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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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    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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Take $n=1$, let $B$ be a "fat Cantor set" (a compact nowhere-dense set which has positive measure) in $[0,1]$, and $f(x) = m(B \cap [0,x])$. Then f' = 1 at almost every point of $B$, but f' is not continuous at such points because f' = 0 on $B^c$ which is dense. So f' is not Riemann integrable (and neither is $\sqrt{1+(f')^2}$).

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    Sweet! Thanks for your help!2012-04-11