Theorem 5.15.114388B85DDD67

Proof. Observe that sets of the form (5.15.12) are $\mathcal{F}$-open.² By theorem 5.15.10, it suffices to show its complement in $\Omega $ is thin at each point of (5.15.12), which is obvious.

Now consider $x\in \Omega $ and an $\mathcal{F}$-open neighborhood $V\subseteq \Omega $ of $x$. We want to find a set of the form (5.15.12) contained in $V$ and containing $x$.

Write $E=\Omega \setminus V$. In case $x\in \operatorname{Int}V$, there is nothing to prove. So we may assume that $x\in \overline{E}$. By theorem 5.15.10, $E$ is thin at $x$. By definition, there is an open neighborhood $U\subseteq \Omega $ of $x$ and $\varphi \in \mathrm{PSH}(U)$ such that

We may assume that $\varphi |_{E\cap U}\leq 0<\varphi (x)$. Then the set $\{ y\in U:\varphi (y)>0\} $ suffices for our purpose. $\square$