To contextualize my question, recall that Lipschitz (i.e. $C^{0,1}$) functions by Rademacher's theorem are differentiable almost everywhere. It's interesting to note that this is sharp (and apparently generic in some sense according to Wikipedia) in view of the Weistrass function being $C^{0,\alpha}$ for every $\alpha<1$, but being differentiable nowhere. Therefore, up to a set of Lebesgue measure $0$, $$\text{Lipshitz}=\text{differentiable}.$$ Thus, for a Lipschitz function $f$, set $Df$ to be all points where the derivative is well-defined (otherwise, set $Df=\infty$). Now, there is a theorem that tells us Lipschitz functions admit a $C^1$-approximation in the following sense. For any Lipschitz function $f$ and every $\epsilon>0$, there is a $C^1$ function $g$ such that $$\mathscr{L}^n(\{f\neq g\}\cup\{Df\neq Dg\}\cup \{Df=\infty\})<\epsilon.$$ Thus, up to sets of arbitrary small but positive error, $$\text{Lipschitz} = C^1.$$ I'm interested in knowing if the a.e. equality of $\text{Lipshitz}=\text{differentiable}$ is sharp. In other words, is there an intermediate class $K$ such that $$\text{differentiable}\subset K\subset C^1,$$ and for every $f$ Lipschitz, there exists some $g\in K$ such that $f=g$ almost everywhere.
Here's a non-answer. If the underlying domain is $\mathbb{R}^n$ or $\Omega\subset \mathbb{R}^n$ a bounded domain with Lipschitz boundary, then Morrey's inequality tells us for $p>n$,
$$W^{1,p}\subset C^{0,\alpha}.$$
Thus, if we expect $K$ to be a Soboelv space $W^{1,p}$ then necessarily $p\leq n$ under these mild domain restrictions.
Sorry I made a mistake in the above. Instead, if you try to work out for which $p>n$ such that $W^{1,p}\subset C^{1,\alpha}$, then you find by Sobolev embedding that $p<0$. So none of those are actually ruled out.