Lemma 5.29.19505602F4CB54

Proof. In the decreasing case $u$ is a common lower bound for the sequence and its limit; in the increasing case $u_1$ is one. Choose a Hermitian form $\Omega \geq \theta $ and truncate all functions at $-T$. The truncations are bounded, monotone and $\Omega $-plurisubharmonic, so the convergence in capacity of locally bounded monotone sequences, recalled in section 5.27, applies to them on the balls of a finite cover. By quasi-continuity of the common lower bound, there is an open set of arbitrarily small capacity containing its pole set such that the restriction of the lower bound to the compact complement is continuous. There it is finite and hence bounded below, so the capacity of the sublevel set of the lower bound at height $-T$ tends to zero as $T\to \infty $. Since the functions agree with their truncations outside this sublevel set, letting $T\to \infty $ proves the lemma. $\square$