In general, it is well known that, on the real line, say on $[0,1]$, if a function $f$ is of (pointwise) bounded variation, meaning that $$ \sum_{i=1}^n |f(x_i)-f(x_{i-1})| <+\infty $$ for every partition ${x_i}_{0}^n$ of $[0,1]$, then $f$ can be written as the difference of two monotone functions, hence it is differentiable a.e. w.r.t. the Lebesgue measure.

I am wondering if the same is true for $BV$ functions in $\mathbb R^d$ for $d \ge 2$.

Of course, the right definition of $BV$ in $d$-dimensional domains passes through the theory of distributions: $f \colon \Omega \subset \mathbb R^d \to \mathbb R$ is in $BV(\Omega)$ if it is an $L^1$ function whose distirbutional gradient $Df$ can be represented by a finite Radon measure (see here).

Question 1.Let $f \colon \Omega \subset \mathbb R^d \to \mathbb R$ be in $BV(\Omega)$. Is it true that $f$ is differentiable a.e. with respect to the Lebesgue measure?

What I know is that they are approximately differentiable a.e. (in view of Calderon-Zygmund Theorem) so an approximate differential exists a.e. but I am not aware of any link between the approximate differentiability and the pointwise a.e. one.

## Addendum.

In view of Mizar's answer, it seems that the answer to **Q1** is negative, as it has been exhibited a $BV$ function which does not have even a continuous a.e. representative (in $L^1$).

While checking the details of the answer I received, I would like to ask another version of question above (do not know if still meaningful or not).

Question 2.Let $f \colon \Omega \subset \mathbb R^d \to \mathbb R$ be in $BV(\Omega)$. Assume further $f \in C^0(\Omega)$ i.e. it is continuous. Is it true that $f$ is differentiable a.e. with respect to the Lebesgue measure?