Theorem 10.31.499A2FDB1A65B5

Proof. By the scaling used in theorem 10.31.38, it suffices to take $\lambda =1$. For positive integers $j$, let

These are nonzero finite non-pluripolar measures, and they increase with $j$. Only the weights are truncated; the functions $\max (u,-j)$ and $\max (v,-j)$ need not be $\theta $-plurisubharmonic. Theorem 10.31.44 gives $a_ j\in \mathcal E_\varphi $ with $\mu _{a_ j}=\mathrm{e}^{a_ j}\tau _ j$. Since $\tau _ j$ increases, exponential order implies that $a_ j$ decreases.

Set $r=P_\theta (u,v)\in \mathcal E_\varphi $. On $\{ a_ j<r\} $, one has $r\leq u,v$ and

with the same inequality for $v$. Consequently

Also $a_ j\leq a_1\leq \varphi +\sup _ Xa_1$. The one-sided first-moment criterion of proposition 10.31.8, with fixed reference $r$, bounds $\sup _ Xa_ j$ from below. Thus the decreasing sequence has an $L^1$ limit $a\not\equiv -\infty $, and the same criterion puts $a$ in $\mathcal E_\varphi $.

We identify the equation before comparing $a$ with the inputs. The rooftop

lies below $a_ j$, $a$, $u$ and $v$.

For $T>0$, let $\chi _ T=g_ T(q)$, with $g_ T$ as in the proof of lemma 10.31.35: a continuous nondecreasing cutoff that vanishes where $q\leq -2T$, equals one where $q\geq -T$, and increases to one off the pole set of $q$ as $T\to \infty $. On $\{ \chi _ T>0\} $ both inputs exceed $-2T$. Hence, for $j>2T$,

The factors $\chi _ T\mathrm{e}^{-u}$ and $\chi _ T\mathrm{e}^{-v}$ are bounded by $\mathrm{e}^{2T}$. Since $a_ j\downarrow a$ and $a_ j\leq \sup _ Xa_1$, dominated convergence against the fixed measure $\mu _ u+\mu _ v$ gives convergence in total variation of the right-hand sides to the corresponding expression with $a$.

For the left-hand side, $a_ j\to a$ in capacity by lemma 5.29.19. Choose a Hermitian form $\Omega \geq \theta $, and replace $a_ j$ by $b_ j=\max (a_ j,-2T)$ before forming bounded products. The $b_ j$ are uniformly bounded $\Omega $-plurisubharmonic functions converging in capacity. The expansion

and bounded capacity continuity apply against the fixed quasi-continuous weight $\chi _ T$. Where $\chi _ T\ne 0$ no potential is truncated, so locality gives convergence to $\chi _ T\mu _ a$. In this way the possibly signed bounded product is handled before the cutoff is inserted.

Thus (10.31.50) holds in the limit. Let $T\to \infty $. Both sides ignore the poles of $q$, so monotone convergence gives

Since $a\in \mathcal E_\varphi $, this also shows that the weighted right-hand side is finite.

For $\varepsilon >0$, the equation implies

Strict domination, theorem 13.19.1, gives $u\geq a-\varepsilon $, and letting $\varepsilon \downarrow 0$ gives $u\geq a$. Similarly $v\geq a$, so $\mu _ a\leq \mu _ u+\mu _ v$. Two solutions have the same exponentially weighted measure, so corollary 13.19.3, applied in both directions, gives uniqueness. $\square$