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

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.

share|cite|improve this question
This does not answer your question, but a special case of it:…. – Derek Allums Dec 19 '12 at 14:40
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

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


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.

share|cite|improve this answer
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

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.