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

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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).

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