PomeLemma 5.40.7.4798FCFAC293 Let $\rho \in \mathrm{PSH}(X,\theta )$ have analytic singularities with smooth remainder and satisfy $\theta +\mathrm{dd}^{\mathrm{c}}\rho \geq \varepsilon \omega $ for some $\varepsilon >0$. Set $\Omega =\{ \rho >-\infty \} $. If $\varphi \in \mathrm{PSH}(X,\theta )$ has minimal singularities and is continuous on $\Omega $, then $\varphi -\rho $ is continuous, bounded below, and proper on $\Omega $: all its sublevel sets are compact in $\Omega $.
Lemma 5.40.74798FCFAC293
Proof. Adding constants to $\rho $ and $\varphi $ does not affect the assertions, so we may assume that $\rho \leq 0$ and $\sup _ X\varphi =0$. Since $\varphi $ has minimal singularities, there is a constant $C$ with $\varphi \geq V_\theta -C$. Choose $M\geq 0$ with $\theta \geq -M\omega $. If $a\in (0,1)$ is sufficiently close to one, then
\[ \theta +\mathrm{dd}^{\mathrm{c}}(a\rho ) =a(\theta +\mathrm{dd}^{\mathrm{c}}\rho )+(1-a)\theta \geq \bigl (a\varepsilon -(1-a)M\bigr )\omega >0. \]
Thus $a\rho $ is a competitor for $V_\theta $, and therefore
\[ 0\geq \varphi \geq V_\theta -C\geq a\rho -C, \]
and hence
By hypothesis $\varphi $ is continuous on $\Omega $, and $\rho $ is smooth there because it has analytic singularities with smooth remainder. Hence $\varphi -\rho $ is continuous on $\Omega $, and by (5.40.8) it is bounded below. Since $\rho \to -\infty $ at $X\setminus \Omega $, the sets $\{ \rho \geq -c\} $ are compact subsets of $\Omega $, and (5.40.8) places every sublevel set of $\varphi -\rho $ in one of them. Being closed in $\Omega $, each sublevel set is therefore compact. $\square$
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