Theorem 6.9.5 (Siu)946909269108

Proof. Lelong numbers vanish on components of dimension zero, so we may assume that $X$ is connected of dimension $n\geq 1$. We describe $E_ c(\varphi )$ as an intersection of analytic sets defined by the orders of the ideals $\mathcal I(m\varphi )$ and then show that such an intersection is analytic. Let $\mathfrak a$ be a coherent ideal and $k$ an integer. The set of points $y$ where $\operatorname {ord}_ y\mathfrak a\geq k$ (with $\operatorname {ord}_ y\mathfrak a=\infty $ if $\mathfrak a_ y=0$) is analytic. Indeed, let $g_1,\ldots ,g_ r$ generate $\mathfrak a$ on a coordinate neighborhood $W$. A germ has order at least $k$ at $y$ if and only if its derivatives of order less than $k$ vanish at $y$, since these derivatives are multiples of its Taylor coefficients, by (5.2.1). Hence for $y\in W$ we have $\operatorname {ord}_ y\mathfrak a\geq k$ if and only if $\partial ^\alpha g_ i(y)=0$ for all $i$ and all $|\alpha |<k$: the condition is necessary because the $g_ i$ belong to $\mathfrak a_ y$, and sufficient because, by the Leibniz rule, every element $\sum _ ih_ ig_ i$ then vanishes to order at least $k$ at $y$. This is a finite system of holomorphic equations, and for $k\leq 0$ the set is all of $W$. By (6.9.2),

Indeed, if $\nu (\varphi ,x)\geq c$, the lower bound in (6.9.2) gives the integer inequality $\operatorname {ord}_ x\mathcal I(m\varphi )>mc-n$, that is, $\operatorname {ord}_ x\mathcal I(m\varphi )\geq \lfloor mc-n\rfloor +1$, for every $m$. If $\nu (\varphi ,x)<c$, the upper bound gives $\operatorname {ord}_ x\mathcal I(m\varphi )\leq m\nu (\varphi ,x)\leq mc-n<\lfloor mc-n\rfloor +1$ as soon as $m(c-\nu (\varphi ,x))\geq n$.

Thus $E_ c(\varphi )$ is an intersection of analytic subsets of $X$. An arbitrary intersection $A=\bigcap _\lambda A_\lambda $ of analytic subsets of $X$ is analytic. Since $A$ is closed, it suffices to show that each $p\in A$ has a neighborhood $W$ such that $A\cap W$ is the intersection of $W$ with finitely many of the $A_\lambda $. Let $I_\lambda \subset \mathcal O_{X,p}$ be the ideal generated by the germs at $p$ of local defining functions of $A_\lambda $. By proposition 9.2.4 the sum $\sum _\lambda I_\lambda $ is finitely generated, so it equals $I_{\lambda _1}+\cdots +I_{\lambda _ q}$ for finitely many indices; put $B=A_{\lambda _1}\cap \cdots \cap A_{\lambda _ q}$. Every local defining function of each $A_\lambda $ is then a combination of defining functions of $A_{\lambda _1},\ldots ,A_{\lambda _ q}$ near $p$, so $A_\lambda $ contains a neighborhood of $p$ in $B$. This neighborhood may depend on $\lambda $, and we now remove this dependence.

By corollary 9.3.16, the irreducible components of the analytic set $B$ form a locally finite family; so some neighborhood $N$ of $p$ meets only finitely many of them. Let $B_1,\ldots ,B_ l$ be those that contain $p$. The union of the other components meeting $N$ is closed and does not contain $p$, so there is an open neighborhood $W'\subset N$ of $p$ with $B\cap W'\subset B_1\cup \cdots \cup B_ l$. Fix $\lambda $ and $i$. The regular locus $(B_ i)_{\mathrm{reg}}$ is a connected manifold, dense in $B_ i$ (corollary 9.3.16), and its analytic subset $A_\lambda \cap (B_ i)_{\mathrm{reg}}$ has nonempty interior in it, because $A_\lambda $ contains a neighborhood of $p$ in $B_ i$ and $p$ lies in the closure of $(B_ i)_{\mathrm{reg}}$. By lemma 9.2.8, $A_\lambda \cap (B_ i)_{\mathrm{reg}}$ is therefore all of $(B_ i)_{\mathrm{reg}}$, and $B_ i\subset A_\lambda $ because $A_\lambda $ is closed. Therefore $B\cap W'\subset A_\lambda $ for every $\lambda $, and $A\cap W'=B\cap W'$. Applied to (6.9.6), this shows that $E_ c(\varphi )$ is analytic.

By proposition 6.7.1, $\nu (\varphi ,x)=0$ at every point where $\varphi (x)>-\infty $. Thus $E_ c(\varphi )$ is contained in the pole set of $\varphi $, which has measure zero because $\varphi $ is locally integrable (theorem 5.9.1), and therefore has no interior points; so $E_ c(\varphi )\neq X$.

Finally, let $T$ be a closed positive $(1,1)$-current on $X$. Every point of $X$ at which $X$ has positive dimension has a coordinate ball $U$ on which $T=\mathrm{dd}^{\mathrm{c}}u$ with $u$ plurisubharmonic, and $\nu (T,y)=\nu (u,y)$ for $y\in U$ (definition 6.7.4). By what we have proved, applied to the weight $u$ on $U$, the set $\{ y\in U:\nu (T,y)\geq c\} $ is an analytic subset of $U$ with empty interior. Since the components of $X$ of dimension zero do not meet $\{ y:\nu (T,y)\geq c\} $, and analyticity is a local property, this set is an analytic subset of $X$ whose intersection with every component of positive dimension is proper. $\square$