Theorem 21.5.2F4072D9ED060

Proof. Write $\mu _ j=\operatorname {MA}_{\mathbb R}(F_ j)$ and $\mu =\operatorname {MA}_{\mathbb R}(F)$. Steps 1 and 2 compare the masses of compact and of open sets, and Step 3 deduces vague convergence.

Step 1: compact sets. Let $K$ be compact. If $x\in K$ and $0\neq p\in \partial F_ j(x)$ (in particular $n\geq 1$), choose $e\in N_{\mathbb R}$ with $|e|=1$ and $\langle p,e\rangle =|p|$; the subgradient inequality at $x$, evaluated at $x+e$, gives $|p|\leq F_ j(x+e)-F_ j(x)\leq 2\sup \{ |F_ j(y)|:\operatorname {dist}(y,K)\leq 1\} $. By local uniform convergence, the sets $\partial F_ j(K)$ are therefore contained in a fixed ball for all large $j$. If $x_ j\in K$, $p_ j\in \partial F_ j(x_ j)$, $x_ j\to x$ and $p_ j\to p$, then letting $j\to \infty $ in the subgradient inequalities gives $p\in \partial F(x)$. The set $\partial F(K)$ is compact, being bounded and closed by proposition 21.2.4(1), since $K$ is compact. Consequently, for every $\varepsilon >0$, the set $\partial F_ j(K)$ lies in the closed $\varepsilon $-neighborhood $(\partial F(K))^\varepsilon $ of $\partial F(K)$ for all large $j$. Otherwise, for infinitely many $j$ there would be points $x_ j\in K$ and $p_ j\in \partial F_ j(x_ j)$ at distance at least $\varepsilon $ from $\partial F(K)$; as $K$ is compact and the $p_ j$ are bounded, a subsequence of $(x_ j,p_ j)$ would converge to some $(x,p)$ with $p\in \partial F(x)$, which is impossible. As $\varepsilon \downarrow 0$, these neighborhoods decrease to $\partial F(K)$, and continuity from above of Lebesgue measure on sets of finite measure gives

Step 2: open sets. Let $U$ be open and let $G=F^*$, with $D$ as in theorem 21.5.1. For $p\in \partial F(U)\cap D$, the point $x=\nabla G(p)$ lies in $U$ and is the only point with $p\in \partial F(x)$, so $x$ is the unique minimum point of $F-\langle p,\cdot \rangle $. Choose $r>0$ with $\overline{B(x,r)}\subset U$. Since the sphere $\{ y:|y-x|=r\} $ is compact, $F-\langle p,\cdot \rangle $ exceeds its minimum by some $\delta >0$ on it. For large $j$ we have $|F_ j-F|<\delta /3$ on $\overline{B(x,r)}$, so the minimum of $F_ j-\langle p,\cdot \rangle $ on $\overline{B(x,r)}$ is attained at a point $x_ j\in B(x,r)$. A local minimum of a convex function is a global minimum, because the function is convex along each segment starting at that point; hence $p\in \partial F_ j(x_ j)\subset \partial F_ j(U)$ for all large $j$. Since $\partial F(U)\setminus D$ is a null set, Fatou's lemma, applied to the indicator functions of the sets $\partial F_ j(U)$, gives

Step 3: vague convergence. Let $f\geq 0$ be continuous with compact support $S$. By Step 1 there is $C$ with $\mu _ j(S)\leq C$ for all large $j$. For $t>0$, the set $\{ f\geq t\} $ is compact and $\{ f>t\} $ is open. By Tonelli's theorem, $\int f\, \mathrm d\nu =\int _0^\infty \nu (\{ f>t\} )\, \mathrm dt=\int _0^\infty \nu (\{ f\geq t\} )\, \mathrm dt$ for every positive measure $\nu $. Hence Step 2 and Fatou's lemma give

and Step 1 and Fatou's lemma, applied to the nonnegative functions $t\mapsto C-\mu _ j(\{ f\geq t\} )$ on $(0,\max f]$, give

Hence $\int f\, \mathrm d\mu _ j\to \int f\, \mathrm d\mu $. A general continuous function with compact support is the difference of its positive and negative parts, which proves vague convergence.

Full mass. Let $F\in \mathcal P(N_{\mathbb R},P)$. Then $K_ F\subset P$ by proposition 21.4.2, and the total mass of $\operatorname {MA}_{\mathbb R}(F)$ is $n!\operatorname {vol}(K_ F)$ by theorem 21.5.1. Suppose that $P$ has nonempty interior. If $K_ F=P$, the two masses agree. If $K_ F\neq P$, choose $p\in P\setminus K_ F$ and a ball $B(p,r)$ disjoint from the closed set $K_ F$. Since $P$ is the closure of its interior by lemma 21.2.3(4), there is an interior point $q$ of $P$ in $B(p,r/2)$, and a small ball about $q$ lies in $P\setminus K_ F$. Hence $\operatorname {vol}(K_ F)<\operatorname {vol}(P)$. This proves the equivalence when $P$ has nonempty interior. If $P$ has empty interior, then $n>0$, both $\operatorname {vol}(K_ F)$ and $\operatorname {vol}(P)$ vanish, and $\mathcal E(N_{\mathbb R},P)=\mathcal P(N_{\mathbb R},P)$ by definition, so the equivalence holds as well.

Finally, if $F\leq H+C$, then $F^*\geq H^*-C$, so $\operatorname {dom}F^*\subset \operatorname {dom}H^*$ and $K_ F\subset K_ H$. The inequality between the total masses follows from theorem 21.5.1. $\square$