Pome9BB432CC1FE6Corollary 12.6.25.9BB432CC1FE6 If $A\geq 0$ is singular, then every bounded $u\in \mathrm{PSH}(X,\theta _ A)$ satisfies
For every $u\in \mathrm{PSH}(X,\theta _ A)$, bounded or not, the non-pluripolar top power of $\theta _ A+\mathrm{dd}^{\mathrm{c}}u$ is also zero. Thus the nonzero semipositive boundary forms $\theta _ A$ of proposition 12.6.5 have zero non-pluripolar top mass.
Proof. For $\varepsilon >0$, the same bounded $u$ is $\theta _{A+\varepsilon I}$-plurisubharmonic, so the preceding proposition bounds its mass by $H\det (A+\varepsilon I)$, which tends to zero with $\varepsilon $. By the multilinear construction of bounded products in proposition 5.28.2,
is a polynomial in $\varepsilon $ whose coefficients are mixed products of finite mass. Its total mass therefore converges to $\int _ X(\theta _ A+\mathrm{dd}^{\mathrm{c}}u)^3$ as $\varepsilon \downarrow 0$, so this mass is zero, and positivity makes the measure zero.
For unbounded $u$, we subtract a constant so that $u\leq 0$. Since $\theta _ A\geq 0$, the truncations $u_ k=\max (u,-k)$ are bounded and $\theta _ A$-plurisubharmonic, so their top products vanish by the bounded case. Restricting them to $\{ u>-k\} $ and passing to the limit that defines the non-pluripolar product gives zero as well. $\square$
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