Proposition 18.7.1B727D2CB45A6

Proof. Set $p=P_\theta [u]_{\mathcal I}$ and $q_ j=P_{\theta +\epsilon _ jh}[a_ j]_{\mathcal I}$. Projection and the ideal order give

The first inequality holds because $p$ is a competitor in the larger background, and the second follows from proposition 18.4.6, since $p\sim _{\mathcal I}u\preceq _{\mathcal I}a_ j$. All $q_ j$ have supremum zero and belong to $\mathrm{PSH}(X,\theta +Ch)$ for a fixed $C\geq \sup _ j\epsilon _ j$. By normalized compactness, every subsequence therefore has a further subsequence converging in $L^1$. Let $q$ be such a limit. Passing to the limit in the current inequalities gives $q\in \mathrm{PSH}(X,\theta )$, and lemma 5.24.3 gives $p\leq q\leq 0$.

Fix a local modification $\pi :Z\to U$ and a prime divisor $E\subset Z$. By lemma 5.43.19, the chosen subsequence still converges in $L^1_{\mathrm{loc}}$ after pullback. Since $q_ j\sim _{\mathcal I}a_ j$, the two functions have the same divisorial Lelong numbers, and the pointwise semicontinuity in lemma 5.38.7 yields

Indeed, at every point of $E$ the pullback of each $q_ j$ has Lelong number at least $\nu _ E(q_ j)$; it remains to apply the pointwise inequality at a point of $E$ where the pullback of $q$ has Lelong number $\nu _ E(q)$. On the other hand, $p\leq q$ gives $\nu _ E(q)\leq \nu _ E(p)=\nu _ E(u)$. Thus $q$ and $u$ have the same Lelong numbers along all local divisors. Blowing up a point of any local modification recovers the pointwise Lelong number there as a generic divisorial number; on a curve the point is already a divisor. By theorem 5.37.3, $q\sim _{\mathcal I}u$.

Consequently $q$ is a competitor for $p$, so $q\leq p$. Every cluster value is therefore $p$, which proves convergence of the whole sequence. For analytic $a_ j$, proposition 18.4.8 identifies the ordinary and ideal envelopes, and the last assertion follows from uniqueness of $L^1$ limits and lemma 5.24.3. $\square$