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 $\mathbb{R}^n$?
Constructing non-Lebesgue-measurable sets in $\mathbb{R}^n$ from ones in $\mathbb{R}$
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measure-theory
2 Answers
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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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0Yes; maybe "almost every $y$" is more correct. – 2011-10-31
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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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0in general it should be true for $A\times B$ being non-measurable given $A$ and $B$ are non measurable – 2011-10-28