Pome

Lemma 21.11.578DE662F4EFE6

Lemma 21.11.57.8DE662F4EFE6 Let $F\in \mathcal{E}(N_{\mathbb {R}},P)$ and set $G=F^*$. Then the map $\nabla G$ pushes forward $n!\, \mathrm{d}\operatorname{vol}|_{P}$ to $\operatorname{MA}_{\mathbb R}(F)$. Equivalently, for every bounded Borel function $h$ on $N_{\mathbb R}$,

\[ \int _{N_{\mathbb R}} h\, \operatorname{MA}_{\mathbb R}(F)= n!\int _ P h(\nabla G(m))\, \mathrm{d}\operatorname{vol}(m). \]

Proof. If $\operatorname{vol}P=0$, the assertion is clear from proposition 21.11.38 and the definition of $\operatorname{MA}_{\mathbb R}$. We may therefore assume that $P$ has non-empty interior.

Since $F\in \mathcal P(N_{\mathbb R},P)$, we have $\operatorname{Dom}G\subseteq P$ by proposition 21.11.38. On the other hand, by the definition of $\mathcal E(N_{\mathbb R},P)$ and proposition 21.11.34,

\[ P=\overline{\nabla F(N_{\mathbb R})}\subseteq \overline{\operatorname{Dom}G}. \]

Hence $\overline{\operatorname{Dom}G}=P$. Since $\operatorname{Dom}G$ is convex, $G$ is finite on $\operatorname{Int}P$ and is equal to $\infty $ outside $P$. Let $D$ be the set of points in $\operatorname{Int}P$ at which $G$ is differentiable. By [Theorem 25.5, Roc70], $P\setminus D$ has measure zero. By [Theorem 23.5, Roc70], $m\in \nabla F(n)$ if and only if $n\in \nabla G(m)$. Hence for any Borel set $E\subseteq N_{\mathbb R}$,

\[ \{ m\in D:\nabla G(m)\in E\} =D\cap \nabla F(E). \]

It follows from definition 21.11.50 that

\[ n!\operatorname{vol}\{ m\in P:\nabla G(m)\in E\} =n!\operatorname{vol}\nabla F(E)=\operatorname{MA}_{\mathbb R}(F)(E). \]

Our assertions follow. $\square$

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