Theorem 11.8.1B0762464DCDC

Proof. Suppose $C_{\delta _ j}\leq M$ and put $\kappa =(M+1)^{-1}$. Normalizing any smooth strictly positive $\mathrm{dd}^{\mathrm{c}}$-closed complementary form gives

The generalized Lamari criterion, theorem 11.6.1, supplies $v_ j$ with

Since $B_{\delta _ j}\geq 0$, the potentials belong to the fixed normalized $\omega $-plurisubharmonic family. By theorem 5.25.2, a subsequence converges in $L^1$ to $v\in \mathrm{PSH}(X,\omega )$. The transverse approximate identity converges to $\chi [\zeta =0]$, so passing in distributions gives

On a smaller chart meeting the inner hypersurface, add a smooth $\rho $ with $\mathrm{dd}^{\mathrm{c}}\rho \geq \omega $. The closed positive current $\mathrm{dd}^{\mathrm{c}}(v+\rho )$ dominates $\kappa [\zeta =0]$ where $\chi =1$. Hence the Lelong number of $v$ is at least $\kappa $ at every point of $U'\times \{ 0\} $. By Siu's theorem in 5.34.2, the set $Z=\{ \nu (v,\cdot )\geq \kappa \} $ is analytic and has dimension at most $n-1$. Smooth local corrections do not change its definition. The compact analytic set $Z$ has finitely many global irreducible components. Their intersections with the connected inner hypersurface are analytic subsets covering it. At least one intersection is the whole hypersurface, since finitely many proper analytic subsets cannot cover an open polydisk. That component has dimension $n-1$ and is the required global hypersurface. If no such hypersurface exists, there is no bounded sequence $C_{\delta _ j}$ with $\delta _ j\downarrow 0$, which proves the asserted divergence. $\square$