Zaremba's punctured ball

Counterexample

Let $n \ge 2$ and let $\Omega = \set{x \in \R^n : 0 < \abs{x} < 1}$ be the unit open ball with its centre removed. Its boundary is the unit sphere together with the point $0$. Let $f = 0$ on the unit sphere and $f(0) = 1$; this $f$ is continuous on $\partial\Omega$, since $0$ is an isolated point of $\partial\Omega$. The Perron–Wiener–Brelot solution of $f$ is $H_f = 0$, so $0$ is not a regular boundary point. Check: the constant $0$ is in the lower class, so $H_f \ge 0$. For $\varepsilon > 0$, let $u_\varepsilon(x) = \varepsilon (\abs{x}^{2-n} - 1)$ if $n \ge 3$ and $u_\varepsilon(x) = \varepsilon \log (1/\abs{x})$ if $n = 2$. This function is harmonic, hence superharmonic, on $\Omega$. It is $0$ on the unit sphere and tends to $+\infty$ at $0$, so it is in the upper class and $H_f \le u_\varepsilon$. Letting $\varepsilon \to 0$ gives $H_f \le 0$, so $H_f = 0$, which does not tend to $f(0) = 1$.