It is known that every convex function $f: \Omega\to \mathbb{R}$, $\Omega$ convex subset of $\mathbb{R}^n$, has a weak derivative of bounded variation $Df\in BV_{loc}(\mathbb{R}^n)$ (e.g. Evans and Gariepy 1992), and therefore a distributional Hessian $\mu_{ij}:=[D_i D_j f]$ which is a non-negative definite Radon measure which can be decomposed as $\mu_{ij}=\mu_{ij}^a+\mu_{ij}^s$ where $\mu^s$ is the singular part and $\mu^a$ is the absolutely continuous part with respect to the Lebesgue measure $\mathcal{L}^n$.

My question is: suppose that $\mu_{ij}^s$ vanishes, does $Df$ admit a continuous representative, and so is $f$ $C^1$?

Here are some comments. For $n=1$ the question is trivial, because $\mu^s=0$ is equivalent to the absolute continuity of $D f$.

For $n>1$ the condition that $Df$ has weak derivatives in $L^1$ does not imply that $Df$ is continuous. Perhaps by adding the convexity of $f$ one can conclude that $f$ is $C^1$?

I also know that the singular part decomposes into Cantor and Jump parts $\mu^s=\mu^c+\mu^j$ (e.g.\ Ambrosio, Fusco, Pallara, 2000), however it is seems to me that as it is defined the Jump set can be empty for discontinuous $Df$. Under convexity of $f$ the condition $\mu^j=0$ is sufficient for the continuous differentiability of $f$?