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Is Lebesgue measure of the boundary of a bounded Lipschitz domain in $\mathbb R^n$ zero? I guess the answer is yes but I can't find a reference for that. Could someone give me a reference for the answer?

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Evans and Gariepy "Measure Theory and fine properties of functions" has what you want, a little disguised maybe: The area formula implies that the graph of a Lipschitz function $\mathbb{R}^{n-1} \to \mathbb{R}$ has Hausdorff dimension $n-1$, so that the $n$ dimensional Hausdorff measure of said graph is $0$, but $H^n=L^n$, where the last is Lebesgue measure. Your result now follows since a Lipschitz domain is locally the graph of a Lipschitz function.

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  • $\begingroup$ Thank you very much for your help! $\endgroup$ – Lovingmath Mar 20 '15 at 4:56

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