Proposition 11.33.22770AB03C0D2

Proof. Suppose that $F\leq G$ almost everywhere and, to reach a contradiction, that $c_\varphi (F)<c_\varphi (G)$. Let $u,v\sim \varphi $ solve the two equations. Since $F\leq G$ and the extended measure is unchanged by adding constants, for every constant $a$,

and the ratio is strictly less than one. As $u\sim \varphi $, the envelope $P_{\varepsilon \theta }(u)$ dominates a translate of $P_{\varepsilon \theta }(\varphi )$ and is therefore not identically $-\infty $; moreover $v+a\preceq u$. The singular strict domination principle, theorem 15.2.2, gives $u\geq v+a$ for every $a$, which is impossible at any point where both potentials are finite. Thus $B_\varphi (F)\geq B_\varphi (G)$.

Replacing $F$ by $F+a$ and the constant by $\mathrm{e}^{-a}c_\varphi (F)$ leaves the right-hand side of (11.33.1) unchanged, so uniqueness of the constant proves the translation rule. Finally, $G$ lies between the constant translates $F+\mathop{\rm ess\, inf}_ X(G-F)$ and $F+\mathop{\rm ess\, sup}_ X(G-F)$; monotonicity and the translation rule give (11.33.3). $\square$