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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? – Martin Argerami Nov 27 '12 at 2:23
@MartinArgerami Yes – Frank Lu Nov 27 '12 at 2:24
A related question on MathOverflow:… – Jonas Meyer Nov 27 '12 at 2:33
And one of the proof sketches there does use the Vitali covering theorem. – Jonas Meyer Nov 27 '12 at 2:42

1 Answer 1

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 cubes – Frank Lu Nov 27 '12 at 2:34
@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 $a_i<b_i$ for all $i$ to get only "nontrivial" cubes, then the problem is nontrivial. (It is the same clarification that took place initially with the MathOverflow question.) – Jonas Meyer Nov 27 '12 at 2:36
@JonasMeyer Here all cubes are nondegenerate. All cubes have positive lebesgue measure. – Frank Lu Nov 27 '12 at 2:40

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