Theorem 13.24.58BB867A0263F

Proof. If $p=P_\theta [p]_{\mathcal I}$, apply theorem 13.24.3 with $u=p$. For the converse it suffices to prove the more general assertion, because analytic ordinary models are ideal models by proposition 18.4.8.

Put $q=P_\theta [p]_{\mathcal I}$ and $\delta =\underline{\operatorname {vol}}_\theta (p)>0$. Choose $s_ j\in (0,1)$ tending to zero with $d_\theta (p_ j,p)<\delta s_ j^ n$, after discarding finitely many indices. For $m_ j=\max (p_ j,p)$ one has

By lemma 13.18.1 with $b=s_ j^{-1}$, there is $h_ j\in \mathrm{PSH}(X,\theta )$ with $(1-s_ j)m_ j+s_ jh_ j\leq p_ j$. Set $H_ j=P_\theta [h_ j]_{\mathcal I}$. Since $m_ j\geq p$, monotonicity, addition and scaling of ideal envelopes (proposition 18.4.6) give

Consequently $\underline{\operatorname {vol}}_\theta (p_ j)\geq (1-s_ j)^ n\underline{\operatorname {vol}}_\theta (q)\geq (1-s_ j)^ n\delta $ by addition and volume monotonicity, since $q\geq p$. The models $p_ j,p$ therefore lie in $\mathcal M_{\delta /2}(\theta )$ for large $j$, so proposition 13.22.4 gives $p_ j\to p$ in $L^1$.

The $H_ j$ have supremum zero, so normalized compactness bounds $\int _ X(-H_ j)\, \mathrm{d}V$ by a constant $C_\theta $. Since $q\leq 0$, the convex inequality above gives $q-p_ j\leq s_ j(q-H_ j)\leq -s_ jH_ j$, and hence

Passing to the $L^1$ limit gives $q\leq p$ almost everywhere, and hence everywhere by lemma 5.24.3. The opposite inequality $p=P_\theta [p]\leq P_\theta [p]_{\mathcal I}=q$ holds by proposition 18.4.4, so $p=q$. $\square$