5.43.17 ModificationsD643D283EFB3

Proof. Let $S\subset X$ be a proper analytic subset over whose complement $\pi $ is an isomorphism, and put $D=\pi ^{-1}(S)$. By lemma 5.43.2, pullback preserves the curvature inequality, and the pullback of a $\theta $-plurisubharmonic function is not identically $-\infty $, since $\pi $ is an isomorphism off $D$. The pullback map is injective because $\pi $ is surjective.

To prove surjectivity, let $b\in \mathrm{PSH}(Y,\pi ^*\theta )$, and define $u=b\circ \pi ^{-1}$ off $S$. On a coordinate neighborhood in $X$, choose a smooth function $\rho $ with $\mathrm{dd}^{\mathrm{c}}\rho \geq \theta $. Then $b+\rho \circ \pi $ is plurisubharmonic on the inverse image of this neighborhood. By properness, its local upper bounds give a local upper bound for $u+\rho $ near $S$, so the removable-singularity theorem extends $u+\rho $ plurisubharmonically across $S$. The resulting quasi-plurisubharmonic extensions of $u$ agree almost everywhere on overlaps, hence everywhere, and glue to a function on $X$.

The current $\theta +\mathrm{dd}^{\mathrm{c}}u$ is positive off $S$. Locally it is the sum of the positive current $\mathrm{dd}^{\mathrm{c}}(u+\rho )$ and the smooth form $\theta -\mathrm{dd}^{\mathrm{c}}\rho $. The smooth form gives $S$ zero mass, so the restriction of $\theta +\mathrm{dd}^{\mathrm{c}}u$ to $S$ is positive as well. Thus $u\in \mathrm{PSH}(X,\theta )$. The plurisubharmonic functions $b+\rho \circ \pi $ and $(u+\rho )\circ \pi $ agree off $D$, hence almost everywhere, hence everywhere. Therefore $b=u\circ \pi $. $\square$

Proof. By lemma 5.25.1, the functions $v_ j$ are uniformly bounded above on relatively compact subsets of $X$. The pullbacks are therefore locally uniformly bounded above and have the common smooth lower curvature bound $-\pi ^*\theta $. After a smooth local correction, the compactness theorem for plurisubharmonic functions shows that every subsequence has a further subsequence that either converges in $L^1_{\mathrm{loc}}$ or tends locally uniformly to $-\infty $ on each connected coordinate region in $Y$.

Every such region meets the complement of the exceptional locus. There $\pi $ is a biholomorphism, and change of variables gives $L^1_{\mathrm{loc}}$ convergence to $v\circ \pi $, which excludes convergence to $-\infty $. Each subsequential limit agrees with $v\circ \pi $ off the exceptional locus, hence almost everywhere, since this analytic set has smooth volume zero. Thus the limit is unique, and applying this argument on a countable cover by coordinate regions proves convergence of the whole sequence in $L^1_{\mathrm{loc}}(Y)$. $\square$

Proof. Choose a proper analytic subset $S\subset X$ such that $\pi $ is an isomorphism outside $D=\pi ^{-1}(S)$. Both $S$ and $D$ have capacity zero. For $\epsilon >0$, choose an open neighborhood $V$ of $D$ of capacity less than $\epsilon $. By properness, there is a neighborhood $U_0$ of $S$ with $\pi ^{-1}(U_0)\subset V$; shrink it to an open set $U$ with $\operatorname {Cap}_ X(U)<\epsilon $. The inverse of $\pi $ is biholomorphic near the compact set $X\setminus U$. A finite cover by nested coordinate charts and the comparison of local and global capacities in section 5.29 give a constant $C_\epsilon $ such that, for every Borel set $E\subset X$,

Apply these inequalities to $E=\{ |v_ j-v|>\delta \} $ for each fixed $\delta >0$, omitting the pluripolar pole sets, and let first $j\to \infty $ and then $\epsilon \downarrow 0$. $\square$