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Let $X$ be a compact metric space (hence separable) and $\mu$ a Borel probability measure.

Given an open set $A$ and $r,\epsilon>0$ $\ $does there exist a finite set of disjoint open balls $\left\{ B_{i}\right\} $ contained in $A$ and of radius smaller than $r$ , so that $\mu(\cup B_{i})\geq\mu(A)-\epsilon.$

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It's not easy to fill up all the volume of a box by regular and arbitrary small spheres.. –  Berci Feb 15 '13 at 1:44
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@Berci, true, but if the spheres are allowed to be of different and arbitrarily small sizes then you should be able to get within $\epsilon$ with a finite number of them. –  Gerry Myerson Feb 15 '13 at 1:57
    
Yes, but.. how? –  Berci Feb 15 '13 at 2:02
    
Robert: It is an interesting problem. Do you have any thoughts to share on it? Where did it come from? –  Jonas Meyer Feb 15 '13 at 4:20
    
Buckets filled with sand certainly convince me that the answer is yes. At least in Euclidean space. –  Michael Greinecker Feb 15 '13 at 8:59

1 Answer 1

Recall

Lemma (Finite Vitali covering lemma) Let $(X,d)$ be a metric space, $\{B(a_j,r_j),j\in [K]\}$ a finite collection of open balls. We can find a subset $J$ of $[K]$ such that the balls $B(a_j,r_j),j\in J$ are disjoint and $$\bigcup_{i\in [K]}B(a_i,r_i)\subset \bigcup_{j\in J}B(a_j,3r_j).$$

A proof is given page 41 in the book Ergodic Theory: with a view towards Number Theory, Einsiedler M., Ward T.

For each $a\in A$, fix $r_a<r/3$ such that $B(a,3r_a)\subset A$. As $X$ is separable, we can, by Lindelöf property, extract from the cover $\{B(a,r_a),a\in A\}$ of $A$ a countable subcover $\{B(a_j,r_j),j\in \Bbb N\}$. Now take $N$ such that $\mu(A)-\mu\left(\bigcup_{j=0}^NB(a_j,r_j)\right)<\varepsilon$. Then we conclude by finite Vitali covering lemma.

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I love the title of the book you mention: Ergodic theory: with a view towards ergodic theory. –  Omar Antolín-Camarena Feb 17 '13 at 16:39
    
@Davide Thanks. –  Robert Feb 17 '13 at 19:01
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@DavideGiraudo actually there is something that confuses me. The Vitali covering lemma says that the balls with radius r are disjoint not the ones with radius 3r. How do we know the small balls take up almost all the measure? –  Robert Feb 22 '13 at 22:27
    
@Robert You are perfectly right, my argument is wrong. Please unaccept my answer, then I will be able to delete it. –  Davide Giraudo Feb 22 '13 at 22:35
    
@DavideGiraudo Ok. But I think you were on the right track. It seems that you can do a Vitali cover with small radius, but you have to add the extra hypothesis that the measure is doubling. –  Robert Feb 23 '13 at 0:30

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