Proposition 16.25.34338492503DC

Proof. Write $p=P_\theta (f)$ and choose $C$ with $|f|\leq C$. Then $V_\theta -C$ is a competitor for $f$, and every competitor $v$ satisfies $v-C\leq 0$, hence $v\leq V_\theta +C$; thus $V_\theta -C\leq p\leq V_\theta +C$. Consequently $p$ has minimal singularities, and its mass is that of $V_\theta $, namely $\operatorname {vol}(\{ \theta \} )>0$, by definition 13.9.1, proposition 13.9.2. A potential lies below $f$ quasi-everywhere if and only if it lies below $p$: one direction is the maximality of $p$, the other holds because $p\leq f$ quasi-everywhere. Hence the two constrained families coincide, and $w=P_\theta [u](p)$. Since $u\preceq p$, this is the relative envelope of definition 16.22.13, by proposition 16.25.2. The measure bound in proposition 16.22.14, which applies because $p$ has positive mass, gives the first inequality. By theorem 16.13.4, $\mu _ p$ is carried by $\{ p=f\} $, so $\mu _ w$ is carried by $\{ w=p\} \cap \{ p=f\} \subset \{ w=f\} $. Finally, $f-w$ is nonnegative quasi-everywhere and vanishes $\mu _ w$-almost everywhere, which gives the integral identity. $\square$