Corollary 13.9.6D91589CC5A89

Proof. After adding a constant to $\rho $ we may assume $\rho \leq u$. For $0<t<1$ put $q_ t=(1-t)v+t\rho $. Then $u\geq q_ t-O(1)$, and $\{ u<q_ t\} \subset \{ u<v\} $, because $u<(1-t)v+t\rho \leq (1-t)v+tu$ forces $u<v$. Multilinearity and proposition 13.9.5 therefore give

As $t\downarrow 0$, every point where $u<v$ and $\rho >-\infty $ eventually lies in $\{ u<q_ t\} $, so Fatou's lemma, applied along a sequence $t\downarrow 0$, gives $\mu _\rho (\{ u<v\} \cap \{ \rho >-\infty \} )=0$. Since $\mu _\rho $ gives no mass to the pole set of $\rho $, this proves the first assertion.

Now let the class be big and $u$ have minimal singularities. The class contains a Kähler current $\theta +\mathrm{dd}^{\mathrm{c}}\rho \geq \delta \omega $ with $\delta >0$. Then $u\geq \rho -O(1)$ by minimality, and $\mu _\rho \geq \delta ^ n\omega ^ n$, because multilinearity expands $\langle (\theta +\mathrm{dd}^{\mathrm{c}}\rho )^ n\rangle =\langle (\delta \omega +(\theta +\mathrm{dd}^{\mathrm{c}}\rho -\delta \omega ))^ n\rangle $ into positive terms, one of which is $\delta ^ n\omega ^ n$. The first assertion gives $u\geq v$ almost everywhere with respect to $\omega ^ n$, that is, almost everywhere for Lebesgue measure in charts. Near each point, $u$ and $v$ become plurisubharmonic after adding a common smooth local potential $h$ of $\theta $, and then $\max \{ u,v\} +h$ and $u+h$ are plurisubharmonic functions that agree almost everywhere; by theorem 5.9.1, theorem 3.4.12 they agree everywhere, that is, $u\geq v$ everywhere. $\square$