Proof.
Observe that $\varphi \leq 0$ by proposition 5.17.3.
Step 1. We first assume that $I\cong \mathbb {Z}_{>0}$ as directed sets. Namely, $(\varphi _ i)_{i>0}$ is a sequence.
For each $k>0$, let
\[ \psi _ k=\operatorname *{sup}_{j\geq k}{}\! {}^{*} \varphi _ j. \]
Then $(\psi _ k)_{k}$ is a decreasing sequence in $\mathrm{PSH}(X,\theta )$ with limit $\varphi $, see corollary 5.14.8 for example.
Thanks to 13.13.14, for each $k>0$, we have
\[ \mathds {1}_{\{ \varphi _ k=\psi _ k\} }\, \theta _{\varphi _ k}^ n\leq \mathds {1}_{\{ \varphi _ k=\psi _ k\} }\, \theta _{\psi _ k}^ n. \]
Multiplying both sides with $\mathds {1}_{\{ \varphi _ k=0\} }$, we find that
\[ \mathds {1}_{\{ \varphi _ k=0\} }\, \theta _{\varphi _ k}^ n\leq \mathds {1}_{\{ \varphi _ k=0\} }\, \theta _{\psi _ k}^ n. \]
Take $b,C>0$. For each $k>0$, we have
\[ \begin{split} \int _{\{ \varphi _ k=0\} } \theta _{\varphi _ k}^ n\leq \int _{\{ \varphi _ k=0\} } \theta _{\psi _ k}^ n\leq \int _{\{ \psi _ k=0\} } \theta _{\psi _ k}^ n\\ =\int _{\{ \psi _ k=0\} } \theta _{\psi _ k\lor (V_{\theta }-C)}^ n\leq \int _ X \mathrm{e}^{b\psi _ k}\, \theta _{\psi _ k\lor (V_{\theta }-C)}^ n, \end{split} \]
where the equality part follows from proposition 13.11.4 1.
Letting $k\to \infty $ with the help of theorem 13.13.9, we conclude that
\[ \varlimsup _{k\to \infty }\int _{\{ \varphi _ k=0\} } \theta _{\varphi _ k}^ n\leq \int _ X \mathrm{e}^{b\varphi }\, \theta _{\varphi \lor (V_{\theta }-C)}^ n. \]
Letting $b\to \infty $ and then $C\to \infty $, we conclude (13.13.17).
Step 2. We now handle the general net case.
Fix a volume form on $X$, so that we can talk about the $L^1$-norm on $X$ in the sequel.
Suppose that (13.13.17) fails. Then there is $\delta >0$ such that, for every $i_0\in I$, one can find $i\geq i_0$ satisfying
\[ \int _{\{ \varphi _ i=0\} }\theta _{\varphi _ i}^ n>\int _{\{ \varphi =0\} }\theta _{\varphi }^ n+\delta . \]
Since $\varphi _ i\to \varphi $ in $L^1$, for every $k>0$ there is $a_ k\in I$ such that
\[ \| \varphi _ i-\varphi \| _{L^1}<k^{-1},\quad i\geq a_ k. \]
We can then recursively choose an increasing sequence $(i_ k)_{k>0}$ in $I$ with $i_ k\geq a_ k$ and
\[ \int _{\{ \varphi _{i_ k}=0\} }\theta _{\varphi _{i_ k}}^ n>\int _{\{ \varphi =0\} }\theta _{\varphi }^ n+\delta . \]
This sequence converges to $\varphi $ in $L^1$, contradicting Step 1.
$\square$
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