PomeProposition 16.28.55.9BBE1DE257CB If a smooth real $(1,1)$-form $\theta $ is big, every bounded $u\in \mathrm{PSH}(X,\theta )$ has strictly positive Monge–Ampère mass.
Proposition 16.28.559BBE1DE257CB
Proof. Choose $\rho \in \mathrm{PSH}(X,\theta )$ and $a>0$ with $\theta _\rho \geq a\omega $, normalize $\rho \leq 0$, and put
\[ p=P_{a\omega }(u-\rho ). \]
The obstacle is finite almost everywhere and bounded below, so lemma 5.40.2 shows that $p$ is bounded. The obstacle is quasi-continuous off the pluripolar pole set of $\rho $. The contact theorem, theorem 16.28.5, applied in the Hermitian background $a\omega $, shows that $\mathrm{MA}_{a\omega }(p)$ is carried by $D=\{ p+\rho =u\} $, outside the pole set of $\rho $.
By lemma 5.24.3, $p+\rho \leq u$ everywhere, and
\[ \theta +\mathrm{dd}^{\mathrm{c}}(p+\rho ) =(a\omega +\mathrm{dd}^{\mathrm{c}}p)+(\theta _\rho -a\omega ) \geq a\omega +\mathrm{dd}^{\mathrm{c}}p\geq 0. \]
On common finite truncation levels, positivity of the mixed products and ordered contact give
\[ \mathbf1_ D\mathrm{MA}_{a\omega }(p) \leq \mathbf1_ D\mathrm{MA}_\theta (p+\rho ) \leq \mathbf1_ D\mathrm{MA}_\theta (u). \]
Exhausting by these levels proves the same inequalities for the extended measures. Since $p$ is bounded, its Monge–Ampère mass with respect to the Hermitian form $a\omega $ is positive by corollary 14.19.11. Integrating the inequalities proves the assertion. $\square$
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