I would like to know if, given $f\in W^{1,2}(\mathbb{R}^n,\mathbb{R})$, it is true that we can always cover $graph(f)\subset\mathbb{R}^{n+1}$ with the images of countably many Lipschitz maps $g_k:\mathbb{R}^n\to\mathbb{R}^{n+1}$ and a set $E$ with $\mathscr{H}^n(E)=0$, i.e. $$graph(f)\subset \bigcup_{k\in\mathbb{N}} g_k(\mathbb{R}^n) \cup E$$ (which is the definition of $graph(f)$ being countably rectifiable). Any hint, reference or additional conditions for this to be true are very much appreciated.

$\DeclareMathOperator{\graph}{\operatorname{graph}}$ I would like to know if, given $f\in W^{1,2}(\mathbb{R}^n,\mathbb{R})$, it is true that we can always cover $\graph(f)\subset\mathbb{R}^{n+1}$ with the images of countably many Lipschitz maps $g_k:\mathbb{R}^n\to\mathbb{R}^{n+1}$ and a set $E$ with $\mathscr{H}^n(E)=0$, i.e. $$\graph(f)\subset \bigcup_{k\in\mathbb{N}} g_k(\mathbb{R}^n) \cup E$$ (which is the definition of $\graph(f)$ being countably rectifiable). Any hint, reference or additional conditions for this to be true are very much appreciated.

I would like to know if, given $f\in W^{1,2}(\mathbb{R}^n,\mathbb{R})$, it is true that we can always cover $graph(f)\subset\mathbb{R}^{n+1}$ with the images of countably many Lipschitz maps $g_k:\mathbb{R}^n\to\mathbb{R}^{n+1}$, i.e. $$graph(f)\subset \bigcup_{k\in\mathbb{N}} g_k(\mathbb{R}^n)$$ (which is the definition of $graph(f)$ being countably rectifiable). Any hint, reference or additional conditions for this to be true are very much appreciated.

I would like to know if, given $f\in W^{1,2}(\mathbb{R}^n,\mathbb{R})$, it is true that we can always cover $graph(f)\subset\mathbb{R}^{n+1}$ with the images of countably many Lipschitz maps $g_k:\mathbb{R}^n\to\mathbb{R}^{n+1}$ and a set $E$ with $\mathscr{H}^n(E)=0$, i.e. $$graph(f)\subset \bigcup_{k\in\mathbb{N}} g_k(\mathbb{R}^n) \cup E$$ (which is the definition of $graph(f)$ being countably rectifiable). Any hint, reference or additional conditions for this to be true are very much appreciated.