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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?

Math1000
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1 Answers1

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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.

Jose27
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