Consider a continuous function $f\colon \mathbb R\to \mathbb R$. Let $F(x,y):=\mathrm{sign\,}(y-f(x))$.
Is it true that for a.e. $x\in \mathbb R$ the point $(x,f(x))$ is not a point of $L^1$ approximate continuity of $F$?
(Recall that a point $\xi\in \mathbb R^2$ is called a point of $L^1$ approximate continuity of a function $F\colon \mathbb R^2\to\mathbb R$ if $\lim_{r\to 0} \frac{1}{r^2}\int_{B(\xi,r)}|F(\eta)-w|\,d\eta = 0$ for some $w\in \mathbb R$)
Remark. If $f$ is $C^1$ then the graph of $f$ cannot contain $L^1$ approximate continuity points of $F$. But if $f$ is not differentiable then some points of its graph can be points of approximate continuity of $F$ (for instance if $f(x)=\sqrt{|x|}$ then $(0,0)$ is such point).