Corollary 6.3.5FFB7E3647C0B

Proof. Let $\Omega \subset \mathbb C^ n$ and $\Omega '\subset \mathbb C^ m$ be pseudoconvex, with nonnegative continuous plurisubharmonic exhaustion functions $\rho $ and $\rho '$. The function $(z,w)\mapsto \rho (z)+\rho '(w)$ is continuous and plurisubharmonic on $\Omega \times \Omega '$, by proposition 5.9.3 applied to the two projections and by proposition 5.9.5. Its sublevel set $\{ \le c\} $ is a closed subset of the compact set $\{ \rho \le c\} \times \{ \rho '\le c\} $, so $\Omega \times \Omega '$ is pseudoconvex.

Next let $s$ be as in the statement and $c\in \mathbb R$. The function $x\mapsto 1/(c-x)$ is convex and increasing on $(-\infty ,c)$, so the complex Hessian matrix of $1/(c-s)$ on the open set $\{ s<c\} $, namely $(c-s)^{-2}(s_{j\bar k})+2(c-s)^{-3}\bigl (\partial _{z_ j}s\, \overline{\partial _{z_ k}s}\bigr )$ by the chain rule, is positive semidefinite. Hence $|z|^2+1/(c-s)$ is a smooth plurisubharmonic function on $\{ s<c\} $, by proposition 5.9.2(1). For $C\le 0$ its sublevel set $\{ \le C\} $ is empty. For $C>0$ this sublevel set is closed in $\{ s<c\} $ and contained in the set $\{ s\le c-1/C\} $, which is a compact subset of $\{ s<c\} $; hence it is compact, and $\{ s<c\} $ is pseudoconvex. Moreover $\{ s<c\} $ is contained in the compact set $\{ s\le c\} $, so it is relatively compact in $\Omega $.

Finally, if $\Omega $ is pseudoconvex, theorem 6.3.4 provides a smooth plurisubharmonic exhaustion function $s$, and the sets $\Omega _ j=\{ s<j\} $, $j\ge 1$, are pseudoconvex, relatively compact in $\Omega $, increasing, and have union $\Omega $. (Some of them may be empty or disconnected; definition 6.3.1 allows arbitrary open sets.) $\square$