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I'm working on the Vitali Covering Lemma. I'd like to see a dimostration of the statement in the title.
I'm looking for a dimostration about the fact that the arbitrary union of set ( with non-epmty interior ) is measurable Lebesgue in $R^n$ using this Lemma.
Paper,links to other works will be good anyway,doesn't matter if you're not directly answering. I've browsed also past question without useful results.Hope somebody could help.

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This does not answer your question, but a special case of it: mathoverflow.net/questions/43721/…. –  Derek Allums Dec 19 '12 at 14:40
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Perhaps first use VCL to show the arbitrary union of closed balls is measurable. At least that one is true... –  GEdgar Dec 19 '12 at 15:03
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1 Answer 1

No, it is not true. Any non-measurable set is a union of its one-point subsets.

Edit:

For the revised question (with non-empty interiors), the result is still false. Consider the simplest case of Lebesgue measure in $\mathbb{R}$, and let $V$ be a non-measurable subset of $[0,1]$, and consider the sets (for each $v\in V$):

$$U_v = \{v\} \cup [2,3]$$

Then each set $U_v$ has non-empty interior, but the union of all the sets $U_v$ is just $V \cup [2,3]$, which is non-measurable, since $V$ is non-measurable.

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I'm sorry I made a mistake while traslating the statement in english. The sets are not just non-empty (for which your answer works) but with interior non-empty (so excluding the singletons for exemple) –  Laura Dec 19 '12 at 14:27
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