10.31 Big backgrounds under bounded massC2AE04961A4E

Proof. We follow the proofs of lemma 15.37.13, theorem 15.37.19 and check their ingredients in the present background. The envelope identity and Hartogs' lemma give $u_ j\leq \varphi +C$. Finite rooftops and stretching envelopes of members of $\mathcal E_\varphi $ stay in $\mathcal E_\varphi $ by proposition 16.28.64, lemma 16.28.63, and their masses are at least $\ell $ by (10.31.1). These facts replace the corresponding ones for semipositive backgrounds.

After a common translation, assume $u_ j,u\leq \varphi \leq 0$. Using lemma 5.44.2, extract once and for all a subsequence along which $u_ j\to u$ almost everywhere for smooth volume and

For $p_{j,N}=P_\theta (u_ j,\ldots ,u_ N)$, partitioning the contact set according to the first obstacle touched gives

In the proof of the extraction lemma, replace $\kappa $ by $\ell $. For each fixed $B\geq 1$, the stretched rooftops $r_{j,N}=Q_ B(p_{j,N},\varphi )$ satisfy

The exponential test (15.37.15) therefore bounds $\sup _ Xr_{j,N}$ from below, independently of $N$, once $j$ is sufficiently large. The construction combining the tail with the finitely many earlier terms, with $B=2b$, uses only convex combinations and finite rooftops that are not identically $-\infty $. It provides every defining competitor for

Since $h\leq u\leq \varphi $, upward stability gives $u\in \mathcal E_\varphi $.

For the continuity statements, put $p_ j=P_\theta (\inf _{k\geq j}u_ k)$ and $p=(\lim _ jp_ j)^*$. The monotone limits $p_{j,N}\downarrow p_ j$ and $p_ j\uparrow p$ have the common lower bound $h\not\equiv -\infty $, so they hold in capacity by lemma 5.29.19. All the limits lie in $\mathcal E_\varphi $, and $p\leq u$. Applying lemma 13.14.8 twice to (10.31.6) gives

Hence $\mu _ p(\{ p<u\} )=0$, and zero domination (theorem 13.19.1 with coefficient zero) gives $p=u$. The upper envelopes $(\sup _{k\geq j}u_ k)^*$ decrease to $u$ by Hartogs' lemma, and squeezing the $u_ j$ between them and $p_ j$ proves convergence in capacity.

Finally, the fixed finite dominating measure gives

uniformly in $j$, and theorem 13.15.23 proves weak convergence to $\mu _ u\leq \nu $. Starting with an arbitrary subsequence, we obtain a further subsequence with both convergences and the same limit. Hence the conclusions hold for the original sequence. $\square$

Proof. We first treat the case $u_ j\leq w$. Since $\overline{\operatorname {vol}}_\theta (w)=M$, we may choose $w'\sim w$ with $\mu _{w'}(X)>M-\ell /2$. Translate $w'$ so that $w'\leq w$, and put $p_ j=P_\theta (u_ j,w')$. These rooftops belong to $\mathcal E_\varphi $. Contact support and ordered contact imply

Indeed, the integrand vanishes on the contact part $\{ p_ j=w'\} $, and on the remaining contact part $p_ j=u_ j<w'\leq w$.

We shall use the following estimate for maxima. If $a\in \mathcal E_\varphi $, $a\leq w'$, and $t>0$, then

To see this, put $m=\max (a,w'-t)$. It has the ordinary singularity type of $w'$, so $\mu _ m(X)\leq M$. On $\{ a>w'-t\} $, locality gives $\mu _ m=\mu _ a$. On the complementary set, where $m=w'-t$, ordered contact gives $\mu _ m\geq \mu _{w'}$. Integrating over the two sets proves (10.31.10).

Suppose that a subsequence had $\sup _ Xu_ j\to -\infty $; then also $\sup _ Xp_ j\to -\infty $. Splitting the mass of $p_ j$ at the level $w'-s$, $s>0$, and using (10.31.9) and (10.31.10), we obtain

For fixed $s$, the middle term tends to zero as $j\to \infty $, since $w'$ is finite off a pluripolar set and $\mu _{w'}$ is finite and non-pluripolar. Letting $j\to \infty $ and then $s\to \infty $ gives $\ell \leq M-\mu _{w'}(X)<\ell /2$, a contradiction. This proves the first assertion.

Now suppose $u_ j\to u$ in $L^1$. Then $u\leq w\preceq \varphi $. Fix $b>1$ and suppose that $Q_ b(u,w)\equiv -\infty $. Since $w-w'$ is bounded, translating competitors by a constant shows that $Q_ b(u,w')\equiv -\infty $ as well. Keep the same $p_ j$ and set

We claim that $\sup _ Xh_ j\to -\infty $. Otherwise, since $h_ j\leq p_ j\leq w'$, theorem 5.25.2 would give a subsequence converging in $L^1$ to some $h\not\equiv -\infty $. Passing to the limit in

first almost everywhere and then everywhere by lemma 5.24.3, would give a competitor for $Q_ b(u,w')$, a contradiction.

On the contact set of $h_ j$ one has $p_ j=b^{-1}h_ j+(1-b^{-1})w'$. Thus $h_ j\leq w'-bs$ implies $p_ j\leq w'-s$ there. Contact support and (10.31.9) therefore imply

Applying (10.31.10) with $a=h_ j$ and $t=bs$ on the complementary set, we obtain

Let first $j\to \infty $ and then $s\to \infty $. The choice of $w'$ again gives $\ell <\ell /2$, a contradiction. Therefore $Q_ b(u,w)\not\equiv -\infty $ for every $b>1$. The resulting convex inequalities and theorem 13.17.1, proposition 13.17.2 give

Letting $b\to \infty $ shows that the upper volumes of $u$ and $\varphi $ are equal; hence $u\in \mathcal E_\varphi $.

For the last assertion put $q_ j=P_\theta (u_ j,w)$. Contact gives

The case already treated bounds $\sup _ Xq_ j$ from below, and therefore also $\sup _ Xu_ j\geq \sup _ Xq_ j$. If $u_ j\to u$ in $L^1$, then, since $q_ j\leq w$, a subsequence of the $q_ j$ converges in $L^1$ to some $q\not\equiv -\infty $, and the case already treated gives $q\in \mathcal E_\varphi $. Since $q\leq u\leq \varphi +C$, upward stability gives $u\in \mathcal E_\varphi $ as well. $\square$

Proof. Start with any subsequence and extract a further subsequence along which the integrals in the hypothesis are at most $2^{-j}$ and $u_ j\to u$ almost everywhere for smooth volume. The envelope identity and Hartogs' lemma give a uniform bound $u_ j,u\leq \varphi +C$. For $N\geq j$ let $p_{j,N}=P_\theta (u_ j,\ldots ,u_ N)$. Partitioning the contact set according to the first obstacle touched gives

Apply proposition 10.31.8 with $w=u$, in its one-sided form. It bounds $\sup _ Xp_{j,N}$ from below, independently of $N$. Consequently the decreasing sequence has a limit

that is not identically $-\infty $, and the same proposition puts $p_ j$ in $\mathcal E_\varphi $. The increasing sequence $p_ j$ has a regularized limit $p$, which belongs to $\mathcal E_\varphi $ by upward stability, since $p_1\leq p\leq u\preceq \varphi $. The inequality $p\leq u$ follows from the almost-everywhere convergence of the extracted sequence.

Both $p_{j,N}\downarrow p_ j$ and $p_ j\uparrow p$ hold in capacity by lemma 5.29.19, since $p_1\not\equiv -\infty $ is a common lower bound. Apply lemma 13.14.8 to (10.31.12), with the bounded weights $\min (1,|p_{j,N}-u|)$ and then $\min (1,|p_ j-u|)$. These weights converge in capacity: discard the deep sublevel sets of $p_1$ and $u$, and use that $\min (1,|x-y|)$ is Lipschitz on finite values. The result is

Hence $\mu _ p(\{ p<u\} )=0$, and zero domination gives $p=u$. The upper envelopes $(\sup _{k\geq j}u_ k)^*$ decrease to $u$. Squeezing the extracted $u_ j$ between these envelopes and $p_ j$ proves convergence in capacity. Since the initial subsequence was arbitrary, the conclusion holds for the original sequence. $\square$

Proof. The difference $u-V$ is bounded, so $V-C\leq b\leq u\leq V$ for some constant $C$. Hence $p\sim V$. The obstacle is quasi-continuous: away from sets of arbitrarily small capacity, $u$ and $V$ are finite and continuous, and their difference has a fixed bounded range. The general contact theorem theorem 15.8.6 therefore applies.

Choose a strict analytic barrier $\rho $ and translate it so that $\rho \leq V$; put $\Omega =X\setminus \{ \rho =-\infty \} $. On this open set $u$, $V$ and $p$ are locally bounded. Let $\tau =\chi ^{-1}$; then $\tau $ is convex and $0<\tau '\leq 1$. By lemma 5.28.13, the function

is $\theta $-plurisubharmonic on $\Omega $ and satisfies

Since $p\leq w\leq V$ on $\Omega $, the removability lemma lemma 5.24.7 extends $w$ across the analytic set $X\setminus \Omega $, preserving its smooth lower curvature bound.

The inequality $p\leq b$ implies $w\leq u$, with equality on $D=\{ p=b\} $. Ordered contact gives

On $D$ the derivative $\tau '(p-V)$ equals $1/\chi '(u-V)$. Together with contact support for $p$, this proves the claim on $\Omega $. The analytic set $X\setminus \Omega $ is not charged by the non-pluripolar measures involved. $\square$

Proof. From the distribution of $V-u$ under $\mu _ u$ we build a concave weight adapted to $u$, apply the preceding lemma, and use $A$ to bound the resulting envelope against $\nu $. Throughout, $c$ and $C$ denote positive constants depending only on $n,\ell _ V,M_ V$.

The normalization gives $u\leq V$. Put $x=V-u\geq 0$ and

with $T=0$ if this set is empty. Since $u\sim V$, $T$ is finite. Zero domination, applied to $u$ and $V-t$ for $t>T$, shows that $u\geq V-T$; thus $x\leq T$ off the pole set. If $T\leq 2$ the claim is immediate. We may therefore suppose $T>2$; in particular $B>0$.

Fix a real number $s$ with $0<s<T-2$. For $t\geq 0$, define

and set

The tail $\mu _ u(\{ x>t\} )$ in the first denominator has a positive lower bound on $[0,T-s]$. Hence $F_ s$ is locally Lipschitz, and $\chi _ s$ has the regularity and concavity required in the lemma.

Tonelli's theorem gives

since $\int _0^\infty (1+t)^{-3/2}\, \mathrm{d}t=2$ and the tail is at most $M_ V$ on the second interval. Chebyshev's inequality gives $\mu _ u(\{ x>t\} )\leq B/t$. Integrating first for $F_ s$ and then for $W_ s$, we obtain

Indeed, $g_ s(t)\geq cB^{-1}t^{-1/2}$ for $t\geq 1$ in this interval, so $F_ s(t)\geq c(1+B)^{-1}t^{1/2}$. For $t\geq 2$ integrate this estimate over $[t/2,t]$; for $1\leq t\leq 2$ use $W_ s(t)\geq t$.

Let $p_ s=P_\theta (V-W_ s(x))$, the envelope of the preceding lemma for $\chi =\chi _ s$. The lemma, the bound $W_ s(t)\leq tF_ s(t)^{1/N}$ and the Cauchy–Schwarz inequality give, since $N=2n+1$,

Write $a_ s=\sup _ Xp_ s\leq 0$. By the definition of $V$, $p_ s-a_ s\leq V$, so $V-p_ s\geq -a_ s$. Since $\mu _{p_ s}(X)\geq \ell _ V$, the last estimate yields $\ell _ V\sqrt{-a_ s}\leq C\sqrt B$, and hence $-a_ s\leq CB$.

The definition of $A$ now gives

Combining this with (10.31.17), and then letting $s\downarrow 0$ for each fixed $1\leq t<T$, we obtain

There is no tail for $t\geq T$. The layer-cake formula therefore gives

because $1<r<\beta $. This implies the stated bound.

Finally $-u=-V+(V-u)$, where $-V\in L^1(\nu )$ because $A<\infty $. The first summand is fixed and integrable, and the second has a uniform $L^ r(\nu )$ bound. This proves uniform integrability; no $L^ r$ bound for $V$ itself is claimed. $\square$

Proof. The limit $v$ is not identically $-\infty $ and satisfies $\sup _ Xv=0$, and $v_ j,v\leq V\leq 0$. By lemma 5.44.12, $\sup _ jA_\theta (\nu _ j)<\infty $, which gives the uniform estimate

In particular the relative first moments $\int _ X(V-v_ j)\mu _{v_ j}$ are bounded, and proposition 10.31.8, with fixed reference $w=V$, puts $v$ in $\mathcal E_ V$.

By proposition 10.31.15, applied to the fixed measure $\nu $, the $v_ j$ are uniformly integrable with respect to $\nu $. For each fixed $T$, the bounded functions $\max (v_ j,-T)$ have a common lower curvature bound in a Hermitian background $\Omega \geq \theta $ and converge in $L^1$ for smooth volume to $\max (v,-T)$; hence lemma 5.44.2 gives their convergence in $L^1(\nu )$. These truncations need not be $\theta $-plurisubharmonic; they are used only in the larger Hermitian background. Uniform integrability, followed by $T\to \infty $, gives $v_ j\to v$ in $L^1(\nu )$, and the limit is integrable because $A_\theta (\nu )<\infty $.

The smoothing lemma, in the form (5.44.13), now yields

Therefore the self-measure convergence theorem theorem 10.31.11 gives convergence in capacity. Finally, (10.31.20) gives the sublevel estimate

and the masses of the $\mu _{v_ j}$ are bounded by $M_ V$. The tail criterion theorem 13.15.23 therefore gives the asserted weak convergence. $\square$

Proof. First suppose that $\mu $ is a probability measure with $A_\theta (\mu )<\infty $, and let $\mu _ j$ be its smooth probability approximations. The theorems for the minimal ordinary type, theorem 15.16.2, theorem 15.16.7, apply to their smooth, nonzero densities.

For the normalized equation, they give $v_ j\sim V$ with $\sup _ Xv_ j=0$ and $\mu _{v_ j}=c_ j\mu _ j$. Since $\mu _ j$ is a probability measure, the mass bounds $\ell _ V\leq \mu _{v_ j}(X)\leq M_ V$ give $\ell _ V\leq c_ j\leq M_ V$. After extraction, $c_ j\to c>0$ and $v_ j\to v$ in $L^1(X)$. The preceding proposition applies, so $v\in \mathcal E_ V$ and $\mu _{v_ j}\rightharpoonup \mu _ v$. The right-hand sides converge weakly to $c\mu $, which proves the equation.

For the positive-exponent equation, take $u_ j\sim V$ satisfying $\mu _{u_ j}=\mathrm{e}^{\lambda u_ j}\mu _ j$, and write $a_ j=\sup _ Xu_ j$ and $v_ j=u_ j-a_ j$. The mass bounds and Jensen's inequality give

and $\sup _ jA_\theta (\mu _ j)<\infty $ by lemma 5.44.12. Thus $a_ j$ is bounded above and below. After extraction, $a_ j\to a$, $v_ j\to v$ in $L^1(X)$, and $\mu _{v_ j}\leq C\mu _ j$. The preceding proposition gives $v\in \mathcal E_ V$, convergence in $L^1(\mu )$ and weak convergence of the Monge–Ampère measures. Set $u=v+a$. By the smoothing lemma, the right-hand sides converge weakly to $\mathrm{e}^{\lambda u}\mu $. This proves the equation, and $u\in \mathcal E_ V$.

The same conclusions hold for every nonzero finite measure whose normalization to a probability measure has finite $A_\theta $, since scaling the measure only changes the positive scalar, or translates the solution of the exponential equation by a constant.

Now let $\mu $ be an arbitrary finite non-pluripolar measure of positive mass. The bounded-carrier decomposition lemma 15.24.4, applied to $\omega $, writes

For $\mu _ k=\min (f,k)\nu $, we have $\mu _ k(X)>0$. By lemma 5.44.10, the normalization of $\mu _ k$ has finite $A_\theta $ for each fixed $k$.

For the normalized equation, the preceding case gives $v_ k\in \mathcal E_ V$ with $\sup _ Xv_ k=0$ and $\mu _{v_ k}=c_ k\mu _ k$, where

After extraction, $c_ k\to c>0$ and $v_ k$ converges in $L^1$ to some $v\not\equiv -\infty $. All the measures $\mu _{v_ k}$ are bounded by the fixed finite non-pluripolar measure $M_ V\mu /\mu _1(X)$. The fixed-dominator theorem theorem 10.31.5 gives $v\in \mathcal E_ V$ and weak convergence to $\mu _ v$. Since $c_ k\mu _ k\to c\mu $ in total variation, $\mu _ v=c\mu $.

For the positive-exponent equation, let $u_ k\in \mathcal E_ V$ solve $\mu _{u_ k}=\mathrm{e}^{\lambda u_ k}\mu _ k$. Since $\mu _ k$ increases with $k$, exponential order (corollary 13.19.3) gives $u_{k+1}\leq u_ k$. Also

The lower bound on the suprema shows that the decreasing limit $u$ is not identically $-\infty $. The same fixed-dominator theorem puts $u$ in $\mathcal E_ V$ and gives weak convergence of the left-hand sides. Dominated convergence with respect to $\nu $, with the integrable bound $\mathrm{e}^{\lambda \sup _ Xu_1}f$, gives convergence of the right-hand sides to $\mathrm{e}^{\lambda u}\mu $ in total variation.

Finally, corollary 13.19.3 with reference $V$, applied in both directions, gives uniqueness of the exponential solution and of the positive scalar. $\square$

Proof. Choose $\rho \in \mathrm{PSH}(X,\theta )$ with $\theta _\rho \geq a\omega $ for some $a>0$. Then $\rho \in \mathrm{PSH}(X,\theta -a\omega )$. Volume addition (proposition 13.17.2), followed by monotonicity in the singularities (theorem 13.17.1), gives

In the addition step the big background is $a\omega $, and the added positive current is $(\theta -a\omega )_\rho $. Bounded mass gives finiteness of the upper volume. $\square$

Proof. First suppose $z\leq \eta $; recall also that $\varphi \leq \eta \leq 0$. For $b>1$ sufficiently close to one, lemma 13.20.1 gives

which is not identically $-\infty $. Set $a=1-b^{-1}$. The stretching lemma lemma 16.28.63, now applied in $\mathcal E_\eta $, gives $z_{1/a}=Q_{1/a}(z,\eta )\in \mathcal E_\eta $. The potential

lies below both $\varphi $ and $z$. Indeed, $z_{1/a}\leq \eta $ and $(1-a)h_ b+a\eta \leq \varphi $ give the first bound, while $h_ b\leq \eta $ and $a z_{1/a}+(1-a)\eta \leq z$ give the second. Thus $p=P_\theta (\varphi ,z)$ is not identically $-\infty $. Volume addition and monotonicity give

As $b\downarrow 1$, we have $a\downarrow 0$, and the first term tends to the last by lemma 13.20.1. The upper volumes of $p$ and $\varphi $ are therefore equal. Since $p\leq \varphi $, theorem 13.18.5 gives $p\in \mathcal E_\varphi $.

For general $z$, the envelope identity in $\mathcal E_\eta $ gives $z-\sup _ Xz\leq \eta $. Set $S=\max (\sup _ Xz,0)$. The preceding case gives $P_\theta (\varphi ,z-S)\in \mathcal E_\varphi $, and

Upward stability finishes the proof. $\square$

Proof. For $C\geq 0$ put $p_ C=P_\theta (\varphi +C,z)$. The preceding proposition gives $p_0\in \mathcal E_\varphi $. Since $p_0\leq p_ C\preceq \varphi $, upward stability gives $p_ C\in \mathcal E_\varphi $. The envelope identity, together with $p_ C\leq z$, yields the uniform bound

Consequently the increasing regularized limit $Q$ is not identically $-\infty $, belongs to $\mathcal E_\varphi $ and is at most $z$. The monotone convergence holds in capacity by lemma 5.29.19.

Finally, lemma 13.19.4, applied with $u=\varphi $ and $w=z$, gives

Proof. By theorem 10.31.22, there is $z\in \mathcal E_ V$ with $\mu _ z=\mathrm{e}^ z\nu $. Since $V$ is an ordinary model and $V\geq \varphi $, the preceding proposition applies with $\eta =V$ and gives $Q=P_\theta [\varphi ](z)\in \mathcal E_\varphi $. On the contact set $\{ Q=z\} $,

The weights are finite off the pole sets, which carry no mass. Thus $\mathrm{e}^{-Q}\mu _ Q\leq \nu $, and this measure is nonzero because $\mu _ Q(X)\geq \ell >0$. $\square$

Proof. We first construct the lower models $\psi _ j$, then a common lower bound for rooftops of the $u_ l$ in the classes $\mathcal E_{\psi _ j}$, and finally treat domination on all of $X$.

By Hartogs' lemma, $u_ j$, $u$ and $\chi $ have a common upper bound; translate them to be nonpositive. The envelope identity then gives $u_ j\leq \varphi _ j$. Since $\varphi _ j\to \varphi $ in $L^1$ by proposition 13.22.4, it follows that $u\leq \varphi $. Put $\Lambda =\nu +\mu _\chi $. Using lemma 5.44.2, extract a subsequence along which $u_ j\to u$ almost everywhere for smooth volume and

Apply theorem 13.23.3 to extract further, keeping these estimates, and obtain increasing models $w_ j\leq \varphi _ j,\varphi $ with lower volume at least $\delta /2$ and $d_\theta (w_ j,\varphi )\to 0$. For $l\geq j$ we have $w_ j\leq w_ l\leq \varphi _ l$. Hence $\psi _ j$ is not identically $-\infty $, and

The upper bound follows by passing to the $L^1$ limit in $\psi _ j\leq \varphi _ l$ as $l\to \infty $. Monotonicity of the ordinary envelope gives $P_\theta [\psi _ j]\leq \varphi _ l$ for all $l\geq j$, so $\psi _ j$ is an ordinary model. It increases with $j$, has lower volume at least $\delta /2$, and

Set $g_ j=P_\theta [\psi _ j](\chi )$. Proposition 10.31.26, with $(\varphi ,\eta )$ replaced by $(\psi _ j,\varphi )$, gives $g_ j\in \mathcal E_{\psi _ j}$, $g_ j\leq \chi $ and $\mu _{g_ j}\leq \mu _\chi $. Moreover $g_ j$ increases with $j$. For $N\geq j$, consider

For each $l\geq j$, the projection $P_\theta [\psi _ j](u_ l)$ belongs to $\mathcal E_{\psi _ j}$ by the same proposition, now with upper reference $\varphi _ l$. The finite rooftop of these projections and $g_ j$ lies below $q_{j,N}$. Since $q_{j,N}\leq g_ j$, the finite-rooftop proposition and upward stability give $q_{j,N}\in \mathcal E_{\psi _ j}$, and hence $\mu _{q_{j,N}}(X)\geq \delta /2$.

Partition the contact set of $q_{j,N}$ into disjoint pieces, one for each obstacle. On a piece where $q_{j,N}=u_ l$, we have $u_ l\leq g_ j\leq \chi $, so the hypothesis bounds the measure there by $\nu $. On the piece where $q_{j,N}=g_ j$, the measure is bounded by $\mu _{g_ j}\leq \mu _\chi $. Consequently $\mu _{q_{j,N}}\leq \Lambda $. Put $F_ j=\min (\mathrm{e}^ u,\mathrm{e}^{g_ j})$. On the $u_ l$ piece,

On the $g_ j$ piece, $g_ j=q_{j,N}\leq u_ N$, so the error is $(\mathrm{e}^{g_ j}-\mathrm{e}^ u)_+\leq |\mathrm{e}^{u_ N}-\mathrm{e}^ u|$. The $u_ N$ piece and the $g_ j$ piece are controlled by the two summands $\nu $ and $\mu _\chi $ of $\Lambda $, so together they contribute at most $I_ N$. Thus

Since $F_ j\geq F_1>0$ outside a pluripolar set, we may choose $\epsilon >0$ with $\Lambda (\{ F_1\leq \epsilon \} )<\delta /4$. The lower mass bound and the domination by $\Lambda $ give

For every sufficiently large $j$, this bounds $\sup _ Xq_{j,N}$ from below, uniformly in $N$. Hence the decreasing limit $q_ j$ as $N\to \infty $ is not identically $-\infty $. Applying theorem 10.31.5 in the fixed class $\mathcal E_{\psi _ j}$, with dominating measure $\Lambda $, gives $q_ j\in \mathcal E_{\psi _ j}$. Now

so $h_ j\in \mathcal E_{\psi _ j}$ by upward stability. We discard the finitely many earlier indices. Since $h_ j\leq u\leq \varphi $, monotonicity and (10.31.31) show that $u$ and $\varphi $ have the same upper volume, and theorem 13.18.5 gives $u\in \mathcal E_\varphi $.

Suppose now that $\mu _{u_ j}\leq \nu $ on $X$. The finite rooftops $p_{j,N}=P_\theta (u_ j,\ldots ,u_ N)$ are not identically $-\infty $, since they lie above $h_ j$. Partitioning their contact sets gives

For fixed $j$, they decrease to $h_ j$ as $N\to \infty $. These lower envelopes increase to a regularized limit $h\leq u\leq \varphi $, and their upper volumes tend to that of $\varphi $; hence $h\in \mathcal E_\varphi $. Both monotone limits hold in capacity by lemma 5.29.19. Applying lemma 13.14.8 twice gives

Hence $\mu _ h(\{ h<u\} )=0$, and zero domination gives $h=u$. The upper envelopes $(\sup _{k\geq j}u_ k)^*$ decrease to $u$ by Hartogs' lemma and the almost-everywhere convergence, and squeezing gives convergence in capacity. Finally,

uniformly in $j$, and the tail criterion theorem 13.15.23 gives weak convergence of the measures. Since the extraction can start from any subsequence, both conclusions hold for the original sequence. $\square$

Proof. By lemma 5.24.3, the lower bound passes to the limit, so $h\leq a\preceq \varphi $, and upward stability gives $a\in \mathcal E_\varphi $. Start with any subsequence and extract a further one along which the integrals in the hypothesis are at most $2^{-j}$ and $a_ j\to a$ almost everywhere for smooth volume. The finite rooftops

are not identically $-\infty $ because $h$ is a competitor. Partitioning their contact sets according to the first obstacle touched gives

No relative-class assertion about these finite rooftops is needed. Their decreasing limits $p_ j=P_\theta (\inf _{k\geq j}a_ k)$ are at least $h$, and the almost-everywhere convergence gives $p_ j\leq a$. Hence the $p_ j$ and their increasing regularized limit $p$ lie in $\mathcal E_\varphi $.

The limits in $N$ and then in $j$ hold in capacity by lemma 5.29.19, since $h$ is a common lower bound. Apply the weighted lower semicontinuity lemma 13.14.8 twice, to the bounded weights $\min (1,|p_{j,N}-a|)\mathrm{e}^{\lambda a}$ and their limits. These weights converge in capacity: discard deep sublevel sets of $h$ and $a$, and use continuity on the remaining finite ranges. We obtain

The exponential is positive off a pluripolar set, which $\mu _ p$ does not charge. Hence $\mu _ p(\{ p<a\} )=0$, and zero domination gives $p=a$. The upper envelopes $(\sup _{k\geq j}a_ k)^*$ decrease to $a$, and squeezing proves convergence in capacity along this subsequence. Since the initial subsequence was arbitrary, the convergence holds for the original sequence. $\square$

Proof. Since $u_ j$ and $v_ j$ decrease to $u$ and $v$, the bounds give $h\leq a\leq \min (u,v)\preceq \varphi $, so $a\in \mathcal E_\varphi $. After discarding finitely many indices we may assume $t_ j\geq 0$, so all potentials $a_ j,a$ are nonpositive. Let $E_ j$ be the intersection of the sets $\{ u>\sigma -t_ j\} $ and $\{ v>\sigma -t_ j\} $, taken only for those inputs that are truncated; a constant sequence imposes no condition. On $E_ j$, locality gives $\mu _{u_ j}=\mu _ u$ and $\mu _{v_ j}=\mu _ v$. Since $a,a_ j\leq u_ j,v_ j$, we have

If $x\notin E_ j$, then one truncated input and its truncation are at most $-t_ j$ at $x$, so $a_ j(x),a(x)\leq -t_ j$. Moreover $|a_ j-a|\mathrm{e}^{(a+a_ j)/4}$ is bounded by an absolute constant, since $a_ j,a\leq 0$. Thus

The assumed uniform mass bound is used only in the last term. The fixed measure $\mu _ u+\mu _ v$ is finite by bounded mass. Lemma 5.44.2, applied to $\mathrm{e}^{a_ j}$ and $\mathrm{e}^ a$, shows that $a_ j\to a$ in measure for $\mu _ u+\mu _ v$, since $a$ is finite almost everywhere for this non-pluripolar measure. The integrand in the first term of (10.31.36) is uniformly bounded and tends to zero in measure, so its integral tends to zero. Lemma 10.31.34 now gives convergence in capacity. When a subsequence is taken, the corresponding truncation levels $t_ j$ are kept.

To identify the equation, fix $T>0$ and choose a continuous nondecreasing function $g_ T:[-\infty ,0]\to [0,1]$ that vanishes on $[-\infty ,-2T]$ and equals one on $[-T,0]$, such that $g_ T\uparrow 1$ at finite points as $T\to \infty $. Put $\chi _ T=g_ T(h)$. Since $h\leq a\leq u,v$, both $u$ and $v$ exceed $-2T$ on $\{ \chi _ T>0\} $. Therefore, once $t_ j>2T$, every truncated input agrees there with its limit, and

Choose a Hermitian form $\Omega \geq \theta $. On the left-hand side, replace $a_ j$ by $\max (a_ j,-2T)$, which does not change $\chi _ T\mu _{a_ j}$ because $\chi _ T$ vanishes where $a_ j\leq -2T$, and expand the bounded product as

Bounded capacity continuity with the fixed quasi-continuous weight $\chi _ T$ therefore gives convergence to $\chi _ T\mu _ a$, after testing against a continuous function. The uncut product may be signed, but its terms have uniformly capacity-controlled total variations.

On the right-hand side, $\chi _ T \mathrm{e}^{-u}$ and $\chi _ T \mathrm{e}^{-v}$ are bounded by $\mathrm{e}^{2T}$. Fixed-measure exponential convergence therefore gives convergence in total variation to $\chi _ T\mathrm{e}^ a(\mathrm{e}^{-u}\mu _ u+\mathrm{e}^{-v}\mu _ v)$. Let $T\to \infty $. The limiting right-hand side is at most $\mu _ u+\mu _ v$, because $a\leq u,v$. Both sides ignore the pole set of $h$, so the cutoff identities prove the global equation and the finiteness of its right-hand side. $\square$

Proof. The proof has four steps: two inputs of minimal singularity type, two inputs of the ordinary type of $\varphi $, one input of that type, and finally an input with a bounded carrier measure.

It suffices to prove the result for $\lambda =1$. Indeed, replace $(\theta ,\varphi ,u,v)$ by $(\lambda \theta ,\lambda \varphi ,\lambda u,\lambda v)$: envelopes and relative classes scale with the potentials, volumes and Monge–Ampère measures scale by $\lambda ^ n$, and a bounded-carrier domination remains one after changing its constant. Dividing the resulting solution by $\lambda $ gives the desired equation. Similarly, a common downward translation of $u$, $v$ and the solution allows us to assume $u,v\leq \varphi \leq 0$; by the envelope identity, these inequalities hold after a sufficiently large common translation. We restore the translation at the end.

Step 1: inputs of minimal singularity type. First take $\varphi =V$ and $u,v\sim V$. Truncating only the weights, define the finite nonzero measures

The functions $\max (u,-j)$ need not be $\theta $-plurisubharmonic; they occur only as weights. Theorem 10.31.22 gives $a_ j\in \mathcal E_ V$ with $\mu _{a_ j}=\mathrm{e}^{a_ j}\tau _ j$.

Choose $D\geq \tfrac 12\| u-v\| _\infty +n\log 2$, where the difference is taken off the common poles, and put $w=(u+v)/2-D$. Positivity of mixed products gives

Exponential order in $\mathcal E_ V$ (corollary 13.19.3) gives $a_ j\geq w$. Since $\tau _ j$ increases with $j$, the same order shows that $a_ j$ decreases. The envelope identity gives $a_1\leq V+\sup _ Xa_1$, so $a_1\sim V$ by the lower bound $w$. Consequently

The decreasing limit $a\geq w$ is not identically $-\infty $ and has the ordinary type of $V$. Fixed-dominator continuity (theorem 10.31.5) gives weak convergence of the left-hand sides, and the right-hand sides converge in total variation by dominated convergence with respect to $\mu _ u+\mu _ v$. Therefore

Comparing this equation with each input measure, exponential order now gives $a\leq u,v$. This comparison uses the limiting equation; the truncated weights alone would not give it.

Step 2: two inputs of the ordinary type of $\varphi $. Suppose $u,v\sim \varphi $, and put $u_ j=\max (u,V-j)$ and $v_ j=\max (v,V-j)$. Step 1 gives solutions $a_ j\sim V$ with

The differences $u_ j-v_ j$ satisfy the same uniform bound as $u-v$. With $D$ as above, $w_ j=(u_ j+v_ j)/2-D$ satisfies the reverse inequality for this equation, and exponential order gives

The fixed lower bound $w=(u+v)/2-D$ has the ordinary type of $\varphi $, and hence lies in $\mathcal E_\varphi $. Extract an $L^1$ limit $a\not\equiv -\infty $; it satisfies $w\leq a\leq \min (u,v)$, so $a\sim \varphi $. The input masses are at most $M_ V$. Lemma 10.31.35, with $\sigma =V$ and lower bound $h=w$, proves convergence in capacity and the equation. The lemma applies although the approximating solutions have the type of $V$ and the limit has the type of $\varphi $.

Step 3: one input of the ordinary type of $\varphi $. Now let $u\in \mathcal E_\varphi $ and $v\sim \varphi $, and set $u_ j=\max (u,\varphi -j)$. Step 2 gives $a_ j\sim \varphi $ with

We construct a common lower bound in $\mathcal E_\varphi $. Put $\delta =n\log 2>0$ and

By stretching and contact, $s_ j\leq a_ j$ and

On $D_ j=\{ s_ j\leq u-2\delta \} $, contact implies $a_ j\leq u_ j-\delta $. Since $\mathrm{e}^{-\delta }=2^{-n}$, the $\mu _{u_ j}$ terms cancel in (10.31.39). Also $v\geq \varphi -C\geq u_ j-C$ for a fixed constant $C$. Hence

All measures being subtracted here are finite.

For $N\geq 1$ form the finite rooftops

They belong to $\mathcal E_\varphi $ by the finite-rooftop proposition. On the set $\{ R_ N\leq u-2\delta \} $, partition the contact set according to the first index $i$ with $R_ N=s_ i$. Since $u_ i\geq u_ N$, (10.31.40) gives

The contact pieces are disjoint, so there is no factor $N$. On $\{ q_ N\leq u-2\delta \} $ only the contact part $q_ N=R_ N$ contributes. Since $u\leq u_ N$ and $t \mathrm{e}^{-t/2}\leq 2/\mathrm{e}$ for $t\geq 0$, we obtain

The first term bounds the contribution of the complementary set, including any contact with $u$. The first-moment criterion proposition 10.31.8, with fixed reference $u$, implies that the decreasing sequence $q_ N$ has a limit $h\in \mathcal E_\varphi $, not identically $-\infty $. For each fixed $j$ and every $N\geq j$, we have $q_ N\leq s_ j\leq a_ j$, so $h\leq a_ j$ for every $j$.

The $a_ j$ now have a subsequence converging in $L^1$ to a limit not identically $-\infty $, bounded above by $u$ and $v$. Lemma 10.31.35, with reference $\varphi $, fixed second input $v$ and common lower bound $h$, proves the required equation in $\mathcal E_\varphi $; here the input masses are at most $M$.

Step 4: a bounded carrier measure for the second input. Finally let $u,v\in \mathcal E_\varphi $ and $\mu _ v\leq A\nu $, with $\nu =\mathrm{MA}_\omega (\psi )$ and $\psi $ bounded. Again put $u_ j=\max (u,\varphi -j)$. By Step 3, applied with the inputs interchanged, there are $a_ j\in \mathcal E_\varphi $ satisfying the same equations and bounds as in Step 3.

Let $r=P_\theta (u,v)\in \mathcal E_\varphi $. On $\{ a_ j<r\} $ one has $r\leq u_ j,v$, so the inequality $t \mathrm{e}^{-t}\leq 1/\mathrm{e}$ gives

Since $a_ j\leq \varphi $, the one-sided first-moment criterion gives a uniform lower bound for $\sup _ Xa_ j$. By lemma 5.44.10,

Applying this to $a_ j-\sup _ Xa_ j$ and using the lower bound for the suprema, we obtain

Keep $s_ j=Q_2(a_ j,u_ j)$, $C_ j$ and $D_ j$ from Step 3. The same cancellation in (10.31.39), now using only $a_ j\leq v$, gives

Take one additional rooftop,

On its contact set with $u-2\delta $ its measure is at most $\mu _ u$; on its other contact part, where $q_ j=s_ j\leq u-2\delta $, the preceding inequality applies. Thus

Moreover the integrand $u-2\delta -q_ j$ vanishes on the first contact part. On the other part, the measure is carried by $C_ j$, where $u_ j-s_ j=2(u_ j-a_ j)$. Therefore

which is uniformly bounded by (10.31.41). The first-moment criterion with reference $u-2\delta $ prevents collapse of the $q_ j$. Extract $L^1$ limits of the $a_ j$ and of the $q_ j$, not identically $-\infty $. The fixed-dominator theorem applies to (10.31.42) and, after a further extraction, gives

Since $q_ j\leq s_ j\leq a_ j$, this is a common lower bound for the extracted solutions, and lemma 10.31.35 gives the equation. Its truncation levels are the original indices of the retained subsequence.

In every step the solution lies below $u$ and $v$, so its measure is bounded by $\mu _ u+\mu _ v$. Two solutions have the same exponentially weighted measure, so corollary 13.19.3, applied in both directions, gives uniqueness, even when that weighted measure has infinite mass. Restoring the common translation and the exponent finishes the proof. $\square$

Proof. We first take $\lambda =1$ and suppose that

Corollary 10.31.28 gives $Q\in \mathcal E_\varphi $ such that

is a finite nonzero measure. Consider

The family contains $Q$. For $w\in \mathcal T$, exponential order (corollary 13.19.3) gives $w\leq Q$. Also

These bounds prevent the collapse of any sequence in $\mathcal T$.

Let

and choose $w_ j\in \mathcal T$ whose weighted masses tend to $\alpha $. Extract an $L^1$ limit $w\not\equiv -\infty $. Since

the fixed-dominator theorem theorem 10.31.5 gives $w\in \mathcal E_\varphi $, convergence in capacity, and $\mu _{w_ j}\rightharpoonup \mu _ w$. Put $\nu _ j=\mathrm{e}^{-w_ j}\mu _{w_ j}$. After a further extraction, these finite measures converge weakly to a measure $\nu $, and

We claim that $\mu _ w=\mathrm{e}^ w\nu $. By lemma 5.44.2,

Let $f$ be continuous. Since $\nu _ j\leq \mu $, replacing $\mathrm{e}^{w_ j}$ by $\mathrm{e}^ w$ in $\int _ Xf \mathrm{e}^{w_ j}\, \mathrm{d}\nu _ j$ changes the integral by an amount tending to zero. For the remaining integral, approximate the fixed bounded measurable function $f \mathrm{e}^ w$ in $L^1(\mu )$ by continuous functions. Since $\nu _ j,\nu \leq \mu $, the approximation error is uniform in $j$, while the integrals of each continuous approximation converge by weak convergence. Thus

This proves the claim. The measure $\nu $ does not charge the poles of $w$, so $\mathrm{e}^{-w}\mu _ w=\nu $. Hence $w\in \mathcal T$ attains the maximum $\alpha $.

Suppose that $\mu -\nu $ were nonzero. Then corollary 10.31.28, applied to $\mu -\nu $, would give $v\in \mathcal E_\varphi $ with

In particular $\mu _ v\leq \mathrm{e}^{\sup _ Xv}\mu \leq A \mathrm{e}^{\sup _ Xv}\sigma $, so the bounded-carrier hypothesis of theorem 10.31.38 holds. That theorem, with $\lambda =1$, gives $a\in \mathcal E_\varphi $ with

The right-hand side lies between $\nu _0$ and $\mu $, so $a\in \mathcal T$, but its mass exceeds $\alpha $. This contradicts maximality. Therefore $\nu =\mu $, and $w$ solves the equation.

Now let $\mu $ be any nonzero finite non-pluripolar measure. The bounded-carrier decomposition lemma 15.24.4, applied to $\omega $, writes

Set $f_ k=\min (f,k)$ and $\mu _ k=f_ k\sigma $. These measures are nonzero, and the first part gives $u_ k\in \mathcal E_\varphi $ with $\mu _{u_ k}=\mathrm{e}^{u_ k}\mu _ k$. Since $\mu _ k$ increases with $k$, exponential order shows that $u_ k$ decreases. The uniform bound

shows that the decreasing limit $u$ is not identically $-\infty $. Moreover $\mu _{u_ k}\leq \mathrm{e}^{\sup _ Xu_1}\mu $, so fixed-dominator continuity gives $u\in \mathcal E_\varphi $ and weak convergence of these measures. On the right-hand side, $\mathrm{e}^{u_ k}f_ k\to \mathrm{e}^ u f$ in $L^1(\sigma )$ by dominated convergence, with dominating function $\mathrm{e}^{\sup _ Xu_1}f$. Thus $\mu _ u=\mathrm{e}^ u\mu $.

For general $\lambda >0$, apply the result for exponent one in the background $\lambda \theta $, with model $\lambda \varphi $ and datum $\lambda ^ n\mu $, and divide the resulting potential by $\lambda $. The scaling of the relative class and of the Monge–Ampère measure gives the equation. Uniqueness follows by applying corollary 13.19.3 to two solutions in both directions. $\square$

Proof. For each integer $j\geq 1$, the preceding theorem gives a solution of

Put $S=\sup _ Xu_1$ and $w_ j=u_1+(j-1)S$. Then $w_ j/j\geq u_1$, and hence

Exponential order at exponent $1/j$ gives $u_ j\leq w_ j$. Define

The upper bound just proved and the positive lower volume $\ell $ imply

The envelope identity gives $v_ j\leq \varphi $. Theorem 5.25.2 gives a subsequence $j_ k$ along which $v_{j_ k}\to u$ in $L^1$, where $u\not\equiv -\infty $ and $\sup _ Xu=0$, and $c_{j_ k}\to c>0$. Fixed-dominator continuity gives $u\in \mathcal E_\varphi $ and $\mu _{v_{j_ k}}\rightharpoonup \mu _ u$.

Lemma 5.44.5, which holds for every smooth background, gives $\mathrm{e}^{v_ j/j}\to 1$ in $L^1(\mu )$. Consequently $c_{j_ k}\mathrm{e}^{v_{j_ k}/j_ k}\mu \to c\mu $ in total variation, and the limiting equation is $\mu _ u=c\mu $. Integrating it and using (10.31.1) gives the stated bounds for $c$. The scalar comparison in corollary 13.19.3 gives uniqueness of $c$. $\square$

Proof. Write $M_0=\overline{\operatorname {vol}}_\theta (V_\theta )$. Every solution above has mass between $\delta $ and $M_0$. In the first case let $s_ j=\sup _ Xu_ j$. Then

which is a uniform lower bound since $\mu _ j(X)\to \mu (X)>0$. To bound $s_ j$ from above, suppose that $s_ j\to +\infty $ along a subsequence. By normalized compactness, after a further extraction, $v_ j=u_ j-s_ j\to v$ in $L^1$ with $\sup _ Xv=0$. The fixed-measure lemma lemma 5.44.2 and $f_ j\to f$ in $L^1(\nu )$ imply

The limit is positive because $f\nu $ is nonzero and does not charge the poles of $v$. But the left-hand side equals $\mathrm{e}^{-\lambda s_ j}\mu _{u_ j}(X)\leq M_0\mathrm{e}^{-\lambda s_ j}$, which tends to zero, a contradiction. Thus $s_ j$ is bounded, and the $u_ j$ form a relatively compact family in $L^1$ whose limits are not identically $-\infty $.

Start with any subsequence and extract $u_ j\to u$ in $L^1$. Extract again so that $\sum _ j\| f_ j-f\| _{L^1(\nu )}<\infty $. The function

dominates all retained $f_ j$. If $C$ is a common upper bound for the $u_ j$, then $\mu _{u_ j}\leq \mathrm{e}^{\lambda C}g\nu $. By theorem 10.31.30, $u\in \mathcal E_\varphi $, the potentials converge in capacity, and the measures converge weakly to $\mu _ u$. On the other hand,

Indeed, the error coming from $f_ j-f$ is at most $\mathrm{e}^{\lambda C}\| f_ j-f\| _{L^1(\nu )}$, and lemma 5.44.2 for the fixed measure $f\nu $ controls the remaining term. Hence $\mu _ u=\mathrm{e}^{\lambda u}\mu $, with convergence in total variation. By the uniqueness in theorem 10.31.44, every cluster limit is this solution, and the subsequence criterion gives all three convergences for the original sequence.

In the second case, integrating the equation gives

Since $\mu _ j(X)\to \mu (X)>0$, these scalars stay in a compact subinterval of $(0,+\infty )$. After extraction, normalized compactness gives $u_ j\to u$ in $L^1$ with $\sup _ Xu=0$, and $c_ j\to c'>0$. Extract a summable density error as above. Then $\mu _{u_ j}\leq Cg\nu $ for a fixed $C$, so theorem 10.31.30 gives $u\in \mathcal E_\varphi $, convergence in capacity and weak convergence to $\mu _ u$. Since $c_ jf_ j\to c'f$ in $L^1(\nu )$, the limiting equation is $\mu _ u=c'\mu $, and the uniqueness of the scalar in theorem 10.31.45 gives $c'=c$. Hence the whole sequence of scalars converges to $c$, and $c_ j\mu _ j\to c\mu $ in total variation. Applying the same argument within any $L^1$-convergent subsequence proves its convergence in capacity and the equation for its limit. $\square$

Proof. By the scaling used in theorem 10.31.38, it suffices to take $\lambda =1$. For positive integers $j$, let

These are nonzero finite non-pluripolar measures, and they increase with $j$. Only the weights are truncated; the functions $\max (u,-j)$ and $\max (v,-j)$ need not be $\theta $-plurisubharmonic. Theorem 10.31.44 gives $a_ j\in \mathcal E_\varphi $ with $\mu _{a_ j}=\mathrm{e}^{a_ j}\tau _ j$. Since $\tau _ j$ increases, exponential order implies that $a_ j$ decreases.

Set $r=P_\theta (u,v)\in \mathcal E_\varphi $. On $\{ a_ j<r\} $, one has $r\leq u,v$ and

with the same inequality for $v$. Consequently

Also $a_ j\leq a_1\leq \varphi +\sup _ Xa_1$. The one-sided first-moment criterion of proposition 10.31.8, with fixed reference $r$, bounds $\sup _ Xa_ j$ from below. Thus the decreasing sequence has an $L^1$ limit $a\not\equiv -\infty $, and the same criterion puts $a$ in $\mathcal E_\varphi $.

We identify the equation before comparing $a$ with the inputs. The rooftop

lies below $a_ j$, $a$, $u$ and $v$.

For $T>0$, let $\chi _ T=g_ T(q)$, with $g_ T$ as in the proof of lemma 10.31.35: a continuous nondecreasing cutoff that vanishes where $q\leq -2T$, equals one where $q\geq -T$, and increases to one off the pole set of $q$ as $T\to \infty $. On $\{ \chi _ T>0\} $ both inputs exceed $-2T$. Hence, for $j>2T$,

The factors $\chi _ T\mathrm{e}^{-u}$ and $\chi _ T\mathrm{e}^{-v}$ are bounded by $\mathrm{e}^{2T}$. Since $a_ j\downarrow a$ and $a_ j\leq \sup _ Xa_1$, dominated convergence against the fixed measure $\mu _ u+\mu _ v$ gives convergence in total variation of the right-hand sides to the corresponding expression with $a$.

For the left-hand side, $a_ j\to a$ in capacity by lemma 5.29.19. Choose a Hermitian form $\Omega \geq \theta $, and replace $a_ j$ by $b_ j=\max (a_ j,-2T)$ before forming bounded products. The $b_ j$ are uniformly bounded $\Omega $-plurisubharmonic functions converging in capacity. The expansion

and bounded capacity continuity apply against the fixed quasi-continuous weight $\chi _ T$. Where $\chi _ T\ne 0$ no potential is truncated, so locality gives convergence to $\chi _ T\mu _ a$. In this way the possibly signed bounded product is handled before the cutoff is inserted.

Thus (10.31.50) holds in the limit. Let $T\to \infty $. Both sides ignore the poles of $q$, so monotone convergence gives

Since $a\in \mathcal E_\varphi $, this also shows that the weighted right-hand side is finite.

For $\varepsilon >0$, the equation implies

Strict domination, theorem 13.19.1, gives $u\geq a-\varepsilon $, and letting $\varepsilon \downarrow 0$ gives $u\geq a$. Similarly $v\geq a$, so $\mu _ a\leq \mu _ u+\mu _ v$. Two solutions have the same exponentially weighted measure, so corollary 13.19.3, applied in both directions, gives uniqueness. $\square$

Proof. Put $F_\lambda (u)=\mathrm{e}^{-\lambda u}\mu _ u$, with zero mass on the poles. The theorem says that $F_\lambda (u\star _\lambda v)=F_\lambda (u)+F_\lambda (v)$. By corollary 13.19.3, the map $F_\lambda $ is injective on $\mathcal E_\varphi $. Commutativity and associativity therefore follow from those of addition of positive measures, including measures of infinite total mass. Finally, $F_\lambda (u+c)=\mathrm{e}^{-\lambda c}F_\lambda (u)$, which together with injectivity gives both identities. $\square$

Proof. For $t\leq 0$, the bounds on $\chi '$ give $Lt\leq \chi (t)\leq t$. Thus

The lower potential belongs to $\mathcal E_\varphi $ by lemma 16.28.63, so $p$ is not identically $-\infty $ and belongs to $\mathcal E_\varphi $ by upward stability. The obstacle is finite and quasi-continuous off the poles, so the contact theorem theorem 15.8.6 applies.

Let $\tau =\chi ^{-1}$. It is convex, with $L^{-1}\leq \tau '\leq 1$. We justify the transformation

without assuming that $\varphi $ is locally bounded on some open set. On a coordinate ball choose a smooth function $q$ with $\theta +\mathrm{dd}^{\mathrm{c}}q\geq 0$. The functions

are locally bounded $\theta $-plurisubharmonic functions on the ball, and $p_ m\leq \varphi _ m$. Lemma 5.28.13 applies to $w_ m=\varphi _ m+\tau (p_ m-\varphi _ m)$ and gives

The function $(s,t)\mapsto t+\tau (s-t)$ on $\{ s\leq t\} $ is nondecreasing in each variable. Consequently $w_ m$ decreases as $m$ increases. Since $p\leq p_ m\leq w_ m\leq \varphi _ m$, the limit is not identically $-\infty $ on the ball. Off the poles it equals $w$. The local limits therefore agree on overlaps after regularization and define a global $\theta $-plurisubharmonic function.

On $D_ m=\{ p>q-m\} $, neither $p$ nor $\varphi $ is truncated. Thus $w_ m=w$ there, and plurifine locality transfers the product inequality to $D_ m$. These sets exhaust the complement of the poles of $p$ in the ball. Since all products ignore the discarded pluripolar set,

The obstacle bound $p\leq \varphi +\chi (u-\varphi )$ gives $w\leq u$, with equality on the contact set of $p$. Ordered contact, the identity $\tau '(p-\varphi )=1/\chi '(u-\varphi )$ on that set, and contact support of $\mu _ p$ prove the assertion. $\square$

Proof. As in proposition 10.31.15, we build a concave weight from the distribution of $\varphi -u$ and apply the preceding lemma. The moment bound on $\mu $ now yields a differential inequality for this weight, which bounds the depth of $\varphi -u$ under $\mu _ u$.

The envelope identity gives $u\leq \varphi \leq 0$. Put $x=\varphi -u$, defined off the poles, and $N=2n+1$. Since $x\leq -u$ and $\mu _ u\leq \mu $, Hölder's inequality gives the uniform first-moment bound

Let $H(t)=\mu _ u(\{ x>t\} )$ and $T=\sup \{ t\geq 0:H(t)>0\} $, with $T=0$ if this set is empty; at this point $T$ may be infinite. If $T<\infty $, zero domination applied to $u$ and $\varphi -t$ for $t>T$ gives $u\geq \varphi -T$. It is therefore enough to bound $T$. We may assume $T>1$, and we choose any $S$ with $1<S<T$.

Define, for $t\geq 0$,

and set $W(t)=\int _0^ t F(s)^{1/N}\, \mathrm{d}s$ and $\chi (-t)=-W(t)$. Since $H(S)>0$, $F$ is bounded on the whole half-line, although its bound may depend on $S$ and $u$. It is locally Lipschitz and nondecreasing. Hence $\chi $ has bounded derivative, is increasing and concave, and satisfies the hypotheses of the preceding lemma. Also $W(t)\geq t$ and $W(t)\leq tF(t)^{1/N}$.

Since $H(t)\leq M$, Tonelli's theorem gives a bound independent of $S$ and $u$:

Let $p=P_\theta (\varphi -W(x))\in \mathcal E_\varphi $, the envelope of the preceding lemma for this $\chi $. On its contact set the lemma bounds its measure by $F(x)^{n/N}\mu _ u$. Hence, using $N=2n+1$ and the Cauchy–Schwarz inequality,

The identity $P_\theta [p]=\varphi $ implies $p-\sup _ Xp\leq \varphi $. Since $p\leq \varphi \leq 0$ and $\mu _ p(X)\geq \ell $, the last integral is at least $\ell \sqrt{-\sup _ Xp}$. Consequently

Using $W(x)\leq \varphi -p\leq -p$, and applying the moment hypothesis to $p-\sup _ Xp$, we obtain

where we choose $C_0=1+2^{m-1}(A_ m+D^ m\mu (X))$. This constant is independent of $S$ and $u$.

For $0<t\leq S$, Chebyshev's inequality and $\mu _ u\leq \mu $ give

Since $W'=F^{1/N}$, the derivative of $(W')^{N+1}$ is at least

All functions being differentiated are absolutely continuous on compact intervals. Integrating from $0$ to $t$ and using $W(0)=0$ gives

In particular

with $c_0>0$ independent of $S$ and $u$. Integrating the derivative of $W^{1-\kappa }$ over $[1,S]$ yields

because $W(1)\geq 1$. Since the integral on the left diverges as $S\to \infty $, every $S<T$ is bounded by a constant independent of $S$ and $u$, and hence so is $T$. The zero-domination argument at the beginning of the proof now gives $u\geq \varphi -C$. The case $T\leq 1$, which includes $B=0$, was covered there as well. $\square$

Proof. Lemma 5.26.2, applied in a Hermitian background dominating $\theta $, gives uniform moments of every finite order against $\mathrm{d}V$ for normalized $\theta $-plurisubharmonic functions. Hölder's inequality therefore implies, for every finite $m$,

Choose $m>2n+1$. For the solution of the exponential equation, set $S=\sup _ Xu$ and $v=u-S$. Then $v\in \mathcal E_\varphi $, $\sup _ Xv=0$, and

The preceding proposition, applied to this fixed dominating measure, shows that $v$, and hence $u$, has the ordinary type of $\varphi $. For the normalized equation, apply the proposition to the measure $cf\, \mathrm{d}V$. $\square$