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Let $A$ be a non-Lebesgue measurable set, and let $B=[0,1]\subseteq\mathbb{R}$. Show that $C=A\times B$ is non-measurable.

I try use the regularity of the lebesgue measure, but don´t work, maybe take a set of borel sets and suppose $C$ is measurable give something..

Thanks.

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1 Answer 1

You might use this result. If $C\subseteq [0,1]^2$ is product Borel measurable, then the section sets $$A_x = \{s\in [0,1]| (x,s)\in C\}$$ are measurable.

If $C$ is product Lebesgue mesurable, then the $A_x$ are measurable almost everywhere.

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but A is'nt in [0,1] –  kEoz Jul 11 '12 at 19:13
    
This will work if the first factor is the real line. It is a general property of measure spaces. –  ncmathsadist Jul 11 '12 at 19:20

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