Proof.
We may assume that $\varphi _0^ i,\varphi _1^ i\leq 0$ for all $i\in I$. Then it follows that $\varphi _ t^ i\leq 0$ for all $t\in (0,1)$ and all $i\in I$ by proposition 17.10.7.
We define
\[ \varphi _ t\coloneqq \operatorname *{sup}_{i\in I}{}\! {}^{*} \varphi ^ i_ t\in \mathrm{PSH}(X,\theta ) \]
for all $t\in [0,1]$. Observe that $[0,1] \ni t\mapsto \varphi _ t$ is convex by the same argument leading to (17.10.8).
Let $(\psi _ t)_{t\in (0,1)}$ be the subgeodesic whose complexification $\Phi _{\psi }$ corresponds to $\operatorname *{sup}_{i}{}\! {}^{*} \Phi _{\varphi ^ i}$, where $\Phi _{\varphi ^ i}$ is the complexification of $(\varphi ^ i_ t)_{t\in (0,1)}$. Then clearly, $\varphi _ t\leq \psi _ t$ for each $t\in (0,1)$. On the other hand, by proposition 5.14.5,
\[ \psi _ t=\sup _{i\in I}\varphi ^ i_ t=\varphi _ t\quad \textup{almost everywhere} \]
for almost all $t\in (0,1)$. Therefore, using proposition 5.14.6, we find $\psi _ t=\varphi _ t$ for almost all $t\in (0,1)$. Since both functions are convex in $t$, we conclude that $\psi _ t=\varphi _ t$ for all $t\in (0,1)$.
It remains to argue that $\varphi _ t\xrightarrow {L^1}\varphi _0$ as $t\to 0+$ and $\varphi _ t\xrightarrow {L^1}\varphi _1$ as $t\to 1-$. By symmetry, it suffices to argue the former.
By Choquet's lemma proposition 5.14.3, applied to the uniformly bounded above family of complexifications $(\Phi _{\varphi ^ i})_{i\in I}$ on the strip, there is a countable subset $I'\subset I$ such that
\[ \operatorname *{sup}_{i\in I}{}\! {}^{*}\Phi _{\varphi ^ i}=\operatorname *{sup}_{i\in I'}{}\! {}^{*}\Phi _{\varphi ^ i}. \]
Applying Choquet's lemma proposition 5.14.3 once more to the endpoint families if necessary, and enlarging $I'$, we may also assume
\[ \operatorname *{sup}_{i\in I}{}\! {}^{*}\varphi _0^ i=\operatorname *{sup}_{i\in I'}{}\! {}^{*}\varphi _0^ i,\quad \operatorname *{sup}_{i\in I}{}\! {}^{*}\varphi _1^ i=\operatorname *{sup}_{i\in I'}{}\! {}^{*}\varphi _1^ i. \]
Thus, after replacing $I$ by $I'$, we may assume that $I$ is countable.
We know that for any $t\in (0,1)$ and any $j\in I$,
\[ \varphi ^ j_ t\leq \varphi _ t\leq t\varphi _1+(1-t)\varphi _0. \]
Letting $t\to 0+$, we find that
\[ \varphi _0^ j\leq \varlimsup _{t\to 0+}\varphi _ t\leq \varphi _0 \]
almost everywhere. Since $I$ is countable, we conclude that
BA313A9B1A1F
17.10.14
\begin{equation} \varphi _0=\varlimsup _{t\to 0+}\varphi _ t \end{equation}
almost everywhere.
Fix $i_0\in I$. Recall that by proposition 17.10.7, for each $t\in (0,1)$, we have
\[ \adjustlimits \inf _{t\in (0,1)}\sup _ X \varphi _ t\geq \adjustlimits \inf _{t\in (0,1)}\sup _ X \varphi ^{i_0}_ t\geq \adjustlimits \sup _ X \inf _{t\in (0,1)}\varphi ^{i_0}_ t>-\infty , \]
so the set $\{ \varphi _ t\} _{t\in (0,1)}$ is relatively compact with respect to the $L^1$-topology by proposition 5.17.3. Let $\psi $ be a cluster point as $t\to 0+$. It suffices to show that $\psi =\varphi _0$. By corollary 5.14.8 and (17.10.14), this holds almost everywhere. Therefore, it holds everywhere by proposition 5.14.6.
$\square$
Pome
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