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Suppose $V \subseteq \mathbb{R}$ is non-Lebesgue-measurable in $\mathbb{R}$ (the particular example I have in mind is when $V$ is a Vitali set). Is it necessarily true that $V^n$ is non-Lebesgue-measurable in $\mathbb{R}^n$?

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Yes. Let $n$ and $m$ be positive integers. If $A\subset\mathbb{R}^{n+m}$ is Lebesgue measurable, then, for almost all $y\in\mathbb{R}^m$, the set $$ A_y=\{x\in\mathbb{R}^n:(x,y)\in A\} $$ is Lebesgue measurable in $\mathbb{R}^n$ (this is usually seen before the proof of Fubini's theorem).

Now let $V\subset\mathbb{R}^{n}$ be non-Lebesgue measurable and and $W\subset\mathbb{R}^{m}$ either non measurable or measurable with positive measure and $A=V\times W$. Then $A_y=V$ for all $y\in W$. If $A$ were measurable, then so would be $V$.

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Thank you! By "almost all $y \in \mathbb{R}^m$", do you mean any $y \in \mathbb{R}^m \setminus X$, where $X$ is a null set? –  Ignis Umbrae Oct 30 '11 at 23:56
    
Yes; maybe "almost every $y$" is more correct. –  Julián Aguirre Oct 31 '11 at 10:05

Vitali set is not just non-lebesgue measurable but non-measurable (unless you remove the constraint $\mu(0,1) = 1$ ). your assertion is true for vitali set

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in general it should be true for $A\times B$ being non-measurable given $A$ and $B$ are non measurable –  user24367 Oct 28 '11 at 17:42

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