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?
Best Answer
No. Take a dense countable set $\{x_1,x_2,\dots\}$ in $\mathbb{R}^d$ and a sequence $(r_i)\subseteq\mathbb{R}^+$ such that $\sum_i r_i^{d-1}<\infty$. Then the function $$f=1_{\bigcup_{i=1}^\infty B_{r_i}(x_i)}$$ is in $BV(\mathbb{R}^d)$ (since $|\bigcup B_{r_i}(x_i)|\le C\sum_i r_i^d$ and $f$ is the limit in $L^1$ of the functions $1_{\bigcup_{i=1}^k B_{r_i}(x_i)}$, whose gradients have total variation bounded by $C\sum_i r_i^{d-1}<\infty$).
Now, for any Lebesgue point $x_0$ of $f$ in the closed set $\{f=0\}$, no representative $g$ is continuous at $x_0$ (representative means a function which coincides a.e. with $f$). Indeed, $x_0$ lies in the closure of the open set $\bigcup B_{r_i}(x_i)$, so it belongs to the closure of $\{g=1\}$. On the other hand, since $x_0$ is a Lebesgue point for $f$, it must also belong to the closure of $\{g=0\}$. This shows that $g$ is not even a.e. continuous (since the set $\{f=0\}$ has positive measure).
Addendum.
The answer is still no even assuming $f$ continuous. Below I construct an example where the differentiability of $f$ fails on a Borel set of positive measure.
Choose a countable dense set $\{x_i\}$ in $B_1(0)$ and a sequence $r_i>0$ such that $\sum_i r_i^{d-1}<\infty$ and $\sum_i|B_{r_i}(x_i)|<|B_1(0)|$. In particular, $r_i\to 0$. Using Besicovitch covering theorem, up to a subsequence we can assume that the balls $B_{r_i}(x_i)$ have bounded overlapping (i.e. any point lies in at most $N$ such balls); in doing this, we could lose the density of $\{x_i\}$ but we still have $B_1(0)\subseteq\overline{\cup_i B_{r_i}(x_i)}$.
Let $S:=B_1(0)\setminus\cup_i\overline{B_{r_i}(x_i)}$. We remark that $|S|>0$ and that, for any $N\ge 1$, $S\subseteq\overline{\{x_i\mid i>N\}}$. It follows that we can find a sequence of positive radii $R_i\to 0$ such that $S\subseteq\cup_{i\ge j}B_{R_i}(x_i)$ for all $j\ge 1$: by compactness, we can find $n_1>0$ such that $$S\subseteq\overline{\{x_i\mid i>0\}}\subseteq\cup_{i=1}^{n_1}B_1(x_i),$$ then we use the above remark with $N=n_1$ and we find $n_2>n_1$ such that $$S\subseteq\overline{\{x_i\mid i>n_1\}}\subseteq\cup_{i=n_1+1}^{n_2}B_{1/2}(x_i),$$ and so on.
Now the function $f(x):=\sum_i R_i(1-r_i^{-1}|x-x_i|)^+$ (a superposition of 'traffic cones' with heights $R_i$ placed on our balls $B_{r_i}(x_i)$) is continuous, as it is a uniform limit of continuous functions, thanks to the bounded overlapping. It also lies in $BV(\mathbb{R}^d)$ thanks to the assumption $\sum_i r_i^{d-1}<\infty$.
I claim that $f$ cannot be differentiable at $x$, for any $x\in S$. Indeed, as $x$ is a minimum point for $f$, we would have $\nabla f(x)=0$. But there is a sequence $i_k\to\infty$ with $x\in B_{R_{i_k}}(x_{i_k})$, so $f(x_{i_k})\ge R_{i_k}\ge|x-x_{i_k}|$, which contradicts $\nabla f(x)=0$ (since $x_{i_k}\to x$).