# Is the derivative of a Lipschitz function Riemann integrable?

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?

-
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 $\mathbb R^n$. –  copper.hat Apr 11 '12 at 6:30
Why do you care about it being Riemann integrable? –  copper.hat Apr 11 '12 at 6:31
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. –  Robert Israel Apr 11 '12 at 7:26
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) –  fantastic Apr 11 '12 at 8:03

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 $\sqrt{1+(f')^2}$).