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.