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](https://en.wikipedia.org/wiki/Bounded_variation#BV_functions_of_several_variables)).



> 
> **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. 




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Addendum.
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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?
> 
> 
> 



Are functions of bounded variation a.e. differentiable?
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).




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Addendum.
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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 $x\_{i\_k}\to x$).


