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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}^n(\{ 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).

In general you cannot find a $C^1$ function $g$ that coincides with a Lipschitz $f$ almost everywhere. Here is a counterexample in the lowest dimension, where $m = n = 1$.

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 $C^c$.

If $g$ were a $C^1$ function that coincided (along with their derivatives) with $f$ a.e., then $g' = 0$ a.e. in $C^c$. By continuity of $g'$, this extends to $g' = 0$ in $C^c$. As $C$ has empty interior, this means that $g$ is constant, and cannot coincide with $f$ a.e.

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.

The best thing you can do is the following: for every $\epsilon > 0$ there exists a $C^1$ function $g: U \to \mathbf{R}$ so that \begin{equation} \mathcal{H}^n(\{ 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).

In general you cannot find a $C^1$ function $g$ that coincides with a Lipschitz $f$ almost everywhere. Here is a counterexample in the lowest dimension, where $m = n = 1$.

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 $C^c$.

If $g$ were a $C^1$ function that coincided (along with their derivatives) with $f$ a.e., then $g' = 0$ a.e. in $C^c$. By continuity of $g'$, this extends to $g' = 0$ in $C^c$. As $C$ has empty interior, this means that $g$ is constant, and cannot coincide with $f$ a.e.

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}^n(\{ 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).

In general you cannot find a $C^1$ function $g$ that coincides with a Lipschitz $f$ almost everywhere. Here is a counterexample in the lowest dimension, where $m = n = 1$.

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 $C^c$.

If $g$ were a $C^1$ function that coincided (along with their derivatives) with $f$ a.e., then $g' = 0$ a.e. in $C^c$. By continuity of $g'$, this extends to $g' = 0$ in $C^c$. As $C$ has empty interior, this means that $g$ is constant, and cannot coincide with $f$ a.e.

The best thing you can do is the following: for every $\epsilon > 0$ there exists a $C^1$ function $g: U \to \mathbf{R}$ so that \begin{equation} \mathcal{H}^n(\{ 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).

In general you cannot find a $C^1$ function $g$ that coincides with $f$ almost everywhere. I think something along the following lines should work as a 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 $f: x \mapsto \int_0^x \mathbf{1}_C$; obviously this is $1$-Lipschitz and has $f(1) = \lambda$. Moreover, as $C^c$ is open, one finds that $f' = 0$ on $C^c$.

If $g$ were a $C^1$ function that coincided (along with their derivatives) with $f$ a.e., then $g' = 0$ a.e. in $C^c$. By continuity of $g'$, this extends to $g' = 0$ everywhere $C^c$. From there it follows that $g'$ vanishes identically because $C$ has empty interior. This means that $g$ that would be constant, rendering it impossible for it to coincide with $f$ a.e.

The best thing you can do is the following: for every $\epsilon > 0$ there exists a $C^1$ function $g: U \to \mathbf{R}$ so that \begin{equation} \mathcal{H}^n(\{ 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).

In general you cannot find a $C^1$ function $g$ that coincides with a Lipschitz $f$ almost everywhere. Here is a counterexample in the lowest dimension, where $m = n = 1$.

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 $C^c$.

If $g$ were a $C^1$ function that coincided (along with their derivatives) with $f$ a.e., then $g' = 0$ a.e. in $C^c$. By continuity of $g'$, this extends to $g' = 0$ in $C^c$. As $C$ has empty interior, this means that $g$ is constant, and cannot coincide with $f$ a.e.