Proof.
Translate $\varphi $ below zero. Put $q=P_{\epsilon \theta }(\varphi )\leq \varphi $, and choose a strict big potential $\rho \leq 0$ with $\theta +\mathrm{dd}^{\mathrm{c}}\rho \geq a\omega $, where $a>0$. Then
\[ s=q+(1-\epsilon )\rho \leq \varphi ,\qquad \theta +\mathrm{dd}^{\mathrm{c}}s\geq (1-\epsilon )a\omega . \]
In particular $s$ is $\theta $-plurisubharmonic and not identically $-\infty $. Its pole set is pluripolar, as is that of $\varphi $, so both are negligible for $\mu $.
Cover $X$ by finitely many closed coordinate balls $K_ i$, each contained in a larger coordinate ball $B_ i$. Apply lemma 5.27.10 to $\mu _ i=\mathbf1_{K_ i}\mu $. This gives bounded local plurisubharmonic $v_ i$, normalized by $1\leq v_ i\leq 2$, with
\[ \mu _ i\ll \nu _ i:=(\mathrm{dd}^{\mathrm{c}}v_ i)^ n. \]
Take a smooth cutoff $\chi _ i$ which is zero near $K_ i$ and $-3$ on a boundary collar of $B_ i$. Choose $\delta _ i>0$ so small that
\[ \theta +\mathrm{dd}^{\mathrm{c}}\bigl (s+\delta _ i(v_ i+\chi _ i)\bigr ) \geq \tfrac 12(1-\epsilon )a\omega +\delta _ i\mathrm{dd}^{\mathrm{c}}v_ i. \]
Let $S_ i$ equal $\max (s,s+\delta _ i(v_ i+\chi _ i))$ on $B_ i$ and $s$ elsewhere. The boundary collar makes this a global $\theta $-plurisubharmonic function. Since $\chi _ i\leq 0$ may be arranged,
\[ s\leq S_ i\leq s+2\delta _ i\leq \varphi +2\delta _ i. \]
For every positive integer $m$, set
\[ q_{i,m}=\max (\varphi -m,S_ i)-2\delta _ i. \]
These potentials have the same ordinary type as $\varphi $, and
036B1477ABB4
15.24.2
\begin{equation} \varphi -m-2\delta _ i\leq q_{i,m}\leq \varphi . \end{equation}
On a neighborhood of $K_ i$, $S_ i=s+\delta _ i v_ i$ away from the pole set of $s$. Let
\[ D_{i,m}=\{ s+\delta _ i v_ i>\varphi -m\} \]
in that neighborhood, restricting to points where the potentials are finite. This is a plurifinely open set there. Locality and positive mixed products give
3F73A2DB6267
15.24.3
\begin{equation} \mathbf1_{D_{i,m}}\mathrm{MA}^{\mathrm{ext}}_\theta (q_{i,m}) \geq \delta _ i^ n\mathbf1_{D_{i,m}}\nu _ i. \end{equation}
Here $s$ need not be bounded. After a local smooth correction, the positive product inequality holds at every common finite truncation level of $s$ and the bounded potential $v_ i$. Exhaust the complement of the pole set of $s$. The measure $\nu _ i$ is non-pluripolar, so it loses no mass in that exhaustion. Then restrict the resulting inequality by plurifine locality. This proves (15.24.3) for the extended measures.
For $\mu _ i$-almost every point of $K_ i$, both $s$ and $\varphi $ are finite, and hence that point belongs to $D_{i,m}$ for some $m$. Consequently, a Borel set which is null for every $\mathrm{MA}^{\mathrm{ext}}_\theta (q_{i,m})$ is null for every $\mu _ i$, and therefore for $\mu $.
Enumerate the countable family $q_{i,m}$ as $q_ k$, and write its bounds as $\varphi -C_ k\leq q_ k\leq \varphi $ with $C_ k<\infty $. Choose positive $t_ k$ such that
\[ \sum _ k t_ k<1,\qquad B:=\sum _ k t_ kC_ k<\infty . \]
The finite convex combinations
\[ \psi _ N=\sum _{k=1}^ N t_ kq_ k +\left(1-\sum _{k=1}^ N t_ k\right)\varphi \]
decrease and lie between $\varphi -B$ and $\varphi $. Their limit $\psi $ is $\theta $-plurisubharmonic, satisfies $\psi \geq \varphi -B$, and hence $\psi \sim \varphi $. For any fixed $k$, separate its summand to write
\[ \psi =t_ kq_ k+(1-t_ k)w_ k, \qquad w_ k\in \mathrm{PSH}(X,\theta ),\qquad w_ k\sim \varphi . \]
The remainder is obtained by the same decreasing convex-sum construction. Positive non-pluripolar mixed products give
\[ \mathrm{MA}^{\mathrm{ext}}_\theta (\psi ) \geq t_ k^ n\mathrm{MA}^{\mathrm{ext}}_\theta (q_ k). \]
This is valid also for infinite measures: first restrict away from the common pole set, which is the pole set of $\varphi $ by the ordinary-type bounds, and then exhaust by finite truncations. All measures ignore that pole set. The preceding null-set conclusion now gives $\mu \ll \mathrm{MA}^{\mathrm{ext}}_\theta (\psi )$.
By theorem 13.14.3, the measure $\mathrm{MA}^{\mathrm{ext}}_\theta (\psi )$ is $\sigma $-finite. The Radon–Nikodym theorem applies to this $\sigma $-finite measure and the finite $\mu $. It gives a nonnegative density $f$ with
\[ \int _ X f\, \mathrm{MA}^{\mathrm{ext}}_\theta (\psi )=\mu (X)<\infty , \]
as claimed. If $\mu =0$, one may simply take $\psi =\varphi $ and $f=0$.
$\square$
Pome
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.