The following theorem is well-known in the ordinary analysis textbook:

**Theorem**: Assume the function $f:U\to\Bbb R^n$ is Lipschitz continuous on an open set $U\subset\Bbb R^m$, then $f$ is almost everywhere differentiable on $U$.

My question:

**Question**: Assume the function $f:U\to\Bbb{R}^n$ is Lipschitz continuous on an open set $U\subset\Bbb R^m$. Prove or disprove that $f$ is almost everywhere $C^1$ on $U$.

Basically, this means that the point where $f$ is not differentiable would be a zero-measure closed set and on the open set where $f$ is differentiable, the gradient should be continuous. Is there any counterexample?

Thank for all the comments. This problem is solved now. The key point is that there exists a function $f$ which is differentiable everywhere with bounded derivative but the set consisting of discontinuous point of $f'$ can have positive measure. See Differentiable function with discontinuous derivative

there exists a function$f$which is differentiable everywhere with bounded derivative but the set consisting of discontinuous point of$f'$can have positive measure--- The discontinuity set of $f'$ can even have full measure, meaning that the continuity set has measure zero. In fact,most functions with a bounded derivative("most" in a Baire category sense) are such that their continuity sets have Hausdorff dimension zero (and even this can be strengthened by using gauge functions). $\endgroup$