Proposition 21.11.53F30A95FDA78E

Proof. Since $F\in \mathcal{P}(N_{\mathbb {R}},P)$, the function $F$ is closed and has domain $N_{\mathbb {R}}$. By proposition 21.11.34, $\overline{\nabla F(N_{\mathbb {R}})}=\overline{\operatorname{Dom}F^*}$, so this closure is convex. As $\nabla F(N_{\mathbb {R}})$ contains the relative interior of $\operatorname{Dom}F^*$, the definition of $\operatorname{MA}_{\mathbb {R}}$ shows that (21.11.54) is equivalent to

We first handle the case where $\operatorname{Int}P\neq \varnothing $. By proposition 21.11.38, $\overline{\nabla F(N_{\mathbb {R}})}\subseteq P$. Hence the latter is equivalent to

Indeed, a proper closed subset of the full-dimensional convex body $P$ has complement containing a non-empty relatively open subset of $P$, hence has strictly smaller volume.

Now assume that $\operatorname{Int}P=\varnothing $, then $\operatorname{vol}\overline{\nabla F(N_{\mathbb {R}})}=\operatorname{vol}P=0$ by proposition 21.11.38. The assertion is clear. $\square$