PomeProposition 16.5.54.75FA2659491D Let $\varphi ,\psi \in \mathcal{E}(X,\theta ;\phi )$. Then
\[ P_{\theta }[\varphi ](\psi )=\operatorname *{sup}_{C > 0}{}\! {}^{*} (\varphi +C)\land \psi =\psi . \]
Proposition 16.5.5475FA2659491D
Proof. The first equality follows from the definition of $P_{\theta }[\varphi ](\psi )$. We focus on the second equality.
Since for each $C\geq 0$,
\[ (\varphi \land \psi +C)\land \psi \leq (\varphi +C)\land \psi \leq \psi , \]
we may replace $\varphi $ by $\varphi \land \psi $ (cf. proposition 16.5.49) and assume that $\varphi \leq \psi $.
Let
\[ \gamma \coloneqq \operatorname *{sup}_{C > 0}{}\! {}^{*} (\varphi +C)\land \psi . \]
Observe that $\gamma \leq \psi $, and $\gamma \in \mathcal{E}(X,\theta ;\phi )$ due to proposition 16.5.49 and proposition 16.5.53. Then corollary 16.3.14 guarantees that
\[ \int _{\{ \gamma <\psi \} } \theta _{\gamma }^ n=0. \]
Therefore, we can apply the domination principle theorem 13.13.43 to conclude that $\gamma =\psi $. $\square$
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.