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We now replace a pointwise obstacle by a constraint on singularities: we take the upper envelope of all nonpositive potentials that are more singular than a given potential $u$. Since the constant $C$ in $v\leq u+C$ may depend on the competitor $v$, the envelope can be less singular than $u$; this happens in example 16.22.12 below.
Definition 16.21.1.99168E437529 For $u\in \mathrm{PSH}(X,\theta )$, the singularity envelope of $u$ is
\[ P_\theta [u]=\bigl (\sup \{ v\in \mathrm{PSH}(X,\theta ):v\leq 0, \ v\preceq u\} \bigr )^*. \]
The functions $v$ in this supremum are the competitors for $P_\theta [u]$. A model potential is a potential $\phi $ with $P_\theta [\phi ]=\phi $ and positive mass $\int _ X\langle \theta _\phi ^ n\rangle >0$. Since $u-\sup _ Xu$ is a competitor and all competitors are nonpositive, the singularity envelope is always defined, belongs to $\mathrm{PSH}(X,\theta )$, and satisfies $u-\sup _ Xu\leq P_\theta [u]\leq V_\theta $ and $\sup _ XP_\theta [u]=0$.
Proposition 16.21.2.037CA226D141 Let $u\in \mathrm{PSH}(X,\theta )$.
$P_\theta [u]$ depends only on the singularity type of $u$, and $u\preceq u'$ implies $P_\theta [u]\leq P_\theta [u']$.
The rooftop envelopes $P_\theta (u+k,V_\theta )$ increase with $k$, and
\[ P_\theta [u]=\left(\lim _{k\to \infty } P_\theta (u+k,V_\theta )\right)^*. \]
The singularity envelope preserves all mixed non-pluripolar masses: if $\theta _1,\ldots ,\theta _ p$, $1\leq p\leq n$, are smooth closed real $(1,1)$-forms and $u_ j\in \mathrm{PSH}(X,\theta _ j)$, then
\[ \int _ X\left\langle \bigwedge _ j(\theta _ j+\mathrm{dd}^{\mathrm{c}}P_{\theta _ j}[u_ j])\right\rangle \wedge \omega ^{n-p} =\int _ X\left\langle \bigwedge _ j(\theta _ j+\mathrm{dd}^{\mathrm{c}}u_ j)\right\rangle \wedge \omega ^{n-p}. \]
In particular, $P_\theta [u]$ has the same mass as $u$.
If $u$ has minimal singularities, then $P_\theta [u]=V_\theta $.
Proof.
(1) The family of competitors depends only on the singularity type of $u$, and it increases when $u$ is replaced by a less singular potential.
(2) The rooftop envelopes increase with $k$ because the rooftop envelope is monotone. Every competitor $v$ for $P_\theta [u]$ satisfies $v\leq u+k$ for some integer $k$ and $v\leq V_\theta $, hence $v\leq P_\theta (u+k,V_\theta )$. Conversely, each $P_\theta (u+k,V_\theta )$ is nonpositive and, by proposition 16.11.6, has the singularity type of $u$; so it is a competitor. Hence the two families have the same supremum, which proves the formula.
(3) By proposition 16.11.6, the tuple of rooftop envelopes $P_{\theta _ j}(u_ j+k,V_{\theta _ j})$ has the singularity types of the $u_ j$, so it has the same mixed mass as the tuple $(u_ j)$ by corollary 13.8.8. By (2) and theorem 5.10.2, which shows that an increasing sequence differs from its regularized limit only on a Lebesgue null set, these rooftop envelopes increase almost everywhere to $P_{\theta _ j}[u_ j]$ as $k\to \infty $, and corollary 13.8.9 shows that their mixed masses converge to the mixed mass of the envelopes.
(4) If $u$ has minimal singularities, then $V_\theta \preceq u$, so $V_\theta $ is itself a competitor; it dominates all other competitors, since they are nonpositive.
$\square$
Proposition 16.21.3.5ADA09DD223E Suppose that $\{ \theta \} $ is big, let $u\in \mathrm{PSH}(X,\theta )$, and put $w=P_\theta [u]$. Then
\[ \mu _ w\leq \mathbf1_{\{ w=V_\theta \} }\mu _{V_\theta }, \]
and $\mu _ w$ is carried by $\{ w=0\} $.
Proof.
Put $w_ k=P_\theta (u+k,V_\theta )$ for integers $k$; these are potentials by proposition 16.11.6. By corollary 16.13.5,
\[ \mu _{w_ k}\leq \mathbf1_{\{ w_ k=u+k\} }\mu _ u+\mathbf1_{\{ w_ k=V_\theta \} }\mu _{V_\theta } \leq \mathbf1_{\{ u+k\leq V_\theta \} }\mu _ u +\mathbf1_{\{ w=V_\theta \} }\mu _{V_\theta }; \]
for the second inequality we used $w_ k\leq V_\theta $ and $w_ k\leq w\leq V_\theta $. Since $V_\theta \leq 0$, the sets $\{ u+k\leq V_\theta \} $ decrease to the pole set of $u$, which is $\mu _ u$-null, and the first term tends to zero in total variation by dominated convergence. By proposition 16.21.2, $w_ k$ increases almost everywhere to $w$, so $\mu _{w_ k}\to \mu _ w$ weakly by corollary 13.8.9. Integrating a continuous function $\chi \geq 0$ and letting $k\to \infty $, we obtain $\int _ X\chi \, \mu _ w\leq \int _ X\chi \mathbf1_{\{ w=V_\theta \} }\, \mu _{V_\theta }$, and lemma 16.12.1 gives the asserted inequality. Finally, a potential lies below $0$ quasi-everywhere if and only if it is nonpositive, so $V_\theta =P_\theta (0)$, and theorem 16.13.4 shows that $\mu _{V_\theta }$ is carried by $\{ V_\theta =0\} $. Hence $\mu _ w$ is carried by $\{ w=V_\theta =0\} $.
$\square$
Pome
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