
, it cannot intersect a ball of G_{j}
(R )
2m
for m≠k because
|2k− 2m |
≥ 2. Similar
considerations apply to the balls of G_{j}. ■
Now let f : ℝ^{n}→ ℝ^{n} be a Lipschitz function. This means there is a constant K such that
∥f (x)− f (y)∥ ≤ K ∥x − y∥
where
∥⋅∥
denotes a norm on ℝ^{n}. For example, f
(x)
could equal Ax where A is an n × n matrix. In this
case,
∥Ax − Ay∥ ≤ ∥A∥∥x − y∥
with
∥A∥
the operator norm. Then the following proposition is a fundamental result which says that a
Lipschitz map of a measurable set is a measurable set. This is a remarkable result because it is not even
true if f is only continuous. For example, see Problem 8 on Page 1491.
Proposition 21.2.5Let f : ℝ^{n}→ ℝ^{n}be Lipschitz continuous with Lipschitz constant K. Then ifF is a Lebesgue measurable set, then so is f
(K )
. Also if N is a set of measure zero, then f
(N )
isa set of measure zero.
Proof: Consider the second claim first. Let V be an open set, m_{n}
(V )
< ε, and V ⊇ K. For each point
x of K, there is a closed ball B
(x,r)
⊆ V such that also r < 1. Let ℱ denote this collection of balls and let
{ℋ1,⋅⋅⋅,ℋNn }
be the finite set each of which consists of countably many disjoint balls from ℱ whose
union includes all of N. Denote by f
(ℋk)
the set of f
(B )
where B ∈ℋ_{k}. Each is compact because f is
continuous and B is closed. Thus f
(N )
is contained in the union of the f
(ℋk)
. It follows
that
∑Nn ∑
mn-(f (N)) ≤ mn (f (B ))
k=1B ∈ℋk
Here the bar on the measure denotes the outer measure determined by m_{n}. Now f
is Lebesgue measurable and in fact is a
set of measure zero. This is by completeness of Lebesgue measure. See Proposition 19.9.6 and completeness
of Lebesgue measure.
For the first part, suppose F is Lebesgue measurable. Then there exists H ⊆ F such that H is
the countable union of compact sets and also that m_{n}
(H)
= m_{n}
(F)
. Thus H is measurable
because it is the union of measurable sets. Say m_{n}
(F)
< ∞ first. Then m_{n}
(F ∖ H)
= 0 and so
f
(F ∖ H)
is a measurable set of measure zero. Then f
(F)
= f
(F ∖H )
∪ f
(H )
. The second is
obviously measurable because it is a countable union of compact sets, the continuous image of
a compact set being compact. It follows that f
(F)
is also measurable. In general, consider
F_{k}≡ F ∩