Does finiteness of Hausdorff measure of the graph of a continuous function implies that the function is BV?

Question

Let $\mathcal{H}^k$ denote the $k$-dimensional Hausdorff measure and $2 \le m < n$ be two integers. Suppose $f : \mathbf{R}^m \to \mathbf{R}^{n-m}$ is continuous and define $F : \mathbf{R}^m \to \mathbf{R}^n$ by setting $F(x) = (x,f(x))$ for $x \in \mathbf{R}^m$. Assume that the graph of $f$ has locally finite $\mathcal{H}^m$ measure, i.e.,

$$ \mathcal{H}^m(K \cap \operatorname{image}F) < \infty \quad \text{for any compact set $K \subseteq \mathbf{R}^n$}\,. $$

$\bullet$ Does it follow that $f$ is of bounded variation, i.e., $f \in BV$?

Assume additionally that $f$ has the Lusin N property, i.e.,

$$ \mathcal{H}^m(Z) = 0 \quad \Rightarrow \quad \mathcal{H}^m(F[Z]) = 0 \,. $$

$\bullet$ Does it follow that $f$ is weakly differentiable?

Answer 1

The answer to both questions is YES. For the proof one can employ the characterisation of BV by integration of essential variation on lines (see see 4.5.10 in Federer 1969). I wrote down the details.