PomeProposition 13.15.6.8B51BF8706B7 If $u\in \mathrm{PSH}(X,\omega )$ and $u\leq 0$, then $F_ a(u)$ is a bounded $\omega $-plurisubharmonic function, and
\[ \mathrm{e}^{na u}\mu _ u\leq \mathrm{MA}_\omega (F_ a(u)). \]
Proposition 13.15.68B51BF8706B7
Proof. First suppose that $u$ is bounded. The chain rule (5.27.9) gives
\[ \omega +\mathrm{dd}^{\mathrm{c}}F_ a(u) =\mathrm{e}^{au}(\omega +\mathrm{dd}^{\mathrm{c}}u)+(1-\mathrm{e}^{au})\omega +a\mathrm{e}^{au}\, \mathrm{d}u\wedge \mathrm d^{\mathrm c}u. \]
All three summands are positive, and positivity of their bounded mixed products yields the claimed inequality.
For general $u$, use the truncations $u_ t=\max (u,-t)$. The bounded functions $F_ a(u_ t)$ decrease and take values between zero and $a^{-1}$, so their limit $F_ a(u)$ is a bounded $\omega $-plurisubharmonic function. On the plurifinely open set $D_ t=\{ u>-t\} $, it agrees with $F_ a(u_ t)$. Locality and the bounded case therefore give
\[ \mathrm{e}^{na u}\mathbf1_{D_ t}\mu _ u \leq \mathbf1_{D_ t}\mathrm{MA}_\omega (F_ a(u)). \]
Let $t\to +\infty $. The pole set is not charged by $\mu _ u$, and it has zero mass for the bounded product on the right. $\square$
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