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?




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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](https://math.stackexchange.com/questions/2664789/characterizing-discontinuous-derivatives)



Is there any strengthened version of Rademacher's Theorem or any counterexample?
The best thing you can do is the following: for every $\epsilon > 0$ there exists a $C^1$ function $g: U \to \mathbf{R}^n$ so that
\begin{equation}
\mathcal{H}^m(\{ f \neq g \} \cup \{ Df \neq Dg \} ) < \epsilon.
\end{equation}
This can be deduced from the Whitney extension theorem; you can find a proof in Leon Simon's lecture notes on GMT (Theorem 5.3, pp. 32-33).


*Edit.* It looks like I misread your question, sorry about that. Here is how to adapt the counterexample.


Let $C \subset [0,1]$ be a fat Cantor set. This is closed, has empty interior, and measure $\lambda := \mathcal{H}^1(C) \in (0,1)$. Define $f: x \mapsto \int\_0^x \mathbf{1}\_C$. This is $1$-Lipschitz and has $f(1) = \lambda$; moreover $f' = 0$ on the open complement $C^c$.


Now, if there were an open set $U \subset [0,1]$ of full measure, and along which $f$ is $C^1$, then $f' = 0$ on $U$, because $C$ has empty interior. But then on the one hand $\int\_{[0,1]} f' = \int\_{[0,1] \cap U} f' = 0$, and on the other hand $\int\_{[0,1]} f' = f(1) - f(0) > 0$; this is absurd.


