Lemma 4.4.165D12CB9707B5

Proof. Only the dense case remains. Work in a connected component, choose $a\in \Omega $, and concentric closed balls with $0<r_0<r_1<r_2<r_3$, all contained in $\Omega $. Removing the smallest ball gives a nondense domain. Its pole function $v$ can be harmonically lowered on the annulus $r_0<|x-a|<r_3$, so it is finite harmonic near the two middle spheres. This preserves its true-boundary blow-up and positivity. Write $O=\Omega \setminus \overline{B(a,r_1)}$. This set is connected: any path in $\Omega $ entering the removed ball can be rerouted through the contained annulus. The harmonic measure $Q$ of its inner sphere is strictly below one on $O$. Indeed the original nonpolar boundary contains a regular point, since the nondense Kellogg exception is polar; $Q$ tends to zero there. The strong maximum principle then gives $q=\max _{|x-a|=r_2}Q(x)<1$.

Set $w_0=0$. Given $w_{j-1}$, let $u_ j$ be the ball solution in $B(a,r_2)$ with data $v+w_{j-1}$ on its sphere. Let $w_ j$ be the bounded solution on $O$ with data $u_ j-v$ on the inner sphere and zero on the true boundary. These are continuous boundary data on the nondense domain. For consecutive differences the ball maximum principle and the bound by $Q$ give

Thus the corrections converge in geometric series, uniformly on the respective boundaries and hence throughout their domains. Their limits $u,w$ are harmonic; $w$ is bounded. On both annulus boundary spheres, $u=v+w$, so annular uniqueness gives this equality throughout the overlap. Paste $u$ inside to $v+w$ outside. The result is superharmonic, harmonic near $a$, bounded below, and has the same polar boundary blow-up as $v$. Adding a constant makes it positive. This proves the assertion including finiteness at the prescribed center. Componentwise construction proves the disconnected version.

For bounded harmonic $h$ with zero boundary limits outside a polar $Z$, compare $\pm h+\varepsilon p$ with zero, where $p$ is the positive blow-up function, finite at a prescribed interior point. Compactified boundary comparison applies even on $Z$, and yields $|h(a)|\le \varepsilon p(a)$. Let $\varepsilon \downarrow 0$ and choose $a$ arbitrarily. The same epsilon argument permits polar exceptions in bounded Perron competitors. $\square$