Proof.
Step 1. We first reduce to the case where $\varphi ,\psi \in \mathcal{E}^{\infty }(X,\theta ;\phi )$.
Assume that (16.5.60) is known in that case. For each $C>0$, we let
\[ \varphi _ C=\varphi \lor (\phi -C),\quad \psi _ C=\psi \lor (\phi -C). \]
Then
\[ E_{\theta }^{\phi }(\varphi _ C\lor \psi _ C)+E_{\theta }^{\phi }(\varphi _ C\land \psi _ C)\geq E_{\theta }^{\phi }(\psi _ C)+E_{\theta }^{\phi }(\varphi _ C). \]
As $C\to \infty $, we have
\[ \varphi _ C\searrow \varphi ,\quad \psi _ C\searrow \psi ,\quad \varphi _ C\lor \psi _ C\searrow \varphi \lor \psi , \]
and, by the monotonicity and maximality of the rooftop,
\[ \varphi _ C\land \psi _ C\searrow \varphi \land \psi . \]
Applying proposition 16.5.20 to these four decreasing sequences gives (16.5.60).
Step 2. We assume that $\varphi ,\psi \in \mathcal{E}^{\infty }(X,\theta ;\phi )$.
Thanks to 16.5.57 and the plurifine locality of the non-pluripolar product proposition 13.11.4 1, we have
\[ \begin{aligned} E_{\theta }^{\phi }(\varphi \lor \psi )-E_{\theta }^{\phi }(\varphi )=& \int _0^1\int _ X (\varphi \lor \psi -\varphi )\, \theta _{(1-t)\varphi +t(\varphi \lor \psi )}^ n\, \mathrm{d}t\\ =& \int _0^1 \int _{\{ \psi >\varphi \} } (\psi -\varphi )\, \theta _{(1-t)\varphi +t\psi }^ n\, \mathrm{d}t. \end{aligned} \]
On the other hand,
\[ \begin{aligned} & E_{\theta }^{\phi }(\psi )-E_{\theta }^{\phi }(\varphi \land \psi )\\ =& \int _0^1 \int _ X (\psi -\min \{ \varphi ,\psi \} )\, \theta _{((1-t)\varphi +t\psi )\land \psi }^ n\, \mathrm{d}t\\ \leq & \int _0^1 \int _{\{ \varphi <\psi \} } (\psi -\varphi )\, \theta _{(1-t)\varphi +t\psi }^ n\, \mathrm{d}t& \textup{by \href{../../tag/4CF7246E6E5A/}{16.3.3}}. \end{aligned} \]
Adding these equations together, we conclude (16.5.60).
$\square$
Pome
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