Proposition 22.1.3915E20104530B

Proof. The $\geq $ direction in (22.1.40) is a direct consequence of proposition 22.1.23. It remains to prove the reverse inequality.

Note that (22.1.40) holds when $\operatorname{vol}\Gamma ^ i>0$ for each $i\in I$, as a consequence of proposition 17.15.22, corollary 16.9.12 and theorem 16.9.57.

In particular, for each Kähler form $\omega $ on $X$, we have

For our purpose, we need to show that for any $\epsilon >0$, we can find $\omega $ so that

We shall show that it is possible to choose $\omega $ so that the stronger statement holds:

Equivalently, we need to choose $\omega $ so that for any Kähler form $\omega '$ on $X$ dominated by $\omega $, we have

Fix a Kähler form $\omega _0$ on $X$ and a Kähler form $\Omega $ such that $\Omega \geq \theta +\omega _0$. It is enough to take $\omega =\delta \omega _0$ for some $\delta \in (0,1)$. If $\omega '$ is dominated by $\omega $, then $\theta +\omega '\leq \Omega $. Set

By the transition compatibility and proposition 17.12.19, the potential $\Gamma ^{i,\theta +\omega }_{-\infty }$ has the same $P$-singularity type in the class $\theta +\omega $ as $\Gamma ^{i,\theta +\omega '}_{-\infty }$. Hence its mass can be computed as

Therefore,

In the last inequality we used the facts that $0\leq \omega -\omega '\leq \omega $ and $\theta +\omega '\leq \Omega $, and the monotonicity theorem theorem 13.13.11. We can choose $\omega $ so that the right-hand side is less than $\epsilon /2$. Our assertion then follows. $\square$