Proposition 13.23.21542F41C8A51

Proof. Put $\delta =\underline{\operatorname {vol}}_\theta (u)$, $m_ j=\max (u_ j,u)$ and $s_ j=d_\theta (u_ j,u)$. After discarding finitely many indices, choose $b_ j\to \infty $ with $b_ j>1$ and $b_ j^ ns_ j<\delta $. The upper-volume deficits of both $u_ j$ and $u$ from $m_ j$ are at most $s_ j$, while $m_ j$ has lower volume at least $\delta $. Thus lemma 13.18.1 shows that the stretching envelopes

are defined. Choose $a\in (0,1)$ with $2nMa<\delta $, and set

These potentials lie below $m_ j$, because $G_ j\leq u_ j$ and $H_ j\leq u$. Addition and monotonicity give

Lemma 13.23.1, with $w=m_ j$, therefore shows that $r_ j=P_\theta (g_ j,h_ j)$ is $\theta $-plurisubharmonic. For large $j$, put $t_ j=(ab_ j)^{-1}<1$. Since $r_ j\leq g_ j$ and $r_ j\leq h_ j$, the defining inequalities for $G_ j$ and $H_ j$ show that

The left-hand side is therefore a competitor for $p_ j$, so $p_ j\not\equiv -\infty $. For either volume, addition and monotonicity give

Since $t_ j\to 0$, this proves $\underline{\operatorname {vol}}_\theta (p_ j)\to \underline{\operatorname {vol}}_\theta (u)$; and since $p_ j\leq u$, (13.22.2) gives $d_\theta (p_ j,u)=U(u)-U(p_ j)\to 0$. $\square$