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If $U$ is a bounded open set in $\mathbb R^n$, then is it necessarily that $\partial U$, the boundary of $U$, is a set of measure $0$ with respect to Lebesgue measure?

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Related: mathoverflow.net/questions/52952/… –  JavaMan Dec 3 '12 at 1:49
    
An equivalent question: Comparing the Lebesgue measure of an open set and its closure –  user51558 Dec 3 '12 at 2:52

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No, not even in one dimension. Let $E$ be a fat Cantor set, which is a closed, nowhere dense subset of $[0,1]$ with positive Lebesgue measure. (Indeed, its measure can be arbitrarily close to 1.) If we take $U = [0,1] \setminus E$ then $U$ is bounded and open, yet we have $\partial U = E$ which has positive measure.

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