PomeLemma 5.40.5.44FD5A853A0A Let $\theta $ be big, let $f$ be a bounded function on $X$, and let $u\in \mathrm{PSH}(X,\theta )$ satisfy $u\leq f$ quasi-everywhere. Then there are $u_ t\in \mathrm{PSH}(X,\theta )$, $t>0$, such that $u_ t\leq f$ everywhere and $u_ t\to u$ almost everywhere as $t\downarrow 0$.
Lemma 5.40.544FD5A853A0A
Proof. Choose $\rho \in \mathrm{PSH}(X,\theta )$ with $\theta +\mathrm{dd}^{\mathrm{c}}\rho \geq \kappa \omega $ for some $\kappa >0$. The exceptional set $E=\{ u>f\} $ is pluripolar. The global form of Josefson's theorem on compact complex manifolds gives a quasi-plurisubharmonic function $\chi $ with $\chi =-\infty $ on $E$; see [Theorem 1.1.5, SGPT]. After rescaling and adding a constant, we may assume that $\chi \leq 0$ and $\mathrm{dd}^{\mathrm{c}}\chi \geq -\omega $. Subtract a constant from $\rho $ so that $\rho \leq \inf _ X f$. With $s=t/\kappa $, put
\[ u_ t=\frac{u+t\chi +s\rho }{1+s}. \]
All three functions in the numerator are finite almost everywhere, so $u_ t$ is not identically $-\infty $. Moreover,
\[ \theta +\mathrm{dd}^{\mathrm{c}}u_ t =\frac{\theta +\mathrm{dd}^{\mathrm{c}}u+t\mathrm{dd}^{\mathrm{c}}\chi +s(\theta +\mathrm{dd}^{\mathrm{c}}\rho )}{1+s} \geq 0. \]
Outside $E$, both $u$ and $\rho $ are at most $f$, while $\chi \leq 0$, so $u_ t\leq f$. On $E$ the numerator is $-\infty $. Finally, $u_ t\to u$ wherever all three functions are finite, which is a set of full measure. $\square$
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.