Take the 2-minute tour ×
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.

If $A$ is a non-measurable set in $\mathbb R^n$ (in the sense of Lebesgue), does it necessarily contain a positive measurable subset?

share|improve this question

1 Answer 1

No. Every measurable subset $M$ of a Vitali set in $[0,1]$ is necessarily of measure $0$, by precisely the same argument that shows that if the Vitali set were measurable then it would have measure zero: the rational translates $M+q$ of $M$, with $q\in[-1,1]\cap\mathbb{Q}$ are pairwise disjoint, and contained in $[-1,2]$; so the measure of their union is the sum of their measures and is finite, hence must be zero.

This is easily extended to $\mathbb{R}^n$ for any $n\gt 1$.

(Of course, it is also false that a measurable subset of a nonmeasurable set must have measure zero, since we can let $V$ be a Vitali set contained in $[0,1]$, and take $A=V\cup (-\infty,0)\cup (1,\infty)$. This is not measurable, but contains measurable set of any measure you care to specify).

share|improve this answer

Your Answer

 
discard

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.