Is the support of a Sobolev function a varifold?

Question

$\DeclareMathOperator{\graph}{\operatorname{graph}}$ 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.

Answer 1

Yes if you choose a suitable representative of a Sobolev function.

Lemma. Let $f\in W^{1,p}(\mathbb{R}^n)$, $1\leq p<\infty$. Then for every $\epsilon>0$, there is a Lipschitz function $g:\mathbb{R}^n\to\mathbb{R}$ such that $$ |\{ x\in\mathbb{R}^n:\, f(x)\neq g(x)\}|<\epsilon. $$ The proof follows from the pointwise inequality $$ |f(x)-f(y)|\leq C|x-y|(M|\nabla f|(x)+M|\nabla f|(y)), $$ where $M$ stands for the Hardy-Littlewood maximal function. $\Box$

The lemma implies that you can find countably many Lipschitz functions $g_i:\mathbb{R}^n\to\mathbb{R}$ such that such that the set $$ E:=\mathbb{R}^n\setminus \bigcup_{i=1}^\infty\{x\in\mathbb{R}^n:g_i(x)\neq f(x)\}. $$ has measure zero. If you redefine $f$ on $E$ so that the function $f$ is constant on that set (that alters $f$ on a set of measure zero which is okay, because Sobolev functions are defined a.e.) it follows that graps of the functions $g_i$ and a graph of a constant function cover the graph of $f$.

For more details see for example Lemma on p. 96 in:

P. Hajłasz, Change of variables formula under minimal assumptions. Colloq. Math. 64 (1993), no. 1, 93–101.