Proposition 10.12.24075D8C83DE1

Proof. By the monotonicity just noted,

and each set in this union is analytic by Siu's analyticity theorem (5.34.2).

If $\alpha $ is nef, each family $\alpha [-\varepsilon \omega ]$ contains a smooth current $\theta +\mathrm{dd}^{\mathrm{c}}f$. By minimality, $V_\varepsilon \geq f-\sup _ Xf$, so $V_\varepsilon $ is bounded and all Lelong numbers of $T_{\min ,\varepsilon }$ vanish. Conversely, if the non-nef locus is empty, every $T_{\min ,\varepsilon }$ has zero Lelong numbers. The regularization just recalled then gives a smooth representative of $\alpha $ bounded below by $-2\varepsilon \omega $. Since $\varepsilon >0$ is arbitrary, $\alpha $ is nef. $\square$