Theorem 4.3.10C4F5D8B270EB

Proof. First let $\mu $ have compact support $K$ and $p$ be continuous. It is bounded on a compact neighborhood of $K$ and bounded outside it by the fixed-pole/Harnack bound, hence globally bounded. On $\Omega \setminus K$, $w-p$ is superharmonic and bounded below. At $K$ its lower boundary limits are nonnegative outside the stated polar exception by continuity of $p$ and lsc of $w$. At the ambient boundary the same holds because compact potentials tend to zero outside a polar set. If $d\ge 3$, at infinity $p\to 0$. The polar boundary minimum theorem gives $w\ge p$ off $K$; the inequality holds almost everywhere everywhere, and canonical ball averages give it at every point.

For general compactly supported $\mu $, the finite potential permits the Lusin construction of pairwise disjoint compact restrictions $\mu _ j$ with continuous potentials and $\mu =\sum \mu _ j$. To obtain them, apply Lusin to each residual potential, choose a compact subset of the residual carrier with arbitrarily small residual mass, and use the support-continuity principle; repeat with errors $2^{-j}$. The sum of the first $j$ restrictions has continuous potential, is dominated by $p$ on $K$, and therefore by $w$ everywhere by the first case. Nonnegative monotone convergence gives $p\le w$. Finally exhaust the support of a general Radon measure by increasing compact subsets. Each restriction has finite potential dominated by $p$, so the compact case applies. Monotone convergence finishes the proof. No convergence of signed differences at infinite values is used. $\square$