Theorem 17.5.17777B716B7208

Proof. Put $\mu =\mu _ u=\mu _ v$. Since $u\in \mathcal E(X,\theta ,\phi )$, this is a finite measure of mass $\int _ X\langle \theta _ u^ n\rangle =\int _ X\langle \theta _\phi ^ n\rangle =m$, and it is non-pluripolar, because non-pluripolar products give no mass to pluripolar sets (lemma 13.2.2). Hence the pole sets of $u$ and $v$ are $\mu $-null, and $u-v$ is defined and finite $\mu $-almost everywhere. We first show that $u-v$ is constant $\mu $-almost everywhere.

Suppose not. Let $\lambda $ be the image of $\mu $ under $u-v$, a finite Borel measure on $\mathbb R$ of total mass $m>0$, and put

with $\sup \varnothing =-\infty $ and $\inf \varnothing =+\infty $. The sets $(-\infty ,c)$ increase to $\mathbb R$ as $c\to +\infty $, so $\lambda ((-\infty ,c))\to m>0$; hence $c_-<+\infty $, and similarly $c_+>-\infty $. If $c_->-\infty $, the $\lambda $-null sets $(-\infty ,c)$ with $c<c_-$ increase to $(-\infty ,c_-)$, so $\lambda ((-\infty ,c_-))=0$; similarly $\lambda ((c_+,\infty ))=0$ if $c_+<+\infty $. Hence $c_+<c_-$ is impossible, since $\mathbb R=(-\infty ,c_-)\cup (c_+,\infty )$ would then be $\lambda $-null. If $c_-=c_+$, then $\lambda $ is carried by the point $c_-$, that is, $u-v$ is constant $\mu $-almost everywhere, contrary to our assumption. Thus $c_-<c_+$, and every $c\in (c_-,c_+)$ satisfies $\lambda ((-\infty ,c))>0$ and $\lambda ((c,\infty ))>0$, that is,

The singletons $\{ c\} $ with $\lambda (\{ c\} )>0$ are pairwise disjoint sets of positive measure, so there are only countably many of them by lemma 2.2.5. Since the interval $(c_-,c_+)$ is uncountable, we may choose $c$ in it with $\mu (\{ u-v=c\} )=0$. Replacing $v$ by $v+c$ changes neither the class $\mathcal E(X,\theta ,\phi )$ nor the measure $\mu _ v$, so we may assume $c=0$. Then $\mu (\{ u<v\} )$ and $\mu (\{ u>v\} )$ are positive, and the set $U=\{ u<v\} $ satisfies

Set $q=(m/\mu (U))^{1/n}>1$. The measure $q^ n\mathbf1_ U\mu $ is non-pluripolar and has mass $m$, so theorem 17.5.15 gives $h\in \mathcal E(X,\theta ,\phi )$ with

For $0<t<1$ put $w_ t=(1-t)v+th$. By corollary 17.3.17, $w_ t\in \mathcal E(X,\theta ,\phi )$. Since $\theta _{w_ t}=(1-t)\theta _ v+t\theta _ h$, corollary 13.5.11, applied to the currents $\theta _ v,\theta _ h$, the common measure $\mu $ and the densities $1$ and $q^ n\mathbf1_ U$, gives

Let $U_ t=\{ u<w_ t\} $, and put

The comparison principle for relative full mass, proposition 16.25.4, applied to $u$ and $w_ t$, and the preceding lower bound give

Since $1-d_ t>0$, it follows that

We let $t\downarrow 0$. At every point where $u,v,h$ are finite and $u\neq v$, we have $w_ t=v+t(h-v)$, so $\mathbf1_{U_ t}\to \mathbf1_ U$ there. These points have full $\mu $-measure, because the three pole sets and the level set $\{ u=v\} $ are $\mu $-null. By dominated convergence,

Moreover $b_ t-d_ t=ntq+O(t^2)$ and $1-d_ t=nt+O(t^2)$, so $(b_ t-d_ t)/(1-d_ t)\to q$. Passing to the limit in (17.5.18) gives $q\mu (U)\leq \mu (U)$, which contradicts $q>1$ and $\mu (U)>0$.

Hence, for the original pair, there is a finite constant $C$ such that $u=v+C$ $\mu $-almost everywhere. Both $u$ and $v+C$ belong to $\mathcal E(X,\theta ,\phi )$, and their common measure $\mu $ gives zero mass to $\{ u<v+C\} $ and to $\{ v+C<u\} $. The domination principle (theorem 16.22.10), applied in both directions, gives $u=v+C$ everywhere, including at the common poles. This proves the first assertion.

For the second, theorem 17.5.15 provides a solution with supremum zero. If $u$ and $v$ are two such solutions, then $u=v+C$, and comparing suprema gives $C=0$. $\square$