Corollary 10.29.698BD3A283B8E

Proof. Continuous functions on a compact manifold are bounded, so any two of them are equivalent, and their Lelong numbers and multiplier ideals are those of a bounded function. Since $\theta \geq 0$, the constant $0$ belongs to $\mathrm{PSH}(X,\theta )$, so $V_\theta =0$, and $P_\theta [u]=P_\theta [0]=V_\theta $. By theorem 10.29.4, locally $\theta +\mathrm{dd}^{\mathrm{c}}u=\mathrm{dd}^{\mathrm{c}}f(y)$ with $f$ continuous and depending on the $s$ real variables $y_ j$; the Bedford–Taylor products of such a current vanish in degree $s+1\leq n$, because they are limits of the corresponding products for smooth convex approximants of $f$, whose complex Hessians have rank at most $s$. Thus $(\theta +\mathrm{dd}^{\mathrm{c}}u)^ n=0$. The form $\theta $ is smooth and semipositive, hence nef. Suppose that $\rho \in \mathrm{PSH}(X,\theta )$ satisfied $\theta +\mathrm{dd}^{\mathrm{c}}\rho \geq \delta \omega $ with $\delta >0$. The function $\rho $ is continuous, and on a small coordinate ball with coordinates $\zeta $ we may write $\theta =\mathrm{dd}^{\mathrm{c}}a$ with $a$ smooth and $\delta \omega \geq \delta '\mathrm{dd}^{\mathrm{c}}|\zeta |^2$ with $\delta '>0$. Then $a+\rho -\delta '|\zeta |^2$ is plurisubharmonic, and expanding $(\mathrm{dd}^{\mathrm{c}}(a+\rho ))^ n$ as a sum of positive mixed Bedford–Taylor products gives $\mathrm{MA}_\theta (\rho )\geq \delta '^ n(\mathrm{dd}^{\mathrm{c}}|\zeta |^2)^ n\neq 0$ on the ball, a contradiction. Hence $\theta $ is not big. $\square$