Let $u: \Omega\subset \mathbb{R}^N \to \mathbb{R}^M$ be a $BV$ function.



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> Is the Hausdorff dimension of the graph of $u$ equal to $N$? How can we prove it?
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**Update.** 


1. In an [answer to this post](http://tps://mathoverflow.net/a/327310/122620), it has been showed that there exist a representative $\tilde u$ of $u$ such that its graph has Hausdorff dimension equal to $N$.
2. In a subsequent post [If the Hausforff dimension of the graph of a function $u$ is $N$ and $\tilde u = u$ a.e. then $\dim\_H \mathrm{graph} \, \tilde u = N$ too](https://mathoverflow.net/questions/327331/if-the-hausforff-dimension-of-the-graph-of-a-function-u-is-n-and-tilde-u) it has been showed that a function can be zero a.e. and still its graph may have dimension strictly greater than $1$. So probably this question is better formulated in terms of *essential graph* of $u$, which possibly is equivalent to asking for the property to hold for one representative of $u$ (see Question 2 in [Hausdorff dimension of the graph of a BV function (in 1 dimensional setting)](https://mathoverflow.net/questions/327698/hausdorff-dimension-of-the-graph-of-a-bv-function-in-1-dimensional-setting))
3. In the post [Hausdorff dimension of the graph of a BV function (in 1 dimensional setting)](https://mathoverflow.net/questions/327698/hausdorff-dimension-of-the-graph-of-a-bv-function-in-1-dimensional-setting), I've asked about a simpler proof of the result in the one-dimensional setting.



Hausdorff dimension of the graph of a BV function
A partial answer here.


Let us recall the following Lusin type result (see e.g. Theorem 5.34 in *Functions of bounded variation* by L. Ambrosio, N. Fusco and D. Pallara):



> 
> **Theorem 1.** There exists a constant $\kappa>0$ such that for every $u\in BV(\mathbb R^N)$ and $\lambda>0$ there exists a Lipschitz function $v\_\lambda \in \mathrm{Lip}(\mathbb R^N)$ such that
>  $$
> |x\in \mathbb R^N : u(x)\ne v\_\lambda(x)| \le \frac{\kappa}{\lambda} |Du|(\mathbb R^N).
> $$
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Let us denote $D\_\lambda := \{x\in \mathbb R^N : u(x) = v\_\lambda(x)\}$
and $D := \bigcup\_{\lambda \in \mathbb N} D\_\lambda$. It is evident that the complement $D^c = \mathbb R^n \setminus D$ is Lebesgue negligible.


Let us show that the Hausdorff dimension of the graph $\Gamma\_D:=\{(x,u(x)) : x\in D\}$ is $N$. Clearly it is sufficient to show that the Hausdorff dimension of the graph $\Gamma\_{D\_\lambda}:=\{(x,u(x)) : x\in D\_\lambda\}$ is $N$ for each $\lambda\in \mathbb N$. And this is true because $u\colon D\_\lambda \to \mathbb R$ coincides with Lipschitz function $v\_\lambda \colon D\_\lambda \to \mathbb R$. Thus we have proved the following claim:



> 
> **Proposition 1.** If $u$ is a $BV$ function then there exists $\tilde u \in BV$ such that $\tilde u = u$ a.e. and the Hausdorff dimension of the graph of $\tilde u$ is $N$.
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What remains not clear to me at the moment is whether the claim is true for every representative of $u$. But note that the answer to this question does not depend on $BV$ regularity of $u$, and can essentially be formulated as follows:


*Question.* Suppose that $u\colon [0,1]\to [0,1]$ is a measurable function such that $u=0$ a.e. Is it true that the Hausdorff dimension of the graph of $u$ is at most $1$?


