Proposition 17.9.10583958ACA6D

Proof. The function $v-u$ is finite almost everywhere, and $v\preceq u$ gives $v-u\leq C'$ for some constant $C'$; hence $C$ is finite. The inequality $v\leq u+C$ holds almost everywhere, hence everywhere, as recalled in section 16.11. Thus the supremum of $v-u$ over the common finite locus is at most $C$; it is at least $C$ because the common finite locus has full measure. The same argument applies to an infimum, and to the complement of any pluripolar set, which still has full measure. The same remarks apply to $G_ t-u$.

By theorem 17.7.8, $G_ t\leq (1-t)u+tv\leq u+tC$. The affine curve $v-C(1-t)$ is a competitor, since $v-C\leq u$, so $G_ t\geq v-(1-t)C$. On the common finite locus $v-u=(v-G_ t)+(G_ t-u)$, and taking essential suprema gives $C\leq (1-t)C+\operatorname *{ess\, sup}_ X(G_ t-u)$. Together with the first bound, this proves the equality for the supremum.

Now let $u\sim v$. The affine curve $u+ct$ is a competitor, since $u+c\leq v$, so $G_ t-u\geq ct$. Applying the equality for the supremum to the reversed segment, $G_ t(u,v)=G_{1-t}(v,u)$, gives $\operatorname *{ess\, sup}_ X(G_ t-v)=(1-t)\operatorname *{ess\, sup}_ X(u-v)=-(1-t)c$, that is, $\operatorname *{ess\, inf}_ X(v-G_ t)=(1-t)c$. Taking essential infima in $v-u=(v-G_ t)+(G_ t-u)$ gives $c\geq (1-t)c+\operatorname *{ess\, inf}_ X(G_ t-u)$, which proves the opposite bound. $\square$