Proof.
(1) Let $x\in X$, and let $B$ be a coordinate ball around $x$, identified with an open ball in $\mathbb C^ n$. On $B$ the regularization $F^*$ is the regularization of $F|_ B$, so by corollary 7.6.2 the set $L\cap \{ F<F^*\} $ has outer capacity zero in $B$ for every compact set $L\subset B$. Exhausting $B$ by countably many compact sets, we obtain $\operatorname {Cap}_ B^*(B\cap \{ F<F^*\} )=0$. By theorem 5.47.8, $B\cap \{ F<F^*\} $ lies in the pole set of a plurisubharmonic function on $B$. Hence $\{ F<F^*\} $ is pluripolar.
(2) For an increasing sequence, $\lim _ ju_ j=\sup _ ju_ j$, so this is the special case of (1) for the family $(u_ j)$.
(3) Let $E$ be pluripolar, and work in a chart, identified with an open set $\Omega \subset \mathbb C^ n$. Let $B'\Subset B\Subset \Omega $ be concentric open balls. The set $E\cap B$ is pluripolar in $B$, so $\operatorname {Cap}_ B^*(E\cap B)=0$ by theorem 5.47.8.
Let $\mu $ be the trace measure of the product, or the total variation of one of its coefficient measures. The restrictions of $u_1,\ldots ,u_ q$ to $B$ form a family that is uniformly bounded on each compact subset of $B$, since $\overline B\subset \Omega $ is compact; so lemma 7.5.7(1), applied to this family and the compact set $\overline{B'}$, gives a constant $C$ with $\mu (A)\le C\operatorname {Cap}_ B(A)$ for every Borel set $A\subset \overline{B'}$. Given $\varepsilon >0$, choose an open set $G\supset E\cap B$ with $\operatorname {Cap}_ B(G)<\varepsilon $; then the outer measure of $E\cap B'$ satisfies $\mu ^*(E\cap B')\le \mu (G\cap B')\le C\operatorname {Cap}_ B(G\cap B')\le C\operatorname {Cap}_ B(G)<C\varepsilon $. Hence $\mu ^*(E\cap B')=0$. Since $\Omega $ is covered by countably many such balls $B'$, the set $E\cap \Omega $ has outer measure zero for $\mu $. For a Borel set $E$ this means that the product does not charge $E$.
$\square$
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.