Lemma 13.13.47.3AA4A74713F0 For any $\varphi \in \mathrm{PSH}(X,\theta )_{>0}$, there is $\psi \in \mathrm{PSH}(X,\theta )$ such that
B1CBA0B9F2D3
$\theta _{\psi }$ is a Kähler current, and
9583FAA82351
$\psi \leq \varphi $.
In particular, there is an increasing sequence $(\varphi _ i)_{i>0}$ in $\mathrm{PSH}(X,\theta )$ converging almost everywhere to $\varphi $ such that $\theta _{\varphi _ i}$ is a Kähler current for all $i\geq 1$.
Proof.
Using 13.13.30, we can find $\epsilon \in (0,1)$ and $\gamma \in \mathrm{PSH}(X,\theta )$ such that
\[ \epsilon V_{\theta }+(1-\epsilon )\gamma \leq \varphi . \]
We observe that the cohomology class $\{ \theta \} $ is big as a consequence of proposition 13.13.3. Therefore, we can take $\eta \in \mathrm{PSH}(X,\theta )$ such that $\theta _{\eta }$ is a Kähler current and $\eta \leq 0$. Since $\eta \leq V_{\theta }$, we may take
\[ \psi \coloneqq \epsilon \eta +(1-\epsilon )\gamma . \]
Then $\psi $ clearly satisfies 1 and 2.
For the latter claim, it suffices to take
\[ \varphi _ i=\left(1-(i+1)^{-1}\right)\varphi +(i+1)^{-1}\psi \]
for each $i>0$.
$\square$
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.