Theorem 17.5.37F3340D970A43

Proof. By proposition 13.9.2, big classes have positive volume, and every closed positive $(1,1)$-current $T$ satisfies $\int _ X\langle T^ n\rangle \leq \operatorname {vol}(\{ T\} )$. It therefore suffices to show that a pseudo-effective class $\alpha $ with $m=\operatorname {vol}(\alpha )>0$ is big, and we do so by verifying Lamari's criterion (lemma 17.5.28). Write $\alpha =\{ \theta \} $ with $\theta $ smooth and closed, choose $C>0$ such that $\theta +C\omega $ is positive definite, and put

the integrand is a positive multiple of the volume form at every point, by the positivity of mixed discriminants of positive definite matrices (lemma 5.3.3(2),(3)).

For $0<\varepsilon \leq 1$ let $\theta _\varepsilon =\theta +\varepsilon \omega $. Since $\alpha $ is pseudo-effective, there is $q\in \mathrm{PSH}(X,\theta )$ (corollary 10.2.19), and $\theta _\varepsilon +\mathrm{dd}^{\mathrm{c}}q\geq \varepsilon \omega $; thus $\{ \theta _\varepsilon \} =\alpha +\varepsilon \{ \omega \} $ is big. By proposition 16.21.2(4), $P_{\theta _\varepsilon }[V_{\theta _\varepsilon }]=V_{\theta _\varepsilon }$, and the mass of $V_{\theta _\varepsilon }$ is

the inequality holds because $\operatorname {vol}(\alpha +\varepsilon \{ \omega \} )$ decreases to $\operatorname {vol}(\alpha )$ as $\varepsilon \downarrow 0$, by definition 13.9.1, proposition 13.9.2. Hence $V_{\theta _\varepsilon }$ is a model potential of mass $m_\varepsilon $.

Fix a smooth strictly positive real $(n-1,n-1)$-form $\tau $ with $\mathrm{dd}^{\mathrm{c}}\tau =0$, normalized by $\int _ X\omega \wedge \tau =1$. The measure $m_\varepsilon \omega \wedge \tau $ has a smooth density with respect to $\omega ^ n$ and total mass $m_\varepsilon $. By corollary 17.5.26, applied in the big class $\{ \theta _\varepsilon \} $ with the reference $V_{\theta _\varepsilon }$, there are $u_\varepsilon \in \mathrm{PSH}(X,\theta _\varepsilon )$ and finite constants $B_\varepsilon $ such that

The constants $B_\varepsilon $ need not be bounded uniformly in $\varepsilon $; we only use that $u_\varepsilon $ is locally bounded outside a closed set of measure zero. By lemma 16.13.2, applied to the big class $\{ \theta _\varepsilon \} $, there is a closed pluripolar set $A_\varepsilon \subset X$ such that $V_{\theta _\varepsilon }$ is locally bounded on $X\setminus A_\varepsilon $; being pluripolar, $A_\varepsilon $ has Lebesgue measure zero in every chart (see the remarks after definition 5.47.2). Hence $u_\varepsilon $, which lies between $V_{\theta _\varepsilon }-B_\varepsilon $ and $V_{\theta _\varepsilon }$, is locally bounded on $X\setminus A_\varepsilon $, and so are the local potentials of $S_\varepsilon $, which differ from $u_\varepsilon $ by smooth local potentials of $\theta _\varepsilon $ (lemma 5.4.2).

We may thus apply lemma 17.5.33 to $S_\varepsilon $, with $A_\varepsilon $, $\nu =\omega \wedge \tau $ and $m_\varepsilon $ in place of $A$, $\nu $ and $m$:

We bound the two factors on the left. Since $\mathrm{dd}^{\mathrm{c}}\tau =0$, the first factor equals the cohomological pairing $\int _ X\theta _\varepsilon \wedge \tau =\int _ X\theta \wedge \tau +\varepsilon $. If $n\geq 2$, the form $\theta _\varepsilon +C\omega $ is positive definite, so theorem 13.5.4 gives

the last inequality holds because $\theta +(C+1)\omega =(\theta +(C+\varepsilon )\omega )+(1-\varepsilon )\omega $ and products of positive $(1,1)$-forms are positive. If $n=1$, the second factor equals $\int _ X\omega =C_0$. As the first factor is nonnegative, in both cases

Letting $\varepsilon \downarrow 0$ gives $\int _ X\theta \wedge \tau \geq m/(nC_0)$. Removing the normalization of $\tau $, we conclude that every smooth strictly positive $\mathrm{dd}^{\mathrm{c}}$-closed $(n-1,n-1)$-form $\tau $ satisfies (17.5.30) with $\delta =m/(nC_0)>0$. By lemma 17.5.28, the class $\alpha $ is big. $\square$