Pome4D5FAF80D021Lemma 16.22.2.4D5FAF80D021 Let $\phi ,\tau \in \mathrm{PSH}(X,\theta )$ with $P_\theta [\phi ]=\phi $ and $\tau \preceq \phi $. Then
In particular, if $\tau \leq 0$, then $\tau \leq \phi $.
Proof. The potential $\tau -\sup _ X\tau $ is nonpositive and more singular than $\phi $, so it is a competitor for $P_\theta [\phi ]=\phi $. Hence $\tau -\phi \leq \sup _ X\tau $ wherever $\phi $ is finite. Conversely, $\phi \leq 0$, so $\tau -\phi \geq \tau $ wherever $\phi $ is finite. The pole set of $\phi $ is pluripolar, and removing a pluripolar set does not change the supremum of the quasi-plurisubharmonic function $\tau $: an upper bound for $\tau $ off a Lebesgue null set holds everywhere. This proves the reverse inequality. $\square$
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