Theorem 21.5.1409656CC47D4

Proof. The set $\operatorname {dom}G$ is convex, and its difference from its interior is a null set: if the interior is nonempty, this difference lies in the relative boundary, which is null by lemma 21.2.3(4); otherwise $\operatorname {dom}G$ lies in a proper affine subspace. By proposition 21.2.4(4), applied to the finite convex function $G$ on the interior of its domain, $D$ is a Borel set, the gradient $\nabla G\colon D\to N_{\mathbb R}$ is a Borel map, and the complement of $D$ in the interior of $\operatorname {dom}G$ is a null set. Therefore the complement of $D$ in $\operatorname {dom}G$ is a null set; it is a Borel set, because $\operatorname {dom}G=\bigcup _ k\{ G\leq k\} $ is a countable union of closed sets, $G$ being lower semicontinuous.

Let $p\in D$. Then $\partial G(p)=\{ \nabla G(p)\} $. Indeed, $\partial G(p)$ is nonempty by proposition 21.2.4, and every subgradient of $G$ at $p$ is also a subgradient at $p$ of the restriction of $G$ to the interior of $\operatorname {dom}G$, whose only subgradient at $p$ is $\nabla G(p)$ by proposition 21.2.4(3). Hence theorem 21.3.2 gives, for every $E\subset N_{\mathbb R}$,

Every subgradient of $F$ lies in $\operatorname {dom}G$, so $\partial F(E)$ differs from the Borel set $(\nabla G)^{-1}(E)\cap D$ by a subset of the fixed null set $\operatorname {dom}G\setminus D$. Hence $\partial F(E)$ is measurable for the completed measure, and $\operatorname {vol}(\partial F(E))=\operatorname {vol}((\nabla G)^{-1}(E)\cap D)$. This proves the pushforward formula, which in turn shows that $E\mapsto n!\operatorname {vol}(\partial F(E))$ is countably additive on Borel sets.

If $K$ is compact, $\partial F(K)$ is bounded by proposition 21.2.4(1), and it is closed, since the graph of $\partial F$ is closed and $K$ is compact; so $\partial F(K)$ is compact and has finite volume. Thus the measure is finite on compact sets, and it is a Radon measure by proposition 2.2.2. If $\operatorname {dom}G$ has empty interior and $n>0$, then $D=\varnothing $ and the measure vanishes. In general, the total mass is $n!\operatorname {vol}(D)=n!\operatorname {vol}(\operatorname {dom}G)$, and $\operatorname {vol}(\operatorname {dom}G)=\operatorname {vol}(K_ F)$ because $K_ F\setminus \operatorname {dom}G$ lies in the relative boundary of $\operatorname {dom}G$, or in a proper affine subspace. In dimension zero, $F$ is a constant, $D$ is the single point of $M_{\mathbb R}$, and the formula gives unit mass, with $0!=1$. $\square$