Corollary 5.16.9F3F797E97E9D

Proof. We may assume that $X$ is connected. Write $\varphi =\operatorname *{sup}_{i\in I}{}\! {}^{*} \varphi _ i$. Then $\varphi \in \mathrm{PSH}(X)$ by proposition 5.14.2.

We observe that the $\leq $ inequality is trivial. It remains to argue the reverse inequality.

It follows from proposition 5.14.3 that we may assume that $I$ is countable. When $I$ is finite, this is already proved in proposition 5.16.8. Otherwise, we may further assume that $I=\mathbb {Z}_{>0}$. Thanks to proposition 5.16.8, we may further assume that $(\varphi _ i)_{i\in \mathbb {Z}_{>0}}$ is an increasing sequence. Furthermore, since the problem is local, we may assume that $x=0$, that $X=B_ n(0,1)$ for some $n\in \mathbb {N}$, and that $(\varphi _ i)_ i$ is uniformly bounded from above on some ball $B_ n(0,r_1)$ with $0<r_1<1$. Since the sequence $(\varphi _ i)_ i$ is increasing, the sequence $\nu (\varphi _ i,0)$ is decreasing. Thus the constants in (5.16.6) can be chosen independently of $i$, and we have

for all $x\in B_ n(0,r_1)\setminus \{ 0\} $ and all $i\geq 1$, where $C$ is a constant independent of $i$. In particular, thanks to proposition 5.14.5, for almost all $x\in B_ n(0,r_1)$, we have

Thanks to proposition 5.14.6, the same holds for all $x\neq 0$ near $0$, and hence

This completes the proof. $\square$