Proof.
Since the conclusion is known when $\varphi ,\psi \in \mathcal{E}^{\infty }(X,\theta ;\phi )$ (see (16.5.15)), it suffices to prove that for each $k=0,\ldots ,n$, we have
9D55DDF004D1
16.5.32
\begin{equation} \lim _{C\to \infty } \int _ X \Bigl ( \varphi \lor (\phi -C)-\psi \lor (\phi -C) \Bigr ) \, \theta _{\varphi \lor (\phi -C)}^ k\wedge \theta _{\psi \lor (\phi -C)}^{n-k}=\int _ X (\varphi -\psi )\, \theta _{\varphi }^ k\wedge \theta _{\psi }^{n-k}. \end{equation}
For this purpose, we may assume that $\varphi ,\psi \leq \phi $ and it suffices to establish the following:
989753C108DD
16.5.33
\begin{equation} \lim _{C\to \infty } \int _ X \Bigl ( \varphi \lor (\phi -C)-\phi \Bigr ) \, \theta _{\varphi \lor (\phi -C)}^ k\wedge \theta _{\psi \lor (\phi -C)}^{n-k}=\int _ X (\varphi -\phi )\, \theta _{\varphi }^ k\wedge \theta _{\psi }^{n-k}. \end{equation}
Fix $C>0$. We compute the difference as follows:
\[ \begin{aligned} & \int _ X \Bigl ( \varphi \lor (\phi -C)-\phi \Bigr ) \, \theta _{\varphi \lor (\phi -C)}^ k\wedge \theta _{\psi \lor (\phi -C)}^{n-k}-\int _ X (\varphi -\phi )\, \theta _{\varphi }^ k\wedge \theta _{\psi }^{n-k}\\ =& \int _{\{ \min \{ \varphi ,\psi \} >\phi -C\} }\left( \varphi -\phi \right) \, \theta _{\varphi }^ k\wedge \theta _{\psi }^{n-k}-\int _ X (\varphi -\phi )\, \theta _{\varphi }^ k\wedge \theta _{\psi }^{n-k}\\ & +\int _{\{ \min \{ \varphi ,\psi \} \leq \phi -C\} }\Bigl ( \varphi \lor (\phi -C)-\phi \Bigr ) \, \theta _{\varphi \lor (\phi -C)}^ k\wedge \theta _{\psi \lor (\phi -C)}^{n-k}\\ =& -\int _{\{ \min \{ \varphi ,\psi \} \leq \phi -C\} }\left( \varphi -\phi \right) \, \theta _{\varphi }^ k\wedge \theta _{\psi }^{n-k}+\\ & +\int _{\{ \min \{ \varphi ,\psi \} \leq \phi -C\} }\Bigl ( \varphi \lor (\phi -C)-\phi \Bigr ) \, \theta _{\varphi \lor (\phi -C)}^ k\wedge \theta _{\psi \lor (\phi -C)}^{n-k} \end{aligned} \]
We first handle the first term. In order to show that it converges to $0$ when $C\to \infty $, it suffices to show that
\[ \int _ X\left( \phi -\varphi \right) \, \theta _{\varphi }^ k\wedge \theta _{\psi }^{n-k}<\infty . \]
For this purpose, it suffices to prove a stronger inequality:
0BA9923E2163
16.5.34
\begin{equation} \int _ X\left( \phi -\varphi \right) \, \theta _{(\varphi +\psi )/2}^ n<\infty . \end{equation}
From corollary 16.5.25, we know that $(\varphi +\psi )/2\in \mathcal{E}^1(X,\theta ;\phi )$. Therefore, (16.5.34) follows from 16.5.26.
It remains to handle the second term. It then suffices to prove the following:
F913F2DBE96C
16.5.35
\begin{equation} \begin{split} \lim _{C\to \infty } C\int _{\{ \varphi \leq \phi -C\} } \theta _{\varphi \lor (\phi -C)}^ k\wedge \theta _{\psi \lor (\phi -C)}^{n-k}=& 0,\\ \lim _{C\to \infty } C\int _{\{ \psi \leq \phi -C\} } \theta _{\varphi \lor (\phi -C)}^ k\wedge \theta _{\psi \lor (\phi -C)}^{n-k}=& 0. \end{split} \end{equation}
By symmetry, it suffices to handle the first. Observe the following inclusions:
\[ \{ \varphi \leq \phi -C\} \subseteq \left\{ \varphi \lor (\phi -C)\leq \frac{1}{2}\Bigl ( \psi \lor (\phi -C)+\phi -C \Bigr ) \right\} \subseteq \{ \varphi \leq \phi -C/2\} . \]
By proposition 16.5.5, we find
\[ \begin{aligned} & C\int _{\{ \varphi \leq \phi -C\} } \theta _{\varphi \lor (\phi -C)}^ k\wedge \theta _{\psi \lor (\phi -C)}^{n-k}\\ \leq & C\int _{\left\{ \varphi \lor (\phi -C)\leq \frac{1}{2}\bigl (\psi \lor (\phi -C)+\phi -C\bigr )\right\} } \theta _{\varphi \lor (\phi -C)}^ k\wedge \theta _{\psi \lor (\phi -C)}^{n-k}\\ \leq & 2^{n-k}C\int _{\left\{ \varphi \lor (\phi -C)\leq \frac{1}{2}\bigl (\psi \lor (\phi -C)+\phi -C\bigr )\right\} } \theta _{\varphi \lor (\phi -C)}^ k\wedge \theta _{\frac{1}{2}\bigl (\psi \lor (\phi -C)+\phi -C\bigr )}^{n-k}\\ \leq & 2^{n-k}C\int _{\{ \varphi \lor (\phi -C)\leq \frac{1}{2}\bigl (\psi \lor (\phi -C)+\phi -C\bigr )\} } \theta _{\varphi \lor (\phi -C)}^ n\\ \leq & 2^{n-k}C\int _{\{ \varphi \leq \phi -C/2\} } \theta _{\varphi \lor (\phi -C)}^ n\\ = & 2^{n-k}C\int _{\{ \varphi \leq \phi -C/2\} } \theta _{\varphi }^ n. \end{aligned} \]
Here on the final step, we have applied the following argument:
\[ \begin{aligned} & \int _{\left\{ \varphi \leq \phi -C/2 \right\} } \theta _{\varphi \lor (\phi -C)}^ n & \\ = & \int _ X \theta _{\varphi \lor (\phi -C)}^ n-\int _{\left\{ \varphi > \phi -C/2 \right\} } \theta _{\varphi \lor (\phi -C)}^ n & \\ = & \int _ X \theta _{\varphi }^ n-\int _{\left\{ \varphi > \phi -C/2 \right\} } \theta _{\varphi }^ n & \textup{by proposition \href{../../tag/9E81D36B1290/}{13.11.4} \href{../../tag/9E81D36B1290/#sgpt-1C036AD6}{1}}\\ =& \int _{\left\{ \varphi \leq \phi -C/2 \right\} } \theta _{\varphi }^ n. & \end{aligned} \]
On the other hand,
\[ \varlimsup _{C\to \infty } C\int _{\{ \varphi \leq \phi -C/2 \} } \theta _{\varphi }^ n\leq 2 \varlimsup _{C\to \infty } \int _{\{ \varphi \leq \phi -C \} }(\phi -\varphi )\, \theta _{\varphi }^ n=0, \]
since $\varphi \in \mathcal{E}^1(X,\theta ;\phi )$. We conclude (16.5.35).
$\square$
Pome
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