# Covering theorems#

We will use $$B(x,r)$$ to denote the closed ball in a metric space $$X$$ centred at $$x\in X$$ with radius $$r \geq 0$$. Since the centre and radius of a ball are not uniquely defined by its elements, formally by a “ball” we mean a pair $$(x,r)\in X \times (0,\infty)$$, but in practice we mean the set of its elements.

Lemma 4 ($$5r$$ covering lemma)

Let $$X$$ be a metric space and $$\mathcal B$$ an arbitrary collection of closed balls of uniformly bounded radii. There exists a disjoint sub-collection $$\mathcal B'\subset \mathcal B$$ such that any $$B\in \mathcal B$$ intersects a ball $$B'\in\mathcal B'$$ with

$\operatorname{rad}B'\geq \operatorname{rad}B/2.$

In particular,

$\bigcup_{B\in \mathcal B'}5B \supset \bigcup_{B\in \mathcal B}B.$

Here, $$5B$$ denotes the ball with the same centre as $$B$$ and 5 times the radius.

Proof. For each $$n\in \mathbb Z$$ let

$\mathcal B_n = \{B\in \mathcal B: 2^n \leq \operatorname{rad}B < 2^{n+1}\}.$

Since the balls in $$\mathcal B$$ have uniformly bounded radii, the $$\mathcal B_n$$ are empty for all $$n> N$$, for some $$N\in\mathbb N$$. Let $$\mathcal B'_N$$ be a maximal disjoint sub-collection of $$\mathcal B_N$$. That is, the elements of $$\mathcal B'_N$$ are disjoint elements of $$\mathcal B_N$$ and if $$B\in \mathcal B_N$$, there exists a $$B'\in\mathcal B'_N$$ with $$B\cap B'\neq \emptyset$$. (In general such a maximal collection exists by Zorn’s lemma. See also Exercise 42.) Let $$\mathcal B'_{N-1}$$ be a maximal collection such that $$\mathcal B'_N\cup \mathcal B'_{N-1}$$ is a disjoint collection. Repeat this for each $$i\in \mathbb N$$, obtaining a maximal collection $$\mathcal B'_{N-i}$$ such that $$\mathcal B'_N\cup \ldots \cup \mathcal B'_{N-i}$$ is a disjoint collection, and set $$\mathcal B'=\bigcup_{n\leq N}\mathcal B'_n$$.

Now suppose that $$B\in \mathcal B$$, say $$B\in \mathcal B_n$$. Then by construction there exists $$B'\in \mathcal B'_m$$ for some $$m\geq n$$ with $$B\cap B'\neq \emptyset$$. In particular, $$\operatorname{rad}B'\geq \operatorname{rad}B/2$$.

The final statement of the lemma follows from the triangle inequality.

Definition 15 (Vitali cover)

Let $$X$$ be a metric space and $$S\subset X$$. A Vitali cover of $$S$$ is a collection $$\mathcal B$$ of closed balls such that, for each $$x\in S$$ and each $$\epsilon>0$$, there exists a ball $$B\in \mathcal B$$ with $$\operatorname{rad}B<\epsilon$$ and $$x\in B$$.

Proposition 1

Let $$X$$ be a metric space, $$S\subset X$$ and suppose that $$\mathcal B$$ is a Vitali cover of $$S$$. Then there exists a disjoint $$\mathcal B'\subset \mathcal B$$ such that, for every finite $$I\subset \mathcal B'$$,

$S\setminus \bigcup_{B\in I}B \subset \bigcup_{B\in \mathcal B'\setminus I}5B.$

In particular, if $$\mathcal B'=\{B_1,B_2,\ldots\}$$ is countable (for example, if $$X$$ is separable), then

$S\setminus \bigcup_{i=1}^n B_i \subset \bigcup_{i>n} 5B_i$

for each $$n\in\mathbb N$$.

Proof. Note that we may suppose $$\mathcal B$$ consists of balls with uniformly bounded radii. Let $$\mathcal B'$$ be a disjoint sub-collection of $$\mathcal B$$ obtained from Lemma 4. If $$I\subset \mathcal B'$$ is finite then

$C:=\bigcup_{B\in I}B$

is closed. Therefore, if $$x \in S\setminus C$$, since $$\mathcal B$$ is a Vitali cover of $$S$$, there exists $$B\in \mathcal B$$ with $$x\in B$$ such that $$B \cap C=\emptyset$$. However, $$B$$ must intersect some $$B'\in\mathcal B'$$ with $$\operatorname{rad}B'\geq \operatorname{rad}B/2$$, and so $$x\in 5B'$$. That is, $$x$$ belongs to

$\bigcup_{B\in \mathcal B'\setminus I}5B,$

as required.

Definition 16 (Doubling measure)

A Borel measure $$\mu$$ on a metric space $$X$$ is a doubling measure if there exists a $$C_{\mu}\geq 1$$ such that

$0<\mu(2B)\leq C_{\mu}\mu(B)<\infty$

for all balls $$B\subset X$$.

Remark 4

Note that, for any $$m\geq 2$$,

$\mu(mB)\leq C_{\mu}^{\log_2 m} \mu(B).$

Lebesgue measure is a doubling measure.

Theorem 9 (Vitali covering theorem)

Let $$\mu$$ be a doubling measure on a metric space $$X$$ and let $$\mathcal B$$ be a Vitali cover of a set $$S\subset X$$. There exists a countable disjoint $$\mathcal B'\subset \mathcal B$$ such that

$\mu\left(S\setminus \bigcup_{B\in\mathcal B'}B\right)=0.$

Proof. First note that it suffices to prove the result for $$S$$ bounded, say $$S$$ is contained in some ball $$\tilde B$$. We may also suppose that each $$B\in\mathcal B$$ is a subset of $$2\tilde B$$.

Let $$\mathcal B'$$ be a disjoint sub-collection of $$\mathcal B$$ obtained from Proposition 1. Note that $$\mathcal B'$$ is countable. Indeed, for each $$m\in\mathbb N$$, at most $$m\mu(2\tilde B)$$ balls $$B\in\mathcal B'$$ can satisfy $$\mu(B)>1/m$$.

Enumerate $$\mathcal B'=\{B_1,B_2,\ldots\}$$. Since the $$B_i$$ are disjoint subsets of $$2\tilde B$$,

$\sum_{i>n}\mu(B_i)\to 0.$

By the conclusion of Proposition 1,

$S \setminus \bigcup_{i=1}^n B_i \subset \bigcup_{i>n} 5B_i$

for each $$n\in\mathbb N$$. Since $$\mu$$ is doubling, $$\mu(5B_i) \leq C\mu(B_i)$$ for each $$i\in\mathbb N$$ and so

$\mu\left(S \setminus \bigcup_{i=1}^n B_i\right) \leq C\sum_{i>n} \mu(B_i)\to 0,$

as required.

Definition 17 (Hardy–Littlewood maximal function)

Let $$\mu$$ be a Borel measure on a metric space $$X$$ and $$f\colon X\to\mathbb R$$ $$\mu$$-measurable. Suppose that $$0<\mu(B(x,r))<\infty$$ for all $$x\in X$$ and all $$r>0$$.

Define the Hardy–Littlewood maximal function of $$f$$ by

$Mf(x) = \sup_{r>0} \frac{1}{\mu(B(x,r))} \int_{B(x,r)} |f| \, \mathrm{d}\mu.$

By Example 29, the maximal function is a Borel function.

Theorem 10 (Hardy–Littlewood maximal inequality)

Let $$\mu$$ be a doubling measure on a metric space $$X$$. There exists a $$C>0$$ such that, for any $$f\colon X\to \mathbb R$$ and $$\lambda>0$$,

(15)#$\mu(\{x: Mf(x) > \lambda\}) \leq \frac{C}{\lambda} \int |f|\, \mathrm{d}\mu.$

Proof. For $$\lambda>0$$, let

$S=\{x\in X : Mf(x)>\lambda\}$

and, for each $$R>0$$, let $$S_R$$ be those $$x\in X$$ for which there exists $$0<r<R$$ such that

(16)#$\int_{B(x,r)} |f| \, \mathrm{d}\mu > \lambda \mu(B(x,r)).$

Similarly to Example 29, each $$S_R$$ is a Borel set. Moreover, the $$S_R$$ monotonically increase to $$S$$. For a moment fix $$R>0$$. Let $$\mathcal B$$ be the collection of balls $$B(x,r)$$ with $$x\in S_R$$ and $$0<r<R$$ that satisfy (16) and let $$\mathcal B'$$ satisfy the conclusion of Lemma 4. Then

$\begin{split}\int_{X} |f| \, \mathrm{d}\mu &\geq \sum_{B\in \mathcal B'} \int_B |f| \, \mathrm{d}\mu\\ &> \sum_{B\in \mathcal B'} \lambda \mu(B)\\ &\geq \frac{1}{C_\mu^{\log_2 5}} \sum_{B\in \mathcal B'}\lambda \mu(5B)\\ &\geq \frac{\lambda}{C_\mu^{\log_2 5}} \mu(S_R).\end{split}$

In particular, (15) holds for $$C=C_\mu^{\log_2 5}$$.

Theorem 11 (Lebasgue differentiation theorem)

Let $$\mu$$ be a doubling measure on a metric space $$X$$ and $$f\colon X \to \mathbb R$$ with $$\int f\, \mathrm{d}\mu<\infty$$. For $$\mu$$-a.e. $$x\in X$$,

$\frac{1}{\mu(B(x,r))}\int_{B(x,r)}|f-f(x)|\, \mathrm{d}\mu \to 0$

as $$r\to 0$$. Such an $$x$$ is called a Lebesgue point of $$f$$.

Proof. First note that the theorem is true if $$f$$ is continuous.

Fix $$\epsilon>0$$ and let $$g\colon X \to \mathbb R$$ be continuous with

$\int |f-g|\, \mathrm{d}\mu<\epsilon$

(such a $$g$$ exists by Example 55). Let

$B=\{x\in X: |f(x)-g(x)|\geq\sqrt{\epsilon}\},$

so that $$\mu(B) < \sqrt{\epsilon}$$.

If

$S= \{x: M(f-g) > \sqrt{\epsilon}\},$

then by Theorem 10,

$\mu(S) \leq \frac{C}{\sqrt{\epsilon}}\|f-g\|_1 <C \sqrt{\epsilon}.$

Moreover, if $$x\not\in S$$,

$\frac{1}{\mu(B(x,r))} \int_{B(x,r)} |f-g| \, \mathrm{d}\mu \leq \sqrt{\epsilon}$

for all $$r>0$$. In particular, since $$g$$ is continuous at $$x$$,

$\limsup_{r\to 0} \frac{1}{\mu(B(x,r))} \int_{B(x,r)} |f-g(x)| \, \mathrm{d}\mu \leq \sqrt{\epsilon}.$

Therefore, if $$x\not \in B$$,

$\limsup_{r\to 0} \frac{1}{\mu(B(x,r))} \int_{B(x,r)} |f-f(x)| \, \mathrm{d}\mu \leq 2\sqrt{\epsilon}.$

We are now done; repeat the above for a countable collection of $$\epsilon\to 0$$. The corresponding $$B\cup S$$ monotonically decrease to a set of measure zero. The set of $$x\in X$$ that does not belong to infinitely many of the $$B\cup S$$ has full measure, and for such an $$x$$,

$\lim_{r\to 0} \frac{1}{\mu(B(x,r))} \int_{B(x,r)} |f-f(x)| \, \mathrm{d}\mu =0.$

Corollary 1 (Lebesgue density theorem)

Let $$\mu$$ be a doubling measure on a metric space $$X$$ and let $$S\subset X$$ be $$\mu$$-measurable with $$\mu(S)<\infty$$. Then

$\lim_{r\to 0}\frac{\mu(S\cap B(x,r))}{\mu(B(x,r))}$

equals 1 for $$\mu$$-a.e. $$x\in S$$ and 0 for $$\mu$$-a.e. $$x\not\in S$$. Such an $$x$$ for which the limit equals 1 is called a density point of $$S$$.

## Exercises#

Example 27

Let $$X$$ be a separable metric space. Show that for any collection of balls, there exists a maximal disjoint sub-collection.

Example 28

Show that the $$5r$$ covering Lemma may not be true if the radii are not uniformly bounded.

Example 29

Let $$\mu$$ be a finite Borel measure on a metric space $$(X,d)$$ and let $$x_n\to x \in X$$ such that $$d(x,x_n)$$ is a decreasing sequence. Let $$U(y,r)$$ denote the open ball centred on $$y$$ with radius $$r$$.

1. Show that, for any $$r>0$$,

$U(x,r)\setminus U(x_n,r)$

decreases to the empty set.

2. Deduce that $$y\mapsto \mu(U(y,r))$$ is lower semi-continuous.

3. Give an example to show that $$y\mapsto \mu(U(y,r))$$ may not be continuous.

4. Show that, for any $$y\in X$$,

$\mu(B(y,r))= \lim_{\mathbb Q \ni q\downarrow r} \mu(U(y,q)).$
5. Deduce that, for any $$r>0$$, $$y\mapsto \mu(B(y,r))$$ is a Borel function.

6. Show that the Hardy–Littlewood maximal function is equivalently defined by taking the supremum over all rational $$r>0$$.

7. Deduce that the Hardy–Littlewood maximal function is a Borel function.

Example 30

Let $$\mu$$ be a doubling measure on a metric space $$X$$ and let $$S\subset X$$ with $$\mu(S)<\infty$$. Suppose that there exists a $$\mu$$-measurable $$S'\supset S$$ with $$\mu(S')=\mu(S)$$. Show that

$\lim_{r\to 0}\frac{\mu(S\cap B(x,r))}{\mu(B(x,r))}=1$

for $$\mu$$-a.e. $$x\in S$$.

Example 31

Prove that on $$\mathbb R^n$$, $$\mathcal L^n=c \mathcal H^n$$ for some $$c>0$$. There are two ways to prove this (recall Example 5).

Let $$\mu,\nu$$ be two finite Borel measures on a set $$X$$ with $$\mu \ll \nu$$ and suppose that $$\nu$$ is doubling. Show that the Radon–Nikodym derivative of $$\mu$$ with respect to $$\nu$$ is given by

$\lim_{r\to 0}\frac{\mu(B(x,r))}{\nu(B(x,r))}$

for $$\nu$$-a.e. $$x\in X$$.

Example 32

For $$f=\chi_{[0,1]}$$, show that $$Mf$$ does not have finite integral.