I am looking for an example of a function $f:[0,1]\to\mathbb{R}$ which is in $L^p$ for some $p$ and whose graph is not a $1$-dimensional varifold in $\mathbb{R}^2$, that is such that it is not possible to write
$$
\operatorname{graph}(f)\subset \bigcup\_{n\in\mathbb{N}} g\_k(\mathbb{R}) \cup E
$$
for Lipschitz functions $g\_k:\mathbb{R}\to \mathbb{R}$ and a measurable set $E$ with null $\mathscr{H}^1$ measure.


This is a follow-up to these two questions [Is the support of a Sobolev function a varifold?](https://mathoverflow.net/questions/429464/is-the-support-of-a-sobolev-function-a-varifold) and [Hausdorff dimension of the graph of a BV function](https://mathoverflow.net/questions/327267/hausdorff-dimension-of-the-graph-of-a-bv-function) where it is proved that the graph of a (representative of a) Sobolev function, and more generally of a $BV$ function *is* indeed a varifold. Since varifolds support a notion of differentiability, my intuition suggests that the result must be false for functions that are only in $L^p$ (and that probably the counterexample is very easy to construct).



$L^p$ function whose graph is not a varifold
While writing this question down I realized that its answer is actually pretty trivial, but I am going to keep it here for the benefit of other users. It suffices to take any measurable function $\tilde{f}:[0,1]\to\mathbb{R}$ whose graph is not a varifold and compose it with a diffeomorphism $\phi$ that sends $\mathbb{R}$ to $(0,1)$ to get an $L^\infty$ function $f=\phi\circ \tilde{f}$ with the same property.


As shown in [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/327646#327646) we can in fact even take $f=0$ almost everywhere. By the way, this kind of construction is also exactly the reason why not all representatives of a $BV$ or Sobolev function have rectifiable graphs.


