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I'm working on the Vitali Covering Lemma. I'd like to see a demonstration of the statement in the title.

I'm looking for a demonstration of the fact that an arbitrary union of sets (each with non-empty interior) is Lebesgue measurable in $R^n$ using the Vitali Covering Lemma.

Even just a paper or links to other works are acceptable, and it doesn't matter if you're not directly answering. I've browsed past questions without finding any useful results. Hope somebody can help.

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    Perhaps first use VCL to show the arbitrary union of closed balls is measurable. At least that one is true...2012-12-19

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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)2012-12-19