PomeDD1F66C6842AProposition 5.47.16.DD1F66C6842A Let $X$ be a complex manifold.
Every nonempty plurifine open subset of $X$ contains a Borel set of positive Lebesgue measure in some chart.
The complement of every pluripolar set is dense for the plurifine topology.
For every $u\in \mathrm{PSH}(X)$, the set $\{ u>-\infty \} $ is plurifine open.
Proof. (1) Let $O$ be a nonempty plurifine open set and $a\in O$. By proposition 5.47.15(2), $O$ contains a set $A=B\cap \{ \psi >0\} $ with $a\in A$, where $B$ is a coordinate ball and $\psi $ is a bounded plurisubharmonic function on a neighborhood of $\overline B$. This set is Borel because $\psi $ is upper semicontinuous. If $A$ had measure zero, then $\max \{ \psi ,0\} =0$ almost everywhere on $B$; since both functions are plurisubharmonic on $B$, they would coincide everywhere by the uniqueness in theorem 5.9.1, contradicting $\psi (a)>0$.
(2) A pluripolar set is Lebesgue null in every chart, so by (1) it contains no nonempty plurifine open set. Equivalently, its complement meets every nonempty plurifine open set.
(3) The set $\{ u>-\infty \} =\bigcup _ m\{ u>-m\} $ is a union of subbasic plurifine open sets. $\square$
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