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Suppose $f:\mathbb R^{n}\to \mathbb R$ is a Lipschitz function. Is $\sqrt{1+|\nabla f|^2}$ Riemann (not Lebesgue) integrable on a bounded open set, say a ball?

In $\mathbb R^1$, a function is Riemann integrable on a bounded interval $[a,b]$ iff it is continuous almost everywhere with respect to Lebesgue measure. Do we have an analogue for higher dimension?

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    Rademacher's Theorem says that $f$ is differentiable a.e., but I know of no results that say anything about the continuity properties. The same characterization of Riemann integrable functions applies in $\mathbb R^n$.2012-04-11
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    Why do you care about it being Riemann integrable?2012-04-11
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    That characterization of Riemann integrability also requires the function to be bounded. Your $\sqrt{1+|\nabla f|^2}$ is of course bounded, but you should quote the theorem correctly.2012-04-11
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    Thanks for your reply! Actually I want to prove a claim about discrete approximation to the surface measure for a bounded Lipschitz domain. Maybe it's better for me to post it as another question. If the same characterization of Riemann integrable functions applies in $R^n$, then the result is true for Lipschiz domains whose unit normal vector field is continuous almost everywhere (with respect to the surface measure)2012-04-11

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