Is the restriction of a Sobolev function to some full-measure set continuous?

Question

Motivation: As a consequence of this question, every function $u$ in $W^{1,2}(\mathbb{R}^n,\mathbb{R})$ has a representative $u^*$ such that there is a null set $E\subset \mathbb{R}^n$ with the property that $u^*:\mathbb{R}^n\setminus E\to \mathbb{R}$ has path-connected graph. This would be obvious if for all Sobolev functions $u$ there were a null set $E$ such that $\mathbb{R}^n\setminus E$ is path-connected and $u$ restricted to $\mathbb{R}^n\setminus E$ is continuous. But I have never heard of this latter fact being true, even without the requirement that $\mathbb{R}^n\setminus E$ is path-connected, so I suppose it is false in general. Here I would like to see an explicit counterexample.

Question: I am looking for an explicit function $u\in W^{1,2}(\mathbb{R}^n,\mathbb{R})$ for which there is no null-set $E$ such that $u:\mathbb{R}^n\setminus E\to \mathbb{R}$ is continuous.

Answer 1

Here is an explicit example for $n \geq 5$.

Fix an enumeration $\{x_i\}_{i \in \mathbb N}$ of some dense set in $\mathbb R^n$. Let $\phi$ be a smooth, nonnegative cutoff function that is $1$ on an open neighbourhood of the origin and supported on the closed unit ball. Set $\psi(x) = \frac{\phi(x)}{|x|}$ for nonzero $x$. For $x = 0$ we define it arbitrarily to be $0$.

Now define $u(x) := \sum_{i=0}^\infty \frac{1}{2^i} \psi(x - x_i)$.

Since $|\psi(x)|^2 \sim 1/|x|^2$ and $|\nabla \psi(x)|^2 \sim 1/|x|^4$, each term and hence each partial sum is in $W^{1,2}(\mathbb R^n)$ as soon as $n \geq 5$. The infinite sum can be seen to converge absolutely in $W^{1, 2}$ norm, so $u$ too is in $W^{1,2}(\mathbb R^n)$.

The key property of $u$ is that its essential supremum is $+\infty$ on any open set, so no matter whether you remove a null set, $u$ remains discontinuous everywhere.