Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.