Theorem 13.17.1D7019DA65027

Proof. We first treat the lower volume. After adding a constant to $u_1$ we may assume $u_1\leq u_2$. Choose $v_2\in \mathrm{PSH}(X,\theta _2)$ with $v_2\sim u_2$ and, for real $t>0$, consider the rooftop envelope

Since $u_1\leq u_2\leq v_2+C$ for a constant $C$, the function $u_1-C$ is a competitor; moreover $p_ t\leq u_1+t$. Hence $p_ t\sim u_1$. By the contact theorem (theorem 15.8.6), the measure $\mathrm{MA}_{\theta _1}(p_ t)$ is carried by the union of $\{ p_ t=u_1+t\} \subset \{ u_1\leq v_2-t\} $ and $\{ p_ t=v_2\} $. On the first set, ordered contact in background $\theta _1$ (proposition 13.14.6) bounds it by $\mathrm{MA}_{\theta _1}(u_1)$. On the second set, $p_ t$ is also $\theta _2$-plurisubharmonic and lies below $v_2$; positivity of the mixed products (lemma 13.14.2) and ordered contact in background $\theta _2$ bound it by $\mathrm{MA}_{\theta _2}(v_2)$. Therefore

As $t\to +\infty $, the first measure on the right has total mass tending to zero: the sets $\{ u_1\leq v_2-t\} $ decrease to a subset of the pole set of $u_1$, and $\mathrm{MA}_{\theta _1}(u_1)$ is a finite measure that puts no mass on pluripolar sets. Integrating and letting $t\to +\infty $ gives $\underline{\operatorname {vol}}_{\theta _1}(u_1)\leq \int _ X\mathrm{MA}_{\theta _2}(v_2)$, and taking the infimum over $v_2$ proves the first inequality.

For the upper volume, choose any $v_1\in \mathrm{PSH}(X,\theta _1)$ with $v_1\sim u_1$, and put $w_ t=\max (v_1,u_2-t)$ for $t>0$. This function is $\theta _2$-plurisubharmonic and satisfies $w_ t\sim u_2$. On the set $D_ t=\{ v_1>u_2-t\} $, locality and positivity of the mixed products imply

The sets $D_ t$ exhaust the complement of the pole set of $v_1$, which carries no mass. Integrating and letting $t\to +\infty $ therefore gives $\int _ X\mathrm{MA}_{\theta _1}(v_1)\leq \overline{\operatorname {vol}}_{\theta _2}(u_2)$, and taking the supremum over $v_1$ proves the second inequality. $\square$