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

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

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

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

Update.

1. In an answer to this post, 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 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))

3. In the post 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.

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

Is the Hausdorff dimension of the graph of $u$ equal to $1$? How can we prove it?

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

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