Proposition 10.5.268A6D68513B0

Proof. The positive current $T_0=\theta +\mathrm{dd}^{\mathrm{c}}V_\theta \in \alpha $ gives the inequalities, by the bound preceding the proposition. Suppose that $\alpha $ is big, and choose $S\in \alpha $ with $S\geq \delta \omega $, $\delta >0$. For $\varepsilon >0$ put $T_\varepsilon =\theta +\mathrm{dd}^{\mathrm{c}}V_{\theta +\varepsilon \omega }\geq -\varepsilon \omega $ and $t=\varepsilon /(\delta +\varepsilon )$. Then $(1-t)T_\varepsilon +tS\geq \bigl (-(1-t)\varepsilon +t\delta \bigr )\omega =0$ is a positive current in $\alpha $, whose potential $(1-t)V_{\theta +\varepsilon \omega }+t\varphi _ S$, with $S=\theta +\mathrm{dd}^{\mathrm{c}}\varphi _ S$, is more singular than $V_\theta $. Since Lelong numbers are additive and positively homogeneous (proposition 6.7.9) and do not decrease when a function is replaced by a more singular one, we obtain

As $\varepsilon \downarrow 0$ we have $t\to 0$, and since $\nu (S,x)$ is finite we obtain $\nu (V_\theta ,x)\leq \nu (\alpha ,x)$. For a prime divisor $D$, evaluate the same inequality at a point $x\in D$ at which the Lelong numbers of $V_{\theta +\varepsilon \omega }$ and of $S$ attain their generic values along $D$ (proposition 6.9.7); it gives $\nu (V_\theta ,D)\leq \nu (V_\theta ,x)\leq (1-t)\nu (\alpha ,D)+t\nu (S,D)$, and letting $\varepsilon \downarrow 0$ proves the equality for $D$.

If $\alpha $ is modified nef and $T_\varepsilon $ is as in definition 10.4.11, then $T_\varepsilon =\theta +\mathrm{dd}^{\mathrm{c}}\varphi _\varepsilon $ with $\varphi _\varepsilon \in \mathrm{PSH}(X,\theta +\varepsilon \omega )$, so $\nu (V_{\theta +\varepsilon \omega },D)\leq \nu (\varphi _\varepsilon ,D)=0$ for every $\varepsilon >0$; hence $\nu (\alpha ,D)=0$. Conversely, if $\nu (\alpha ,D)=0$ for every $D$, then the currents $\theta +\mathrm{dd}^{\mathrm{c}}V_{\theta +\varepsilon \omega }$ satisfy the conditions of definition 10.4.11. $\square$