Proposition 12.36.13A78333ED147

Proof. The residual background is

It is positive for $0\leq a<1$ and semipositive for $a=1$, so every coefficient $a\leq 1$ is feasible. For the reverse inequality, consider the metrics $\gamma _ t=\alpha +t\zeta $, $t>0$. They are Gauduchon, since

Using $\zeta ^2=0$, we obtain

and the threshold formula gives $\tau _\omega (H)=1$.

For $a<1$,

Hence the zero potential solves the residual smooth equation, and uniqueness of the scalar gives the volume. At $a=1$, the form $R_1=\zeta $ is nef, because $\zeta +\epsilon \omega >0$ for $\epsilon >0$, and the same computation gives

By theorem 12.11.1, the residual volume at $a=1$ is zero. Moreover $\omega +\mathrm{dd}^{\mathrm{c}}\sigma _ H=[H]+\zeta $, so the non-pluripolar measure of $\sigma _ H$ is $\zeta ^3=0$.

Finally, fix $a<1$ and choose $0<\delta \leq \min \{ 1-a,1\} $, so that $R_ a-\delta \omega \geq 0$. For the smooth functions $V_ j$ of theorem 23.1.1,

Monotonicity of the determinant on positive matrices bounds the mass of this form below by $\delta ^3\int _ X(\omega +\mathrm{dd}^{\mathrm{c}}V_ j)^3$, which tends to infinity. Thus a finite residual volume and an explicit divisorial threshold coexist with unbounded upper mass. $\square$