Theorem 4.4.852535FA416BF

Proof. For a bounded open set, continuous boundary data $\Phi _ a$ have a finite harmonic Perron solution, and $G(a,\cdot )=\Phi _ a-H_{\Phi _ a}$. Indeed $\Phi _ a$ is an upper competitor, so the right side is nonnegative and admissible, hence at least $G$. Writing $G=\Phi _ a+h_ a$, the function $-h_ a$ is a lower competitor: it is bounded above by $\Phi _ a$ and bounded on a pole neighborhood by harmonicity, so bounded above throughout the bounded set. Its boundary upper limits are at most $\Phi _ a$. Thus $-h_ a\le H_{\Phi _ a}$, giving the opposite inequality. At regular points the formula gives zero boundary limit; conversely such a Green function is positive harmonic near a boundary point and supplies a weak local barrier.

For arbitrary Greenian $\Omega $, exhaust by bounded regular open sets $D_ j\Subset \Omega $ containing a fixed pole $a$. Extend $G_{D_ j}(a,\cdot )$ by zero outside $D_ j$. On the complement of a small closed pole ball these are nonnegative continuous subharmonic functions: they are harmonic inside, zero outside, and at a boundary point their value zero is at most every sufficiently small sphere average. They increase to the zero extension $g$ of $G_\Omega (a,\cdot )$, bounded above there by the fixed-pole kernel bound. The checked subharmonic envelope/polar discrepancy theorem says $g=g^*$ outside a polar set. On $\partial \Omega $, $g=0$ and $g^*\ge \limsup _{\Omega \ni x\to z}G_\Omega (a,x)\ge 0$, so that limsup is zero off a polar set.

Choose one pole in each component; there are at most countably many components. The union of their exceptions is polar. Harnack chains in the pole variable make the zero limit uniform on any compact subset of poles in a component. A compact subset meets only finitely many components, so the limit is uniform on every compact subset of $\Omega $. Integration against a finite compactly supported measure proves admissibility with one common exceptional set. The bounded-density statement is then precisely the compact-tail estimate already proved. On a bounded connected set, regularity and the fixed-pole condition are equivalent by the first part, so the irregular set is contained in that pole's polar exception. These arguments remove the temporary admissibility hypothesis in the capacity theorem without using that theorem to prove the boundary discrepancy. $\square$