Lemma 13.20.1EDD01F5B65E5

Proof. Put $K=v_+(\Omega )$. Since $0<\underline{\operatorname {vol}}_\theta (u)\leq K$, we have $1<b_0\leq 2$. After a common translation we may assume $u\leq r\leq 0$. Suppose, to reach a contradiction, that $\mathcal A_ b(u,r)=\varnothing $ for some $1<b\leq b_0$. Set

Here, as for stretching envelopes, a competitor $g$ for $q_ j$ is required to satisfy $g+(b-1)r\leq v_ j$, which is meaningful also at common poles. Off the poles the obstacle $v_ j-(b-1)r=u+(b-1)(u_ j-r)$ lies between $u-(b-1)j$ and $u$. Hence $q_ j$ is $\theta $-plurisubharmonic and $q_ j\sim u$. As $j$ increases, $q_ j$ decreases. A $\theta $-plurisubharmonic decreasing limit would belong to $\mathcal A_ b(u,r)$, so the limit is identically $-\infty $, and $\sup _ Xq_ j\to -\infty $ by theorem 5.25.2.

Let $C_ j=\{ q_ j+(b-1)r=v_ j\} $. By the contact theorem, $\mu _{q_ j}$ is carried by $C_ j$. The current $b\theta +\mathrm{dd}^{\mathrm{c}}(q_ j+(b-1)r)$ equals $(\theta +\mathrm{dd}^{\mathrm{c}}q_ j)+(b-1)(\theta +\mathrm{dd}^{\mathrm{c}}r)\geq \theta +\mathrm{dd}^{\mathrm{c}}q_ j$. Positivity of the mixed products and ordered contact in background $b\theta $ therefore give

On $C_ j$ we have $q_ j=u+(b-1)(u_ j-r)\geq bu$, because $u_ j\geq u$ and $r\leq 0$. Fix $k>0$. For all sufficiently large $j$, $q_ j\leq -bk$ everywhere, so $C_ j\subset \{ u\leq -k\} $, and consequently

To estimate the right-hand side, note that for any $s,t\in \mathrm{PSH}(X,\theta )$ and $0\leq i\leq n$, positivity of the non-pluripolar mixed products and proposition 13.14.16 imply

The second inequality follows by expanding $(\Omega +\mathrm{dd}^{\mathrm{c}}\frac{s+t}2)^ n$; for the possibly unbounded average, apply this expansion to its truncations at constant levels, which remain $\Omega $-plurisubharmonic. Expanding the integrand in (13.20.2), and using $(b-1)^ i\leq b-1$ for $i\geq 1$, gives

The measure $\mu _ u$ is finite and puts no mass on the pole set of $u$, so the first term tends to zero as $k\to \infty $. By the choice of $b_0$, the remaining error is at most half of $\underline{\operatorname {vol}}_\theta (u)>0$, a contradiction. Thus $\mathcal A_ b(u,r)\neq \varnothing $ for $1<b\leq b_0$; for $b=1$ the family contains $u$, and $Q_1(u,r)=u$.

Put $h_0=Q_{b_0}(u,r)$ and, for $1<b\leq b_0$, $t=(b-1)/(b_0-1)\in (0,1]$. The inequality $h_0\leq b_0u-(b_0-1)r$ shows that $(1-t)u+th_0$ is a competitor for $Q_ b(u,r)$, and $Q_ b(u,r)\leq bu-(b-1)r\leq u$. Hence

Addition, homogeneity and volume monotonicity give

Letting $b\downarrow 1$ proves the limit. Undoing the common translation changes neither the volumes nor the nonemptiness of the competitor families. $\square$