PomeCorollary 14.19.12.AB62A0258D5D If $u,v\in \mathrm{PSH}(X,\omega )\cap L^\infty (X)$ and $\lambda >0$, then
\[ \mathrm{e}^{-\lambda v}\omega _ v^ n\geq \mathrm{e}^{-\lambda u}\omega _ u^ n \quad \Longrightarrow \quad v\leq u. \]
Corollary 14.19.12AB62A0258D5D
Proof. For every $t>0$, the assumption gives $\omega _ u^ n\leq \mathrm{e}^{-\lambda t}\omega _ v^ n$ on $\{ u<v-t\} $. Since $\omega _{v-t}=\omega _ v$, theorem 14.19.10, applied to $u$ and $v-t$, gives $u\geq v-t$. Letting $t\downarrow 0$ proves $v\leq u$. $\square$
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