Departments of Mathematics
Montana State University
Fall 2015
Prof. Kevin Wildrick
An introduction to non-smooth analysis and geometry
Lecture 3: Covering theorems
Although it is not obvious, differentiation results are almost always based on covering
theorems.
1. The 5B covering lemma
The following statement is a fact about metric spaces, and does not involve measures.
Given a ball B = B(x, r), we write 5B := B(x, 5r).
Theorem 1.1 (The 5B covering lemma). Let (X, d) be a metric space, and let F be
any non-empty collection of closed balls in (X, d) with uniformly bounded and positive
diameter. Then there is a disjointed subfamily G of F such that
[
[
B⊆
5B.
B∈F
B∈G
If (X, d) is separable, then G is countable.
Proof. We will argue using the Hausdorff maximality principle, (a.k.a. the axiom of
choice). Let Ω denote the set of all disjointed subfamilies B of F with the following
property: if a ball B from F meets any ball from B, then it meets a ball from B whose
radius is more than half that of B.
Let B0 be a ball from F such that
supB∈F radius(B)
< radius(B0 ).
2
Then {B0 } is an element of Ω, showing that Ω is non-empty. We may partially order Ω
by inclusion. If Γ is a totally-ordered subset of Ω, then its union is again an element of
Ω. Hence, by the Hausdorff maximality principle, Ω has some maximal element G.
We now claim that each ball B(x, r) ∈ F meets some ball B(x0 , r0 ) ∈ G. If the claim
holds, then since G is an element of Ω, we may assume that
r
r0 ≥ .
2
Hence, by the triangle inequality
B(x, r) ⊆ B(x0 , r + r + r0 ) ⊆ B(x0 , 5r0 ).
Suppose that the claim fails. Then there is some ball B in F that does not intersect any
ball in G, which implies that G ∪ {B} is an element of Ω. This contradicts the maximality
of G.
To see that G is countable if X is separable, recall that a separable space cannot contain
uncountably many disjoint open balls.
Note that there is quite a bit of flexibility in this proof - the assumption that the balls
were closed was not important, nor do we have to work exactly with balls.
1
2. Doubling Metric Measure Spaces
One intuitive notion of dimension comes from repeatedly decomposing a set of a given
volume into smaller pieces and seeing the rate at which the number of pieces increases.
We can apply this to the metric measure space setting.
Definition 2.1. Let (X, d, µ) be a metric measure space. If there is a constant C ≥ 1
such that for each x ∈ X and r > 0,
µ(B(x, 2r)) ≤ Cµ(B(x, r)),
then we say that (X, d, µ) is doubling.
The corresponding notion of dimension is defined by
Definition 2.2. Let 0 ≤ s < ∞. We say that a metric measure space (X, d, µ) is shomogeneous if there is a constant C ≥ 1 such that whenever x ∈ X and 0 < r ≤ R < ∞,
then
s
µ(B(x, R))
R
≤C
.
µ(B(x, r))
r
Proposition 2.3. A doubling metric measure space is s-homogeneous for some 0 < s <
∞.
Proof. Find k ∈ N such that 2−(k+1) R < r ≤ 2−k R and apply the doubling condition k
times.
Some non-examples of doubling metric measure spaces include
• Hyperbolic space with the standard volume (fails for large balls)
• Let H be a separable Hilbert space, and let X be the union of the line segments
in the coordinate directions minus the origin. Equip X with the distance from H
and one-dimensional Hausdorff measure, i.e., length measure. Then each point of
X looks like the real line with lebesgue measure on sufficiently small scales, but
the measure is not doubling (even when restricted to arbitrarily small balls).
We will try to restrict ourselves to doubling measures in this course; basically every
example we have or will consider will be doubling.
Every doubling metric measure space supports the necessary covering theorems required
for the most basic differentiation results.
Definition 2.4. Let (X, d) be a metric space and let A ⊆ X. A covering F of A by
closed balls is called fine if for every point a ∈ A and > 0, there is some 0 < r < such
that a ∈ B(x, r) ∈ F.
Theorem 2.5 (Vitali’s covering theorem). Let (X, d, µ) be a doubling metric measure
space, and let F be a fine cover of a subset A ⊆ X. Then there is a disjoint subcollection
C ⊆ F such that
!
[
µ A\
B = 0.
B∈C
Remark 2.6. In fact, the assumption that (X, d, µ) is doubling is a bit too strong; as the
proof will (essentially) show, we need only that
µ(B(x, 2r))
lim sup
<∞
µ(B(x, r))
r→0
for µ-almost every x ∈ X. This is much weaker than doubling (for example, both examples
of non-doubling spaces above satisfy this condition), but we will need doubling for many
more things in this course. A metric measure space (X, d, µ) that satisfies the conclusion
of Theorem 2.5 is called a Vitali space.
Proof of Theorem 2.5. We assume for simplicity (and leave the remaining work for the
reader) that A is bounded and that the radius of a ball in the cover F is uniformly
bounded above. Since µ is doubling, it gives finite measure to any bounded set, and so
!
[
µ
B < ∞.
B∈F
Applying the 5B covering lemma, there is a disjointed, countable collection G ⊆ F such
that
[
A⊆
5B.
B∈G
Let us write G = {Bi }i∈N . Since the measure µ is doubling,
!
[
X
X
µ
5Bi ≤
µ(5Bi ) ≤ C 3
µ(Bi ) ≤ C 3 µ
i∈N
i∈N
i∈N
!
[
Bi
< ∞.
i∈N
This implies that
lim
N →∞
X
µ(5Bi ) = 0.
i≥N
It now suffices to show that
A\
N
[
!
⊆
Bi
i=1
5Bi .
i=N +1
If
a∈
∞
[
A\
N
[
!
Bi
,
i=1
then since F is a fine covering by closed balls, we may find x ∈ X and r > 0 such that
a ∈ B(x, r) ∈ F and
!
N
[
B(x, r) ∩
Bi = ∅.
i=1
However, in the proof of the 5B covering lemma, the punchline was that each ball of F is
contained in 5Bj for some j ∈ N. In the case of B(x, r) we necessarily have j > N.