User:Tkojar/Sandbox/Vitali Covering Lemma

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  • Finite version: Let $ B_{1}, \ldots, B_{n}$ be any finite collection of balls contained in d-dimensional $\mathbb{R}^{d}$ (or, more generally, in an arbitrary metric space) then there exists a subcollection $ B_{j_1}, B_{j_2}, \dots, B_{j_m} $ of these balls which are disjoint and satisfy

$B_1 \cup B_2 \cup \cdots \cup B_n \subseteq 3 B_{j_1} \cup 3 B_{j_2} \cup \cdots \cup 3 B_{j_m}$

where $3 B_{j_k}$ denotes the ball with the same center as $B_{j_k}$ but with three times the radius.

  • Infinite version: Let $\set {B_j: j \in J}$ be an arbitrary collection of balls in $\R^d$ (or, more generally, in a separable metric space) such that

$\sup \set {\map {\mathrm {rad} } {B_j} : j \in J } < \infty$

where $\map {\mathrm {rad} } {B_j}$ denotes the radius of the ball $B_j$. Then there exists a countable subcollection

$\set {B_j: j \in J'}, \quad J' \subset J$

of balls from the original collection which are disjoint and satisfy

$\ds \bigcup_{j \mathop \in J} B_j \subseteq \bigcup_{j \mathop \in J'} 5 B_j$

Proof of finite version

Without loss of generality, we assume that the collection of balls is not empty; that is $n > 0$.

Let $B_{j_1}$ be the ball of largest radius.

Inductively, assume that $B_{j_1}, \dots, B_{j_k}$ have been chosen.

If there is some ball in $B_1, \ldots, B_n$ that is disjoint from $B_{j_1} \cup B_{j_2} \cup \cdots \cup B_{j_k}$, let $B_{j_{k + 1} }$ be such ball with maximal radius (breaking ties arbitrarily).

Otherwise, we set $m := k$ and terminate the inductive definition.

Now set $\ds X := \bigcup_{k\mathop = 1}^m 3 B_{j_k}$.

It remains to show that $B_i \subset X$ for every $i = 1, 2, \ldots, n$.

This is clear if $i \in \set {j_1,\dots, j_m}$.

Otherwise, there necessarily is some $k \in \set {1, \ldots, m}$ such that $B_i$ intersects $B_{j_k}$ and the radius of $B_{j_k}$ is at least as large as that of $B_i$.

The triangle inequality then easily implies that $B_i \subset 3 B_{j_k} \subset X$, as needed.

This completes the proof of the finite version.

Proof of infinite version

Let $F$ denote the collection of all balls $B_j$, $j \in J$, that are given in the statement of the covering lemma.

The following result provides a certain disjoint subcollection $G$ of $F$.

If this subcollection $G$ is described as $\set {B_j, j \in J'}$, the property of $G$, stated below, readily proves that

$\bigcup_{j \mathop \in J} B_j \subseteq \bigcup_{j \mathop \in J'} 5 B_j$

Precise form of the covering lemma: Let $F$ be a collection of (nondegenerate) balls in a metric space, with bounded radii. There exists a disjoint subcollection $G$ of $F$ with the following property:

every ball $B$ in $F$ intersects a ball $C$ in $G$ such that $B \subset C$

(Degenerate balls only contain the center; they are excluded from this discussion.)

Let $R$ be the supremum of the radii of balls in $F$.

Consider the partition of $F$ into subcollections $F_n$, $n \ge 0$, consisting of balls $B$; whose radius is in $\hointl {\dfrac R {2^{n + 1} } } {\dfrac R {2^n} }$.

A sequence $G_n$, with $G_n \subset F_n$ is defined inductively as follows.

First, set $H_0 = F_0$ and let $G_0$ be a maximal disjoint subcollection of $H_0$.

Assuming that $G_0, \ldots, G_n$ have been selected, let

$\mathbf H_{n + 1} = \set {B \in \mathbf F_{n + 1}: B \cap C = \O, \forall C \in \mathbf G_0 \cup \mathbf G_1 \cup \ldots \cup \mathbf G_n}$

and let $G_{n + 1}$ be a maximal disjoint subcollection of $H_{n + 1}$.

The subcollection

$\ds \mathbf G := \bigcup_{n \mathop = 0}^\infty \mathbf G_n$

of $F$ satisfies the requirements: $G$ is a disjoint collection, and every ball $B \in F$ intersects a ball $C \in G$ such that $B \subset 5 C$.

Indeed, let $n$ be such that $B$ belongs to $F_n$.

Either $B$ does not belong to $H_n$, which implies $n > 0$ and means that $B$; intersects a ball from the union of $G_0, \ldots, G_{n - 1}$ or $B \in H_n$ and by maximality of $G_n$ $B$ intersects a ball in $G_n$.

In any case, $B$ intersects a ball $C$ that belongs to the union of $G_0, \ldots, G_n$.

Such a ball C has radius $ > \dfrac R {2^{n + 1} }$.

Since the radius of $B$ is $\le \dfrac R {2^n}$, it is less than twice that of $C$ and the conclusion $B\subset C$ follows from the triangle inequality as in the finite version.


Source of Name

This entry was named for Giuseppe Vitali.