Theorem 13.18.5.D73DFBAE1EC1 Suppose $\underline{\operatorname {vol}}_\theta (\varphi )>0$. Then
Every member $u$ of this class also satisfies $\underline{\operatorname {vol}}_\theta (u) =\underline{\operatorname {vol}}_\theta (\varphi )$.
D73DFBAE1EC1Theorem 13.18.5.D73DFBAE1EC1 Suppose $\underline{\operatorname {vol}}_\theta (\varphi )>0$. Then
Every member $u$ of this class also satisfies $\underline{\operatorname {vol}}_\theta (u) =\underline{\operatorname {vol}}_\theta (\varphi )$.
Proof. First let $u\in \mathcal E_{\mathrm{env}}(X,\theta ,\varphi )$. For each $a\in (0,1)$, a member of $\mathcal A_ b(u,\varphi )$ with $b=(1-a)^{-1}$ gives $h\in \mathrm{PSH}(X,\theta )$ such that $a\varphi +(1-a)h\leq u$. By theorem 13.17.1, proposition 13.17.2 and homogeneity, for either volume,
Letting $a\to 1$ proves that both volumes of $u$ equal those of $\varphi $; this direction does not use positive lower volume.
Conversely, let $u\in \mathcal E_{\mathrm{vol}}(X,\theta ,\varphi )$. Then $D=0$ in lemma 13.18.1 with $v=\varphi $, so $\mathcal A_ b(u,\varphi )\neq \varnothing $ for every $b>1$. For $b=1$ the family contains $u$. This is the defining condition of $\mathcal E_{\mathrm{env}}(X,\theta ,\varphi )$. $\square$
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