Lemma 16.13.3175BA4BB6A7C

Proof. Adding a constant to $f$, we may assume that $-C\leq f\leq 0$. Then $V_\theta -C\leq f$, so $V_\theta -C$ is a competitor, while every competitor is nonpositive and hence at most $V_\theta $. Thus $V_\theta -C\leq u\leq V_\theta $, and $u$ has minimal singularities.

Let $A$ be the proper analytic subset of lemma 16.13.2. Since $u$ has minimal singularities, it is locally bounded on $X\setminus A$, and $\mu _ u(A)=0$.

Let $x\in X\setminus A$ with $u(x)<f(x)$, and choose $\delta >0$ with $u(x)<f(x)-3\delta $. Since $u$ is upper semicontinuous and $f$ is lower semicontinuous, there is a coordinate ball $B\ni x$ with $\overline B\subset X\setminus A$ such that $u<u(x)+\delta $ and $f>f(x)-\delta $ on $\overline B$; thus $\sup _{\overline B}u\leq \inf _{\overline B}f-\delta $. Shrinking $B$, we may assume that $B\Subset B'$ for a coordinate ball $B'$ with $\overline{B'}\subset X\setminus A$ and that $\theta =\mathrm{dd}^{\mathrm{c}}h$ near $\overline{B'}$ for a smooth function $h$ (lemma 5.4.2), and then, shrinking $B$ further, that the oscillation $\operatorname {osc}_{\overline B}h=\sup _{\overline B}h-\inf _{\overline B}h$ is less than $\delta $.

We show that $\mu _ u$ vanishes on $B$. Since $u$ is bounded near $\overline{B'}$, corollary 10.3.10 gives $\widetilde u\in \mathrm{PSH}(X,\theta )$ with $\widetilde u\geq u$, $\widetilde u=u$ on $X\setminus B$ and $(\theta +\mathrm{dd}^{\mathrm{c}}\widetilde u)^ n=0$ on $B$. The function $\widetilde u+h$ is plurisubharmonic, hence subharmonic, on $B$, and at every point $\zeta \in \partial B$ its upper limit from $B$ is at most $(\widetilde u+h)(\zeta )=(u+h)(\zeta )\leq \sup _{\overline B}(u+h)$, by upper semicontinuity. The maximum principle (proposition 3.4.7) therefore gives

Hence $\widetilde u\leq f$ quasi-everywhere on $X$, so $\widetilde u$ is a competitor and $\widetilde u\leq u$. Therefore $\widetilde u=u$, and $\mu _ u=(\theta +\mathrm{dd}^{\mathrm{c}}\widetilde u)^ n$ vanishes on $B$.

The set $\{ u<f\} $ is open, because $f-u$ is lower semicontinuous. Covering $\{ u<f\} \setminus A$ by countably many balls as above, we obtain $\mu _ u(\{ u<f\} \setminus A)=0$. Moreover $u\leq f$ quasi-everywhere by proposition 16.11.4, so the set $\{ u>f\} $ is pluripolar. Since $\mu _ u$ gives no mass to $A$ or to $\{ u>f\} $, it is carried by $\{ u=f\} $. $\square$