4.3.8 Sweeping identities and the domination principle1E913881086D

Proof. Work componentwise, choosing the prescribed point in its component and an arbitrary point in every other component outside the polar set. Cover $Z$ in the component by countably many small closed interior balls avoiding that point. The checked bounded-polar globalization with compact cutoffs gives finite measures supported in these balls, whose fundamental potentials are infinite on the covered polar pieces. Their Green potentials have the same poles by the local harmonic-correction identity, and are finite at the chosen point. Weight them so the sum of their masses and their potentials at that point is finite. The sum is a Green potential finite there and infinite on $Z$. Add the constant one for strict positivity, and glue the componentwise functions; superharmonicity is local.

The discrepancy between $R_ Eu$ and its regularization is polar. For $a\notin E$ choose the pole function $p$ for its discrepancy on $E$, finite at $a$. Then $\widehat R_ Eu+\varepsilon p$ is a majorant of $u$ on $E$; evaluating at $a$ and letting $\varepsilon \downarrow 0$ proves equality there. If a polar subset is removed from $E$, adding its $\varepsilon $-pole function to any majorant on the remainder proves the opposite inequality almost everywhere; ball averages make it hold for the regularizations everywhere. For increasing $F_ j$, their increasing regularized reductions have a superharmonic limit $v\le \widehat R_ Fu$. On each $F_ j$ they equal $u$ outside a polar set; the union of these exceptions is polar. Thus $v$ majorizes $u$ on $F$ outside a polar set, and the polar-removal argument gives $v\ge \widehat R_ Fu$.

Put $h=H^ W_ u$, which is finite harmonic because $0\le h\le u$ on $W$, and set $v=h$ on $W$, $v=u$ elsewhere. We justify the pasting even if $u$ has infinite boundary values. For any ball $B\Subset \Omega $, positivity of Perron solutions gives $H^ B_ v\le H^ B_ u\le u$ on $B\setminus W$. On $B\cap W$, compare the boundary values of $H^ B_ v$ and $v$, using the restriction lemma first from $B$ and then from $W$; both domains involved are bounded. This gives $H^ B_ v\le v$ there as well. If $\widehat v$ is the lsc regularization, $H^ B_{\widehat v}\le H^ B_ v\le v$; its harmonic continuity then gives $H^ B_{\widehat v}\le \widehat v$. The ball comparison characterization proves $\widehat v$ superharmonic. On $\partial W$, approximate the lsc boundary function $u$ increasingly by continuous functions. At regular points of $W$, their Perron solutions have the prescribed limits, so $\liminf _ W h\ge u$ there. The irregular set is polar. Hence $\widehat v\ge u$ on $\Omega \setminus W$ outside a polar set, proving $\widehat v\ge \widehat R_{\Omega \setminus W}u$. Conversely any defining majorant is in the upper Perron class for $u$ on $W$, so dominates $v$; regularization gives the reverse inequality. On $W$, $\widehat v=v=h$, proving the formula.

Linearity of integrable boundary solutions proves additivity inside $W$; outside $W$, both reductions equal the original function off a polar set. Ball averages remove that exception from the additive identity. In an exhaustion the reductions decrease by set order. Their interior harmonic limit is nonnegative and at most $u$. Every global harmonic minorant of $u$ is a lower Perron competitor on each $W_ j$, since it is continuous near $\partial W_ j$; it is therefore at most this limit. This identifies the greatest harmonic minorant. The zero characterization is the global Riesz theorem. $\square$

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. $\square$