Proposition 16.24.1.585A4246348B For quasi-plurisubharmonic functions $u,v$ on $X$, the following are equivalent:
$\mathcal I(tv)\subseteq \mathcal I(tu)$ for every real $t>0$;
$\mathcal I(tv)\subseteq \mathcal I(tu)$ for every positive integer $t$;
$\nu (v,E)\geq \nu (u,E)$ for every prime divisor $E$ over $X$.
We write $v\preceq _{\mathcal I}u$ when these conditions hold, and $v\sim _{\mathcal I}u$ when $v\preceq _{\mathcal I}u$ and $u\preceq _{\mathcal I}v$. In particular, equality of the multiplier ideals of all positive integer multiples is equivalent to equality for all positive real multiples, and to equality of all divisorial Lelong numbers.
Proof.
Condition (2) is a special case of (1). Assume (2) and let $E$ be a prime divisor over an open subset $U\subset X$ (possibly $U=X$). The inclusions (2) restrict to $U$, and a smaller ideal has a larger order along $E$, so $\operatorname {ord}_ E\mathcal I(tv)\geq \operatorname {ord}_ E\mathcal I(tu)$ for every positive integer $t$. Dividing by $t$ and letting $t\to \infty $, corollary 6.12.14, applied on $U$, gives $\nu (v,E)\geq \nu (u,E)$; this is (3). Conversely, assume (3), let $t>0$, and let $f$ be a germ of $\mathcal I(tv)$ at a point $x$. The valuative criterion, theorem 6.12.16, gives $\varepsilon >0$ and a neighborhood $V$ of $x$ such that
\[ \operatorname {ord}_ E f+A_ X(E) \geq (1+\varepsilon )t\nu (v,E) \geq (1+\varepsilon )t\nu (u,E) \]
for every prime divisor $E$ over an open neighborhood of $x$ contained in $V$ whose center contains $x$; the second inequality is (3) for these divisors. Since $t\nu (u,E)=\nu (tu,E)$ by proposition 6.7.9, the same criterion, applied to $tu$, shows that $f\in \mathcal I(tu)$; this is (1). Applying the equivalence in both directions proves the assertions about equalities.
$\square$
Pome
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