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Let $\mathcal{K}$ be ,not necessarily countable, a family of compact cubes in $\mathbb{R}^N$. How to show that $\bigcup${$K:K\in\mathcal{K}$} is a Lebesgue measurable set?

Here all cubes are nondegenerate.

I think it may be necessary to use the Vitali's covering Theorem. But I am not sure how to use it. Can someone give some hints?

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    What is a Lebesgue set? You mean Lebesgue-measurable?2012-11-27
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    @MartinArgerami Yes2012-11-27
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    A related question on MathOverflow: http://mathoverflow.net/questions/43721/is-arbitrary-union-of-closed-balls-in-rn-lebesgue-measurable2012-11-27
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    And one of the proof sketches there does use the Vitali covering theorem.2012-11-27

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Do these have nonvoid interiors? If not, you can form an arbitrary subset of $\mathbb{R}^N$ with such a union.

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    these are just regular cubes2012-11-27
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    @Frank Lu: ncmathsadist's point, which I was also going to comment to ask you to clarify, is that if a cube is defined to be a set of the form $[a_1,b_1]\times[a_2,b_2]\times\cdots\times[a_N,b_N]$, you could also get a point by taking $a_i=b_i$ for each $i$. If you assume that $a_i$i$ to get only "nontrivial" cubes, then the problem is nontrivial. (It is the same clarification that took place initially with the MathOverflow question.) – 2012-11-27
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    @JonasMeyer Here all cubes are nondegenerate. All cubes have positive lebesgue measure.2012-11-27