21.5 The real Monge–Ampère measure07CCE1F902ED

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 $0!=1$. $\square$

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. $\square$

Proof. Smooth functions. Let $F$ be smooth with positive definite Hessian. For $x\neq y$,

so $\nabla F$ is injective, and it is a local diffeomorphism by the inverse function theorem. Since $\partial F(E)=\nabla F(E)$ by proposition 21.2.4(3), the change of variables formula gives the asserted identity. Now let $F$ be smooth and convex; its Hessian is positive semidefinite, because the second derivatives of $F$ along lines are nonnegative. Apply the identity just proved to $F+\varepsilon |x|^2/2$, where $|x|$ is the Euclidean norm in the lattice coordinates. As $\varepsilon \downarrow 0$, these functions converge to $F$ locally uniformly and their densities $n!\det (D^2F+\varepsilon I)$ converge to $n!\det (D^2F)$ locally uniformly; comparing the two vague limits, by theorem 21.5.2, proves the identity for $F$.

Functions of class $C^{1,1}_{\mathrm{loc}}$. We work in the coordinates of a basis of $N$. Fix $k$ and $l$, and let $g=\partial _ kF$, which is locally Lipschitz. Let $B=B(x_0,r)$ be a ball, and let $L_ B$ be a Lipschitz constant of $g$ on $\overline{B(x_0,2r)}$. For a nonnegative test function $\chi $ supported in $B$ and $0<|h|<r$, the substitution $x\mapsto x-he_ l$ in one of the two terms gives

and the right side is at most $L_ B\int \chi \, \mathrm dx$ in absolute value. As $h\to 0$, the difference quotients of $\chi $ converge uniformly to $\partial _ l\chi $, with supports in a fixed compact set, so the left side tends to $\int g\, \partial _ l\chi \, \mathrm dx=-\langle \partial _ lg,\chi \rangle $. Hence $|\langle \partial _ l\partial _ kF,\chi \rangle |\leq L_ B\int \chi \, \mathrm dx$, and corollary 2.8.2(2), applied to Lebesgue measure, gives a Borel function $H_{kl}$ that represents the distribution $\partial _ l\partial _ kF$ and satisfies $|H_{kl}|\leq L_ B$ almost everywhere on each such ball $B$. Covering $N_{\mathbb R}$ by countably many such balls and setting $H_{kl}=0$ on the null set of points of some ball of the cover at which the bound fails, we may assume that $H_{kl}$ is bounded on every compact set. This proves that $D^2F=(H_{kl})$ is a locally bounded Borel matrix-valued function. Since distributional derivatives commute, $H_{kl}=H_{lk}$ almost everywhere, by the uniqueness in corollary 2.8.2.

Next, $D^2F(x)$ is positive semidefinite for almost every $x$. Indeed, let $v\in \mathbb R^ n$ and let $\chi $ be a nonnegative test function. Taylor's formula shows that the second difference quotients $h^{-2}(\chi (x+hv)+\chi (x-hv)-2\chi (x))$ converge uniformly to $\sum _{k,l}v_ kv_ l\partial _ k\partial _ l\chi $ as $h\to 0$, so the substitutions $x\mapsto x\mp hv$ give

because $F$ is convex. Thus the function $\sum _{k,l}v_ kv_ lH_{kl}$ defines a positive distribution, so its mollifications are nonnegative by proposition 2.5.3(3); since they converge to it at its Lebesgue points (theorem 2.8.3), it is nonnegative almost everywhere. Applying this to a countable dense set of vectors $v$, and using the continuity of the quadratic form in $v$, we see that $D^2F(x)$ is positive semidefinite for almost every $x$.

We now obtain the formula by mollification. With the mollifiers $\rho _\varepsilon $ of lemma 21.4.3, in the same coordinates, the functions $F_\varepsilon =F*\rho _\varepsilon $ are smooth and convex, and they converge to $F$ locally uniformly, by that lemma. By proposition 2.5.3(1), their second derivatives are $\partial _ k\partial _ lF_\varepsilon =H_{kl}*\rho _\varepsilon $. Hence, for $\varepsilon \leq 1$, the Hessians $D^2F_\varepsilon $ are bounded on each compact set $K$ by the bound of $D^2F$ on the set of points at distance at most $1$ from $K$, and by theorem 2.8.3 they converge to $D^2F(x)$ at every point $x$ that is a Lebesgue point of all the entries $H_{kl}$, that is, almost everywhere. By the smooth case, $\operatorname {MA}_{\mathbb R}(F_\varepsilon )=n!\det (D^2F_\varepsilon )\, \mathrm dx$. Let $\varepsilon $ run through a sequence tending to zero. For $f\in C_ c(N_{\mathbb R})$, dominated convergence gives $n!\int f\det (D^2F_\varepsilon )\, \mathrm dx\to n!\int f\det (D^2F)\, \mathrm dx$, while $\operatorname {MA}_{\mathbb R}(F_\varepsilon )\to \operatorname {MA}_{\mathbb R}(F)$ vaguely by theorem 21.5.2. This proves the formula.

Mixed measures. For smooth convex $F_1,\ldots ,F_ n$ define

where $D$ is the mixed discriminant of section 5.3, applied to the real symmetric, hence Hermitian, Hessian matrices. This measure is positive, because the mixed discriminant of positive semidefinite matrices is nonnegative (lemma 5.3.3(3)), and it is symmetric and multilinear, with $D(A,\ldots ,A)=\det A$ (lemma 5.3.3(1)). Expanding $\operatorname {MA}_{\mathbb R}(\sum _{j\in J}F_ j,\ldots ,\sum _{j\in J}F_ j)$ by multilinearity and using symmetry, we obtain the polarization identity

where the empty sum is the zero function, whose measure is zero for $n>0$ (its only slope is $0$). Indeed, after the expansion, a term $\operatorname {MA}_{\mathbb R}(F_{\sigma (1)},\ldots ,F_{\sigma (n)})$ with $\sigma \colon \{ 1,\ldots ,n\} \to \{ 1,\ldots ,n\} $ appears with total coefficient $\sum _{J\supset \sigma (\{ 1,\ldots ,n\} )}(-1)^{n-|J|}$, which vanishes unless $\sigma $ is a bijection, and the $n!$ bijections all give $\operatorname {MA}_{\mathbb R}(F_1,\ldots ,F_ n)$ by symmetry. Any symmetric, multilinear assignment with the given diagonal values satisfies (21.5.5), by the same expansion; this proves uniqueness.

Now let $F_1,\ldots ,F_ n$ be arbitrary finite convex functions, and let $F_{j,\varepsilon }=F_ j*\rho _\varepsilon $ be their mollifications, which are smooth and convex by lemma 21.4.3. For every $J$, $\sum _{j\in J}F_{j,\varepsilon }\to \sum _{j\in J}F_ j$ locally uniformly as $\varepsilon \to 0$ (lemma 2.5.1(1)). Let $\varepsilon $ run through a sequence tending to zero. By (21.5.5) for the smooth functions $F_{j,\varepsilon }$ and by theorem 21.5.2, for every $f\in C_ c(N_{\mathbb R},\mathbb R)$ the integrals $\int f\, \mathrm d\operatorname {MA}_{\mathbb R}(F_{1,\varepsilon },\ldots ,F_{n,\varepsilon })$ converge to

The functional $\Lambda $ is linear, and it is positive as a limit of integrals against positive measures. By theorem 2.3.3, there is a unique Radon measure $\operatorname {MA}_{\mathbb R}(F_1,\ldots ,F_ n)$ with $\int f\, \mathrm d\operatorname {MA}_{\mathbb R}(F_1,\ldots ,F_ n)=\Lambda (f)$ for every $f$; for smooth $F_ j$ it agrees with the previous definition, by (21.5.5). It is symmetric, because $\Lambda $ is a symmetric function of $F_1,\ldots ,F_ n$. Additivity and positive homogeneity in each argument hold for the smooth mixed measures; since mollification is linear, they pass to the vague limit. For $F_1=\cdots =F_ n=F$, the smooth mixed measures are $n!\det (D^2F_\varepsilon )\, \mathrm dx=\operatorname {MA}_{\mathbb R}(F_\varepsilon )$, which converge vaguely to $\operatorname {MA}_{\mathbb R}(F)$; hence $\operatorname {MA}_{\mathbb R}(F,\ldots ,F)=\operatorname {MA}_{\mathbb R}(F)$. Finally, if $F_ j^{(k)}\to F_ j$ locally uniformly as $k\to \infty $ for each $j$, then $\sum _{j\in J}F_ j^{(k)}\to \sum _{j\in J}F_ j$ locally uniformly for every $J$, and the formula for $\Lambda $ together with theorem 21.5.2 shows that $\operatorname {MA}_{\mathbb R}(F_1^{(k)},\ldots ,F_ n^{(k)})\to \operatorname {MA}_{\mathbb R}(F_1,\ldots ,F_ n)$ vaguely. In dimension zero the mixed measure has no arguments, and it is the unit mass at the point. $\square$