Lemma 16.22.3B36FB6000F3B

Proof. (1) Off the pole set of $u$, the functions $u$ and $v$ are finite and $f=u+(a-1)(u-v)\leq u$, because $u\leq v$; on the pole set, $f=-\infty $. Thus $f$ is a Borel function with $f\leq u$. Let $K$ be a compact subset of a chart. Since $u$ and $v$ are quasi-continuous (lemma 16.11.2(1)), there are open sets of arbitrarily small capacity outside which both are continuous on $K$, as functions with values in $[-\infty ,\infty )$. On the complement in $K$ of such a set, $f$ is continuous at every point where $u$ is finite, because $v\geq u$ is finite near such a point, and at a pole of $u$ it is continuous with value $-\infty $, because $f\leq u$. Hence $f$ is quasi-continuous. Off the pole set of $u$, the inequalities $\eta \leq f$ and $a^{-1}\eta +(1-a^{-1})v\leq u$ are equivalent. Since the pole set is pluripolar, and an inequality between quasi-plurisubharmonic functions that holds quasi-everywhere holds everywhere, this proves the equivalence.

(2) By (1), the potentials satisfying the inequality are exactly the competitors for $f$; since $f\leq u$, they lie below $u$ quasi-everywhere, hence everywhere. Thus $\mathcal A_\theta (f)$ is nonempty, by assumption, and uniformly bounded above, and proposition 16.11.4 shows that $\eta $ is its largest member; in particular $\eta \leq u$ and $\gamma \leq u$. By theorem 16.13.4, $\mu _\eta $ is carried by $\{ \eta =f\} $. At a point of this set at which $u$ is finite, $\eta =au+(1-a)v$ means $\gamma =u$; since $\mu _\eta $ does not charge the pole set of $u$, it is carried by $D$. Moreover $\mu _\gamma \geq a^{-n}\mu _\eta $, and lemma 16.13.1, applied to $\gamma \leq u$, gives $\mathbf1_ D\mu _\gamma \leq \mathbf1_ D\mu _ u$. Hence