Proposition 6.9.73E1F53764C07

Proof. Put $\lambda =\nu (\varphi ,Z)$. For every integer $j\geq 1$, theorem 6.9.5 shows that $Z\cap E_{\lambda +1/j}(\varphi )$ is an analytic subset of $Z$, and it is a proper subset by the definition of $\lambda $. At every point of $Z$ outside the union of these sets, $\nu (\varphi ,x)<\lambda +1/j$ for all $j$, hence $\nu (\varphi ,x)=\lambda $.

We now show that the complement in $Z$ of an arbitrary countable union of proper analytic subsets of $Z$ is dense in $Z$; applied to the sets $Z\cap E_{\lambda +1/j}(\varphi )$, this shows that $\nu (\varphi ,Z)$ is attained on a dense subset of $Z$. The density is clear if $Z$ is a point, for then the proper analytic subsets of $Z$ are empty. Let $\dim Z>0$. By corollary 9.3.16, applied to the family consisting of $Z$ alone, $Z$ is an irreducible component of itself; so its regular locus $Z_{\mathrm{reg}}$ is a connected manifold, dense in $Z$.

Let $B$ be a proper analytic subset of $Z$. Then $B\cap Z_{\mathrm{reg}}$ is an analytic subset of $Z_{\mathrm{reg}}$, and it is proper, since otherwise the closed set $B$ would contain the closure $Z$ of $Z_{\mathrm{reg}}$. By lemma 9.2.8, every point of $B\cap Z_{\mathrm{reg}}$ has a connected open neighborhood $U$ in $Z_{\mathrm{reg}}$ with a holomorphic function $g\not\equiv 0$ on $U$ such that $B\cap U\subset \{ g=0\} $; shrinking $U$ to a coordinate ball, on which $g\not\equiv 0$ by the identity theorem, we see from lemma 5.5.1 that $B\cap U$ is a null set in the chart. Covering a chart $N$ of $Z_{\mathrm{reg}}$ by countably many such balls and by open sets disjoint from $B$, we see that $B\cap N$ is a null set. Consequently, a countable union of proper analytic subsets of $Z$ meets every chart of $Z_{\mathrm{reg}}$ in a null set, which has no interior points. Its complement is therefore dense in $Z_{\mathrm{reg}}$, and hence in $Z$. For a closed positive $(1,1)$-current the same argument applies, since theorem 6.9.5 holds for its Lelong numbers.

For the second assertion, note first that $\varphi \circ \pi $ is a weight on $Y$: by proposition 5.9.3 it is plurisubharmonic or identically $-\infty $ on each component of $Y$, and the second alternative is excluded. Indeed, since $\pi $ is bimeromorphic (section 9.4), there is a nowhere dense analytic subset $A\subset X$ such that $\pi ^{-1}(A)$ is nowhere dense in $Y$ and $\pi $ maps $Y\setminus \pi ^{-1}(A)$ biholomorphically onto $X\setminus A$. If $\varphi \circ \pi $ were identically $-\infty $ on a component $Y_0$ of $Y$, then $\varphi $ would be $-\infty $ on the nonempty open set $\pi (Y_0\setminus \pi ^{-1}(A))$, which is impossible because $\varphi $ is locally integrable.

Now let $E\subset Y$ and $E'\subset Y'$ be prime divisors on smooth models $\pi \colon Y\to X$ and $\pi '\colon Y'\to X$ that represent the same prime divisor over $X$, and let $\rho \colon Y''\to Y$ and $\rho '\colon Y''\to Y'$ be maps from a common smooth model on which the strict transforms of $E$ and $E'$ coincide, say with $E''$. By proposition 9.8.3, $\rho $ is biholomorphic over $Y\setminus S$ for an analytic subset $S\subset Y$ of codimension at least two, so it maps the nonempty open subset $E''\cap \rho ^{-1}(Y\setminus S)$ of $E''$ onto the nonempty open subset $E\setminus S$ of $E$, and $\nu (\varphi \circ \pi \circ \rho ,q)=\nu (\varphi \circ \pi ,\rho (q))$ for $q$ in this set, because Lelong numbers are invariant under biholomorphisms (proposition 6.7.1). By the first part, the generic value along $E$ is attained on a dense subset of $E$, so it equals the infimum of the Lelong numbers over any nonempty open subset of $E$; the same holds for $E''$. Hence $\nu (\varphi \circ \pi ,E)=\nu (\varphi \circ \pi \circ \rho ,E'')$. The same argument for $\rho '$ gives $\nu (\varphi \circ \pi ',E')=\nu (\varphi \circ \pi \circ \rho ,E'')$, since $\pi \circ \rho =\pi '\circ \rho '$. $\square$