Proposition 16.11.4641F3FEC623D

Proof. (1) Cover $X$ by finitely many coordinate balls on which $\theta $ has smooth local potentials $h$ (lemma 5.4.2). Applying theorem 5.10.2 on each ball to the plurisubharmonic functions $v+h$, $v\in \mathcal A_\theta (f)$, and taking the union of the resulting countable subfamilies, we find competitors $v_1,v_2,\ldots $ with $(\sup _ kv_ k)^*=P_\theta (f)$. The finite maxima $w_ k=\max \{ v_1,\ldots ,v_ k\} $ are again competitors, and they increase with $k$. Each $w_ k$ satisfies $w_ k\leq f$ outside a pluripolar set $N_ k$, and $N=\bigcup _ kN_ k$ is pluripolar by theorem 5.47.8. By corollary 5.47.11, $\sup _ kw_ k=P_\theta (f)$ outside a pluripolar set $N'$. Hence $P_\theta (f)\leq f$ outside $N\cup N'$. Since the family is nonempty and uniformly bounded above, $P_\theta (f)$ is $\theta $-plurisubharmonic by theorem 5.10.2. Thus $P_\theta (f)$ is a competitor, and by construction it dominates every competitor.

(2) Let $v$ be a competitor, let $x\in X$, and choose a coordinate ball $B(x,r)$ and a smooth function $h$ with $\mathrm{dd}^{\mathrm{c}}h=\theta $ near its closure. Since $v+h$ is plurisubharmonic and $v\leq f$ outside a Lebesgue null set, the sub-mean value inequality gives

As $r\to 0$, the right side tends to at most $f(x)+h(x)$, because $f+h$ is upper semicontinuous. Hence $v(x)\leq f(x)$.

(3) A modification of $f$ on a pluripolar set does not change $\mathcal A_\theta (f)$, since the union of two pluripolar sets is pluripolar. Moreover $\mathcal A_\theta (f+c)=\mathcal A_\theta (f)+c$, and $\mathcal A_\theta (f)\subset \mathcal A_\theta (g)$ if $f\leq g$ quasi-everywhere. These relations between the families of competitors give the three assertions.

(4) By (3), $\mathcal A_\theta (f)-c\subset \mathcal A_\theta (g)\subset \mathcal A_\theta (f)+c$. Hence $\mathcal A_\theta (g)$ is nonempty and uniformly bounded above, and the two inequalities follow from (3).

(5) By (3) the envelopes $P_\theta (f_ j)$ decrease; if some family $\mathcal A_\theta (f_ j)$ is empty, then so are all later ones and $\mathcal A_\theta (f)$. Let $v$ be the pointwise limit of $P_\theta (f_ j)$. Suppose first that $v(x_0)>-\infty $ for some $x_0$. Then $\sup _ XP_\theta (f_ j)\geq v(x_0)$ for every $j$, so $v\in \mathrm{PSH}(X,\theta )$ by corollary 10.3.6. By (1), $v\leq P_\theta (f_ j)\leq f_ j$ outside a pluripolar set $N_ j$, so $v\leq \inf _ jf_ j=f$ outside the pluripolar set $\bigcup _ jN_ j$. Thus $v$ is a competitor for $f$ and $v\leq P_\theta (f)$; the reverse inequality follows from $P_\theta (f)\leq P_\theta (f_ j)$. Suppose next that $v\equiv -\infty $. A competitor $w$ for $f$ would be a competitor for every $f_ j$, so $w\leq P_\theta (f_ j)$ for all $j$ and $w\leq v$, which is impossible. Hence $\mathcal A_\theta (f)=\varnothing $ and $P_\theta (f)=-\infty =v$. $\square$