# Covering theorems

## Contents

# 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.

\(5r\) 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

In particular,

Here, \(5B\) denotes the ball with the same centre as \(B\) and 5 times the radius.

Proof. For each \(n\in \mathbb Z\) let

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.

(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
\(x\in B\).

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 \(I\subset \mathcal B'\),

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

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

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

as required.

(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

for all balls \(B\subset X\).

Note that, for any \(m\geq 2\),

Lebesgue measure is a doubling measure.

(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

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 \(2\tilde B\),

By the conclusion of Proposition 1,

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

as required.

(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

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

(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 \(\lambda>0\),

Proof. For \(\lambda>0\), let

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

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

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

(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. \(x\in X\),

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

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

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

If

then by Theorem 10,

Moreover, if \(x\not\in S\),

for all \(r>0\). In particular, since \(g\) is continuous at \(x\),

Therefore, if \(x\not \in B\),

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 \(x\),

(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

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#

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

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

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 \(r\).

Show that, for any \(r>0\),

\[U(x,r)\setminus U(x_n,r)\]decreases to the empty set.

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

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

Show that, for any \(y\in X\),

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

Show that the Hardy–Littlewood maximal function is equivalently defined by taking the supremum over all

*rational*\(r>0\).Deduce that the Hardy–Littlewood maximal function is a Borel function.

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

for \(\mu\)-a.e. \(x\in S\).

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

for \(\nu\)-a.e. \(x\in X\).

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