$L^p$ function whose graph is not a varifold

Question

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? and 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).

Answer 1

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 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.