Let $u$ and $v$ be potentials of the same ordinary singularity type. We show that the sum of their weighted measures $\mathrm{e}^{-\lambda u}\mathrm{MA}_\theta (u)+\mathrm{e}^{-\lambda v}\mathrm{MA}_\theta (v)$ is again of the form $\mathrm{e}^{-\lambda a}\mathrm{MA}_\theta (a)$ for a potential $a$ of that type. Truncating $u$ and $v$ gives bounded equations, solved by the bounded subsolution theorem, whose solutions have a common lower barrier. Their exponential error tends to zero even though the approximating masses may grow, so the lower-barrier criterion identifies the limit. This addition step will remove a deficit of mass in the existence proof of section 15.12.
Theorem 15.11.1.7A54D08FEB18 Let $\theta \geq 0$ be big, and suppose $\varphi \in \mathrm{PSH}(X,\theta )$ satisfies $P_{\delta \theta }(\varphi )\not\equiv -\infty $ for some $0<\delta <1$. Let $u,v\in \mathrm{PSH}(X,\theta )$ satisfy $u,v\sim \varphi $, and let $\lambda >0$. If
\[ \mu =\mathrm{e}^{-\lambda u}\mathrm{MA}^{\mathrm{ext}}_\theta (u) +\mathrm{e}^{-\lambda v}\mathrm{MA}^{\mathrm{ext}}_\theta (v) \]
is a finite measure, then there exists $a\in \mathrm{PSH}(X,\theta )$, with $a\sim \varphi $, such that
\[ \mathrm{MA}^{\mathrm{ext}}_\theta (a)=\mathrm{e}^{\lambda a}\mu . \]
Here the weights are evaluated off the pole sets. The solution $a$ is unique in the ordinary singularity type of $\varphi $.
Proof.
Step 1: truncated equations and barriers. Choose $S\geq \max (\sup _ Xu,\sup _ Xv)$. A common translation reduces the theorem to the case $u,v\leq 0$: translating $u$ and $v$ by $-S$ multiplies $\mu $ by $\mathrm{e}^{\lambda S}$, and the resulting solution is then translated back by $S$.
For integers $j\geq 1$, set $u_ j=\max (u,-j)$, $v_ j=\max (v,-j)$ and
\[ \mu _ j=\mathrm{e}^{-\lambda u_ j}\mathrm{MA}_\theta (u_ j) +\mathrm{e}^{-\lambda v_ j}\mathrm{MA}_\theta (v_ j). \]
For $s_ j=(u_ j+v_ j)/2$, positivity of mixed products gives
\[ \mu _ j\leq 2^ n \mathrm{e}^{\lambda j}\mathrm{MA}_\theta (s_ j). \]
Thus $\mu _ j$ has a bounded density with respect to the Monge–Ampère measure of the bounded potential $s_ j$, and $\mu _ j\neq 0$ by proposition 16.28.55. The bounded subsolution theorem theorem 15.10.1 therefore gives bounded $a_ j\in \mathrm{PSH}(X,\theta )$ with $\mathrm{MA}_\theta (a_ j)=\mathrm{e}^{\lambda a_ j}\mu _ j$, and bounded exponential comparison yields
\[ a_ j\leq \min (u_ j,v_ j)\leq 0. \]
Choose $C$ with $|u-v|\leq C$ off the common pole set; then $|u_ j-v_ j|\leq C$. Put
\[ D=C/2+n\lambda ^{-1}\log 2,\qquad b_ j=s_ j-D. \]
Because $|u_ j-v_ j|\leq C$, positivity of mixed products gives
\[ \mathrm{e}^{-\lambda b_ j}\mathrm{MA}_\theta (b_ j) \geq 2^{-n}\mathrm{e}^{\lambda D-\lambda s_ j} \bigl (\mathrm{MA}_\theta (u_ j)+\mathrm{MA}_\theta (v_ j)\bigr ) \geq \mu _ j. \]
Hence $a_ j\geq b_ j\geq w$, where
\[ w=(u+v)/2-D\sim \varphi \]
is a fixed $\theta $-plurisubharmonic function. Extract a subsequence with $a_ j\to a$ in $L^1(X)$. The upper barriers decrease to $\min (u,v)$, so $w\leq a\leq \min (u,v)$ and $a\sim \varphi $.
Step 2: convergence in capacity. Write $E_ j=\{ \min (u,v)>-j\} $. By locality, $\mu _ j=\mu $ on $E_ j$; since $a_ j\leq 0$, it follows that
\[ \int _{E_ j}|\mathrm{e}^{a_ j}-\mathrm{e}^ a|\, \mathrm{MA}_\theta (a_ j) \leq \int _ X|\mathrm{e}^{a_ j}-\mathrm{e}^ a|\, \mathrm{d}\mu \longrightarrow 0 \]
by lemma 5.44.2. On $X\setminus E_ j$ we have $a_ j\leq -j$ and $a\leq -j$, whence $|\mathrm{e}^{a_ j}-\mathrm{e}^ a|\leq \mathrm{e}^{-j}$. Moreover
\[ \mathrm{MA}_\theta (a_ j) =\mathrm{e}^{\lambda (a_ j-u_ j)}\mathrm{MA}_\theta (u_ j) +\mathrm{e}^{\lambda (a_ j-v_ j)}\mathrm{MA}_\theta (v_ j) \leq \mathrm{MA}_\theta (u_ j)+\mathrm{MA}_\theta (v_ j). \]
Choose a Hermitian form $\Omega \geq \theta $. Since $\mathrm{MA}_\theta (u_ j)\leq \mathrm{MA}_\Omega (u_ j)$ and $-j\leq u_ j\leq 0$, and similarly for $v_ j$, the range estimate of proposition 13.15.2, with $r_ n=\max (n-2,0)$, gives
\[ \int _{X\setminus E_ j}|\mathrm{e}^{a_ j}-\mathrm{e}^ a|\, \mathrm{MA}_\theta (a_ j) \leq C'\mathrm{e}^{-j}(1+j)^{r_ n}\longrightarrow 0. \]
The lower-barrier criterion proposition 15.9.3 now gives $a_ j\to a$ in capacity. Note that only a polynomial bound for the masses of the approximating measures was needed.
Step 3: the limiting equation. We identify the equation on sets where $w$ is bounded below, without first proving weak convergence of the measures $\mathrm{MA}_\theta (a_ j)$. For $T>0$, let $g_ T$ be the continuous cutoff from the proof of lemma 13.14.8, and put $\zeta _ T=g_ T(w)$. Where $\zeta _ T>0$, we have $w>-2T$, hence every $a_ j$ and $a$ is above $-2T$. If $j>2T$ and $x\notin E_ j$, the inequalities $w(x)\leq a_ j(x)\leq -j$ force $\zeta _ T(x)=0$. Consequently, for all $j>2T$,
\[ \zeta _ T\mathrm{MA}_\theta (a_ j) =\zeta _ T \mathrm{e}^{\lambda a_ j}\mu . \]
We test this identity against a continuous function. On the left we may replace $a_ j$ by $\max (a_ j,-2T)$ and use continuity of bounded products under convergence in capacity, with the fixed bounded quasi-continuous weight $\zeta _ T$. On the right, the exponentials converge against the fixed measure $\mu $, which gives convergence in total variation. Therefore
\[ \zeta _ T\mathrm{MA}^{\mathrm{ext}}_\theta (a) =\zeta _ T\mathrm{e}^{\lambda a}\mu . \]
As $T\to \infty $, the cutoffs increase to one off the pole set of $w$, which no measure involved charges. This proves the equation and, in particular, shows that $\mathrm{MA}^{\mathrm{ext}}_\theta (a)$ has finite mass. Uniqueness follows from corollary 15.2.3.
$\square$
Pome
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