Proposition 6.12.12966A90DCC98E

Proof. By corollary 6.11.6 there are $\varepsilon >0$ and a neighborhood $V$ of $x$ on which $f$ is defined such that $|f|^2\mathrm e^{-(1+\varepsilon )\varphi }$ is integrable on $V$. Let $E$ be a prime divisor over an open set $U\subset V$, represented on a smooth model $\pi \colon Y\to U$; we may assume that $\operatorname {ord}_ Ef<\infty $. Since $f\in \mathcal I((1+\varepsilon )\varphi )_ y$ for every $y\in U$, proposition 6.12.4, applied to the weight $(1+\varepsilon )\varphi $ on $U$, shows that $|h|^2\mathrm e^{-(1+\varepsilon )\varphi \circ \pi }$, with $h=(f\circ \pi )J_\pi $, is integrable near every point of $Y$. Let $p\in E_{\mathrm{reg}}$. Since orders add under products, the order of $h$ along $E$ at $p$ is $\operatorname {ord}_ Ef+\operatorname {ord}_ EJ_\pi =\operatorname {ord}_ Ef+A_ X(E)-1$. By lemma 6.12.11, applied to the weight $(1+\varepsilon )\varphi \circ \pi $, whose generic Lelong number along $E$ is $(1+\varepsilon )\nu (\varphi ,E)$ by proposition 6.7.9, we obtain (6.12.13). $\square$