PomeProposition 5.15.18.E49CF7960E4C Let $\varphi \in \mathrm{QPSH}(X)$. Then $\varphi $ is $\mathcal{F}$-continuous.
Proposition 5.15.18E49CF7960E4C
Proof. It suffices to show that $\varphi |_{\{ \varphi \neq -\infty \} }$ is $\mathcal{F}$-continuous, since $\{ \varphi =-\infty \} $ is $\mathcal{F}$-closed.
The problem is local, so we may assume that $X\subseteq \mathbb {C}^ n$ is a domain and $\varphi =\psi +g$, where $\psi \in \mathrm{PSH}(X)$ and $g\in C^{\infty }(X)$ and $|g|\leq C$ for some $C>0$. For an open interval $(a,b)\subseteq \mathbb {R}$, it suffices to show that
\[ U\coloneqq \left\{ x\in X:a<\varphi (x)<b\right\} =\left\{ x\in X:a-g(x)<\psi (x)<b-g(x)\right\} \]
is $\mathcal{F}$-open. Take $x\in U$. We can find an open neighborhood $V$ of $x$ in $U$ such that
\[ \sup _{y\in V} \left( a-g(y)\right)<\psi (x)< \inf _{y\in V} \left( b-g(y) \right). \]
Therefore,
\[ \left\{ z\in V:\sup _{y\in V} \left(a-g(y)\right)<\psi (z)<\inf _{y\in V} \left(b-g(y)\right)\right\} \]
is an $\mathcal{F}$-open neighborhood of $x$ in $U$. We conclude that $U$ is $\mathcal{F}$-open. $\square$
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