Lemma 10.31.35F92458E62F62

Proof. Since $u_ j$ and $v_ j$ decrease to $u$ and $v$, the bounds give $h\leq a\leq \min (u,v)\preceq \varphi $, so $a\in \mathcal E_\varphi $. After discarding finitely many indices we may assume $t_ j\geq 0$, so all potentials $a_ j,a$ are nonpositive. Let $E_ j$ be the intersection of the sets $\{ u>\sigma -t_ j\} $ and $\{ v>\sigma -t_ j\} $, taken only for those inputs that are truncated; a constant sequence imposes no condition. On $E_ j$, locality gives $\mu _{u_ j}=\mu _ u$ and $\mu _{v_ j}=\mu _ v$. Since $a,a_ j\leq u_ j,v_ j$, we have

If $x\notin E_ j$, then one truncated input and its truncation are at most $-t_ j$ at $x$, so $a_ j(x),a(x)\leq -t_ j$. Moreover $|a_ j-a|\mathrm{e}^{(a+a_ j)/4}$ is bounded by an absolute constant, since $a_ j,a\leq 0$. Thus

The assumed uniform mass bound is used only in the last term. The fixed measure $\mu _ u+\mu _ v$ is finite by bounded mass. Lemma 5.44.2, applied to $\mathrm{e}^{a_ j}$ and $\mathrm{e}^ a$, shows that $a_ j\to a$ in measure for $\mu _ u+\mu _ v$, since $a$ is finite almost everywhere for this non-pluripolar measure. The integrand in the first term of (10.31.36) is uniformly bounded and tends to zero in measure, so its integral tends to zero. Lemma 10.31.34 now gives convergence in capacity. When a subsequence is taken, the corresponding truncation levels $t_ j$ are kept.

To identify the equation, fix $T>0$ and choose a continuous nondecreasing function $g_ T:[-\infty ,0]\to [0,1]$ that vanishes on $[-\infty ,-2T]$ and equals one on $[-T,0]$, such that $g_ T\uparrow 1$ at finite points as $T\to \infty $. Put $\chi _ T=g_ T(h)$. Since $h\leq a\leq u,v$, both $u$ and $v$ exceed $-2T$ on $\{ \chi _ T>0\} $. Therefore, once $t_ j>2T$, every truncated input agrees there with its limit, and

Choose a Hermitian form $\Omega \geq \theta $. On the left-hand side, replace $a_ j$ by $\max (a_ j,-2T)$, which does not change $\chi _ T\mu _{a_ j}$ because $\chi _ T$ vanishes where $a_ j\leq -2T$, and expand the bounded product as

Bounded capacity continuity with the fixed quasi-continuous weight $\chi _ T$ therefore gives convergence to $\chi _ T\mu _ a$, after testing against a continuous function. The uncut product may be signed, but its terms have uniformly capacity-controlled total variations.

On the right-hand side, $\chi _ T \mathrm{e}^{-u}$ and $\chi _ T \mathrm{e}^{-v}$ are bounded by $\mathrm{e}^{2T}$. Fixed-measure exponential convergence therefore gives convergence in total variation to $\chi _ T\mathrm{e}^ a(\mathrm{e}^{-u}\mu _ u+\mathrm{e}^{-v}\mu _ v)$. Let $T\to \infty $. The limiting right-hand side is at most $\mu _ u+\mu _ v$, because $a\leq u,v$. Both sides ignore the pole set of $h$, so the cutoff identities prove the global equation and the finiteness of its right-hand side. $\square$