The subset of non-measurable set

Question

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

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).

Answer 2

The answer is no.

Let $\mathcal{B}\subset\Bbb{R}$ be a Bernstein set; i.e $\mathcal{B}$ intersects every closed uncountable set but contains none of them.

1. Bernstein set is non measurable.

2. Any measurable subset of a Bernstein set must be a null set ( i.e a set with $0$ outer measure).

See here for the details.

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Let $\mathcal{B}\subset \Bbb{R}$ be a Bernstein set.

Then $\mathcal{B}\cup [2, 3]$ is a non measurable set.

But $[2, 3]$ is a measurable subset of positive measure (in fact measure of $[2, 3]$ is $1$) of the non measurable subset $\mathcal{B}\cup [2, 3]$