Proposition 16.6.300C703BE43035

Proof. Let $\eta =\operatorname *{sup}_{i\in I}{}\! {}^{*} P_{\theta }[\varphi _ i]_{\mathcal{I}}$. Then $\eta \leq P_{\theta }[\varphi ]_{\mathcal{I}}$ as a consequence of 16.6.28.

By corollary 13.13.12, we have

We may assume that for each $i\in I$, $\int _ X\theta _{\varphi _ i}^ n<\int _ X\theta _{\varphi }^ n$, since otherwise $P_{\theta }[\varphi _ i]=P_{\theta }[\varphi ]$ by theorem 16.3.15, and hence $P_{\theta }[\varphi _ i]_{\mathcal{I}}=P_{\theta }[\varphi ]_{\mathcal{I}}$ due to (16.6.19); the same holds for all $j\geq i$, and there is nothing to prove.

So by 13.13.30, we can find a decreasing net $\epsilon _ i\searrow 0$ ($i\in I$) as in (16.4.10) with $\epsilon _ i\in (0,1)$ and $\psi _ i\in \mathrm{PSH}(X,\theta )$ such that for all $i\in I$,

By proposition 16.6.27 and 16.6.28, we have

As above, $\epsilon _ iP_{\theta }[\psi _ i]_{\mathcal{I}}\to 0$ in $L^1$. It follows that $P_{\theta }[\varphi ]_{\mathcal{I}}\leq \eta $ almost everywhere, hence everywhere by proposition 5.14.6. $\square$