User:Tkojar/Sandbox/Vitali Covering Lemma

Theorem

 * 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


 * $\displaystyle B_{1}\cup B_{2}\cup\ldots \cup B_{n}\subseteq 3B_{j_{1}}\cup 3B_{j_{2}}\cup\ldots \cup 3B_{j_{m}}$


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


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


 * $\displaystyle \sup \, \{ \mathrm{rad}(B_j) : j \in J \} <\infty $


 * where $ \mathrm{rad}(B_j) $ denotes the radius of the ball $B_{j}$. Then there exists a countable subcollection


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


 * of balls from the original collection which are disjoint and satisfy


 * $\displaystyle \bigcup_{j\in J} B_{j}\subseteq \bigcup_{j\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 $B_1,\dots,B_n$ 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 $X:=\bigcup_{k=1}^m 3\,B_{j_k}$.

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

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

Otherwise, there necessarily is some $k\in\{1,\dots,m\}$ 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 $\{ B_j, j \in J'\}$, the property of  G, stated below, readily proves that


 * $ \bigcup_{j\in J} B_j \subseteq \bigcup_{j \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  $F_{n}$, $n\geq 0$, consisting of balls B; whose radius is in $(\frac{R}{2^{n+1}},\frac{R}{2^{n}}]$.

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 $G_0 ,..., G_n$ have been selected, let


 * $ \mathbf{H}_{n+1} = \{ B \in \mathbf{F}_{n+1} : \ B \cap C = \emptyset, \ \ \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


 * $\displaystyle \mathbf{G} := \bigcup_{n=0}^\infty \mathbf{G}_n$

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

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 $G_{0},...,G_{n-1}$ 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 $G_{0},...,G_{n}$.

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

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