Proposition 4.2.6498B2C3E2413

Proof. We construct a subharmonic function on $\mathbb R^ d$ that is finite at $x$ and equals $-\infty $ on $A\setminus \{ x\} $ near $x$. Fix $x\in \Omega $ and $\rho >0$ with $\overline{B(x,\rho )}\subset \Omega $. For $k\geq 0$ put $r_ k=2^{-k}\rho $, let $S_ k=\{ y:r_{k+1}\leq |y-x|\leq r_ k\} $ be the closed annulus, and let $A_ k=A\cap S_ k$; these sets cover $A\cap \overline{B(x,\rho )}\setminus \{ x\} $. Each $A_ k$ is polar and has compact closure, so proposition 4.1.11 provides a finite positive measure $\nu _ k'$ of compact support with $E_ d*\nu _ k'=-\infty $ on $A_ k$. We restrict it to the open set $U_ k=\{ y:r_{k+2}<|y-x|<2\rho \} $, which contains $S_ k$, and put $\nu _ k=\nu _ k'|_{U_ k}$. For $y\in A_ k$ and $z\in \operatorname{Supp}\nu _ k'\setminus U_ k$ we have $|y-z|\geq \operatorname {dist}(S_ k,\mathbb R^ d\setminus U_ k)>0$, so $E_ d(y-z)$ is bounded on this set of pairs and

Moreover, $\nu _ k$ is supported in $\overline{U_ k}$, which does not meet $B(x,r_{k+2})$, so $p_ k=E_ d*\nu _ k$ is finite at $x$.

Choose $c_ k>0$ with

and put $\nu =\sum _ kc_ k\nu _ k$, a finite measure supported in $\overline{B(x,2\rho )}$, and $p=E_ d*\nu $. Let $C$ be an upper bound for $E_ d(y-z)$ when $|y-z|\leq 3\rho $. For $y\in \overline{B(x,\rho )}$, monotone convergence applied to the nonnegative functions $C-E_ d(y-\cdot )$ gives

where every term on the right is nonpositive. Hence $p(x)$ is finite, and $p=-\infty $ on $A\cap \overline{B(x,\rho )}\setminus \{ x\} $, because there one term equals $-\infty $. Since $p$ is subharmonic on $\mathbb R^ d$ (proposition 3.4.11), the set

is a fine neighborhood of $x$, and $N\cap A\subset \{ x\} $. If $x\notin A$, this shows that $\Omega \setminus A$ is a fine neighborhood of each of its points, so $A$ is finely closed. If $x\in A$, it shows that $\{ x\} $ is relatively open in $A$ for the fine topology. $\square$