Lemma 6.16.16A23B839899E

Proof. Since the $U_ i$ cover $X$, the maximum is taken over a nonempty set at every point. Fix $x_0\in X$ and choose $i$ with $x_0\in U_ i$. Indices $j$ with $x_0\notin \overline{U'_ j}$ do not occur near $x_0$. If $x_0\in \partial U'_ j$, then $\chi _ j=0$ near $x_0$, because $\chi _ j$ has compact support in $U'_ j$; so near $x_0$ the $j$th term, where it is defined, equals $v_ j\leq v_ i+a=v_ i+a\chi _ i$, since $\chi _ i=1$ on the neighborhood $U_ i$ of $x_0$. Such a term never exceeds the $i$th term and can be omitted. Hence on a small neighborhood of $x_0$, $v$ is the maximum of $v_ j+a\chi _ j$ over the fixed set of indices $j$ with $x_0\in U'_ j$. Each of these functions satisfies $\theta +aC_\chi \omega +\mathrm{dd}^{\mathrm{c}}(v_ j+a\chi _ j)\geq 0$. Adding a local potential of the closed form $\theta +aC_\chi \omega $ on a coordinate ball about $x_0$ (lemma 6.14.2), which makes these functions plurisubharmonic by lemma 6.14.4(2), and applying proposition 5.9.5, we obtain that $v$ is quasi-plurisubharmonic near $x_0$ with $\theta +\mathrm{dd}^{\mathrm{c}}v\geq -aC_\chi \omega $.

On $U_ j$ the $j$th term is $v_ j+a\geq v_ j$, and every term with index $l$ is at most $v_ l+a\leq v_ j+2a$. This proves the two bounds, also at points of the pole sets. $\square$