Are the elements of $H^1_0(\Omega)$ bounded almost everywhere?

Question

Let $\Omega \subset \mathbb{R}^n$ be a smooth, bounded domain, are the elements of $H^1_0(\Omega)$ bounded almost everywhere? Evans has the following example: $$u(x) = \sum_{k = 1}^\infty \frac{1}{2^k}\vert x - r_k\vert^{-\alpha},$$ where $(r_k)_{k \in \mathbb{N}}$ is a countable dense subset of $\Omega = B(0, 1)$ and $0 < \alpha < \frac{n - 2}{2}$, then $u \in H^1(\Omega)$ and $u$ is unbounded on every open subset of $\Omega$. Does $u \in H^1_0(\Omega)$? It seems unlikely because $H^1_0(\Omega)$ is the set of functions that "vanish at the border" while $u$ goes to infinity no matter how close to the border it is.

Answer 1

multiply $u$ by a smooth nonzero function that is compactly supported in $\Omega$ to get an unbounded function in $H^1$ with 0 trace.