Theorem 3.5.1164E880580F0E

Proof. Near $K$ the Green potential differs from its fundamental potential by a harmonic function. Both potentials are locally integrable and have the same distributional Laplacian, so Weyl's lemma gives a harmonic difference. Adding that harmonic function preserves superharmonicity; canonical representatives then identify the equality everywhere, including infinite values. No joint continuity of a cancelled diagonal kernel is assumed.

It suffices to prove the first assertion for $U\mu (x)=\int \Phi (x-y)d\mu (y)$. If $U\mu (a)=\infty $, lsc suffices. Otherwise choose a small ball $B$ about $a$ and split $\mu =\mu '+\mu ''$, with $\mu '=\mu |_ B$, so that the local singular integral $U\mu '(a)<\eta $. In the planar case take all relevant distances less than one so this local kernel is nonnegative. The remainder $U\mu ''$ is harmonic near $a$. If $z_ x\in K$ is a nearest point to $x$, then $z_ x\to a$ as $x\to a$, and $|z_ x-y|\le 2|x-y|$ for $y\in K$. For $d\ge 3$ this gives $U\mu '(x)\le 2^{d-2}U\mu '(z_ x)$. For $d=2$ it gives $U\mu '(x)\le U\mu '(z_ x)+(\log 2)\mu '(B)$. The restriction of $U\mu '$ to $K$ is continuous at $a$, being the difference of the assumed continuous restriction of $U\mu $ and the harmonic remainder. Thus the limsup of the local contribution is at most $2^{d-2}\eta $, or $\eta +(\log 2)\mu (B)$ in dimension two. In that case $\mu (\{ a\} )=0$ by finiteness of the potential, so $\mu (B)\to 0$ as the radius shrinks. The harmonic remainder is continuous; letting $\eta \to 0$ proves continuity.

For the second assertion, Lusin's theorem gives compact $C\subset K$ with lost mass less than $\varepsilon $ and $G\mu |_ C$ finite continuous. On $C$, $G(\mu |_ C)=G\mu -G(\mu |_{K\setminus C})$ is upper semicontinuous, since the second term is lsc; it is also lsc. It is therefore continuous on $C$, and the first assertion gives continuity at every point of $C$. Off $C$ it is harmonic. This proves global finite continuity. $\square$