Vitali's covering lemma is a useful result in analysis, specifically when one is addressing questions about density. Suppose we are working in Rn, n-dimensional Euclidean space, under Lebesgue measure where we denote by |S| the measure of a subset S of Rn. It's probably best to think of |S| as the volume of S. We have the following result:

Vitali's Covering Lemma. If U is an open set in Rn and c is a constant such that c < |U|, then there exist disjoint open balls B1, ..., BN contained in U such that
Σ1≤j≤N|B_j| > 3-nc.

This is one of those situations where the ideas in the proof end up being more useful than the result itself.

Proof of Vitali's Covering Lemma. Since we are considering Rn under the Lebesgue measure, we can get approximate |U| as well as we would like to by taking the measure of larger and larger compact (closed and bounded) subsets of U. Thus there is a compact subset K of U such that |K| > c. Since U is open, for every point x in U there is an open ball around x, say Bx, contained within U. Let C := {Bx : x ∈ U}. Now C forms an open cover of K and since K is compact, it has a finite subcover given, say, by the balls Bx1, ..., Bxm for some positive integer m. Let us take all these balls and put them in the set S. We now perform the following process:

1. If S is not empty, there is a ball of maximum radius in S. We shall take this ball and call it Bj where j-1 is the number of times we have already performed the process; if S is empty, stop.

2. Remove all the balls in S that intersect Bj.

So at step j, we are greedily picking a ball Bj of maximum radius in S out of those balls in S that do not intersect any of the previous balls B1, ..., Bj-1. Since there were only finitely many balls in S when we started the above process, this algorithm must terminate and we end up with some collection of balls B1, ..., BN. Now, suppose B is some ball that was deleted from S at the jth step. Then B had radius at most that of Bj and since B ∩ Bj ≠ ∅, we get that B is contained in the ball centered at the center of Bj of three times the radius of Bj, which we shall denote by 3Bj. Since S was a subcover for K before we started messing with it, we have that the balls 3B1, ..., 3BN also provide a cover for K.

Notice moreover that, by the nature of our algorithm, the balls B1, ..., BN are disjoint.

Now, since the balls 3Bj cover K, their union has measure at least that of K. As a result

Σ1≤j≤N|3Bj| ≥ |K| > c.

But |3Bj| = 3n|Bj| since that's how volume works. Therefore,

Σ1≤j≤N3n|Bj| > c,

and dividing by 3n gives

Σ1≤j≤N|Bj| > 3-nc. QED.

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