Proof of Along an Optimal Map between Absolutely Continuous Measures the Hessians of the Two Convex Potentials are Inverse Matrices

Each result cited is universally quantified over the data in its own statement. A set $Z\in\mathcal{B}(\mathbb{R}^{d})$ is called $\mu$-full if $\mu(Z)=1$, equivalently $\mu(\mathbb{R}^{d}\setminus Z)=0$; a finite intersection of $\mu$-full sets is $\mu$-full, its complement being contained in the union of the complements (claim 4 of Basic Properties of a Measure); likewise for $\nu$.

Step 1: the maps are mutually inverse. Since $T$ is an optimal map from $\mu$ to $\nu$, the coupling $(\mathrm{id},T)_{\#}\mu$ is optimal and $T_{\#}\mu=\nu$; by Brenier's Theorem: Optimal Couplings out of an Absolutely Continuous Measure are Induced by a Unique Map §unique-coupling ($\mu$ being absolutely continuous) every optimal coupling of $\mu$ and $\nu$ equals it, so $(\mu,\nu)$ is uniquely mapped. In the same way, $\nu$ being absolutely continuous, $(\nu,\mu)$ is uniquely mapped with optimal map $S$. By The Optimal Maps of a Uniquely Mapped Pair and of Its Reverse are Mutually Inverse Almost Everywhere §inverse, the set $E_{1}=\{x\in\mathbb{R}^{d}:S(T(x))=x\}$, which belongs to $\mathcal{B}(\mathbb{R}^{d})$ as recorded in that lemma, is $\mu$-full.

Step 2: points of twice differentiability. By The Points of Twice Differentiability of a Convex Function: a Borel Set of Full Measure, and Borel Measurability of the Gradient and Hessian on It §full, read with $d$ in place of $n$, there are $A_{\varphi},A_{\psi}\in\mathcal{B}(\mathbb{R}^{d})$ with $A_{\varphi}\subseteq G$, $A_{\psi}\subseteq G'$, $\lambda_{d}(G\setminus A_{\varphi})=\lambda_{d}(G'\setminus A_{\psi})=0$, such that $\varphi$ is twice differentiable at every point of $A_{\varphi}$ and $\psi$ at every point of $A_{\psi}$. By absolute continuity, $\mu(G\setminus A_{\varphi})=0$ and $\nu(G'\setminus A_{\psi})=0$; as $\mathbb{R}^{d}\setminus A_{\varphi}\subseteq(\mathbb{R}^{d}\setminus G)\cup(G\setminus A_{\varphi})$, the set $A_{\varphi}$ is $\mu$-full, and likewise $A_{\psi}$ is $\nu$-full. Hence $D'\cap A_{\psi}$ is $\nu$-full, and since $T$ is Borel with $T_{\#}\mu=\nu$ (Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pushforward), $E_{2}=T^{-1}(D'\cap A_{\psi})\in\mathcal{B}(\mathbb{R}^{d})$ satisfies $\mu(E_{2})=\nu(D'\cap A_{\psi})=1$.

Step 3: the set $X$. Put $E=D\cap A_{\varphi}\cap E_{1}\cap E_{2}$, a $\mu$-full Borel subset of $D$. By The Lebesgue Density Theorem in $\mathbb{R}^n$ §ae, the set of points of $E$ that are not density points of $E$ is null, hence contained in some $N\in\mathcal{B}(\mathbb{R}^{d})$ with $\lambda_{d}(N)=0$, so $\mu(N)=0$. Put $X=E\setminus N$: then $X\in\mathcal{B}(\mathbb{R}^{d})$, $X\subseteq D$, $\mu(X)=1$, and every point of $X$ is a density point of $E$.

Step 4: the properties at $x\in X$. Let $x\in X$ and $y=T(x)$. Since $x\in E_{1}\cap E_{2}$, we have $y\in D'$ and $S(y)=x$. Since $x\in A_{\varphi}$, $\varphi$ is twice differentiable at $x$ with some first-order coefficient $p$ and Hessian $B=D^{2}\varphi(x)$; by Subgradients near a Point of Twice Differentiability of a Convex Function, and Invariance of the Second-Order Expansion under Lipschitz Truncation §singleton, $\{p\}=\partial_{G}\varphi(x)=\{T(x)\}$ as $x\in D$, so $p=T(x)$. Since $y\in A_{\psi}$, $\psi$ is twice differentiable at $y$ with some first-order coefficient $p'$ and Hessian $B'=D^{2}\psi(y)$, and in the same way $\{p'\}=\partial_{G'}\psi(y)=\{S(y)\}=\{x\}$, so $p'=x$.

We apply Hessians of Two Convex Functions with Mutually Inverse Subgradients are Inverse Matrices at a Density Point §inverse with $n=d$, $U=G$, $U'=G'$, $f=\varphi$, $g=\psi$, the points $x$ and $y$, the matrices $B$ and $B'$, and the set $E\subseteq G$. Its hypothesis on $E$ holds: for $x'\in E$ put $q=T(x')$; then $x'\in D$ gives $q\in\partial_{G}\varphi(x')$, and $x'\in E_{1}\cap E_{2}$ gives $q\in D'\subseteq G'$ and $\partial_{G'}\psi(q)=\{S(q)\}=\{x'\}$, so $x'\in\partial_{G'}\psi(q)$. And $x$ is a density point of $E$. The lemma yields $B'B=I_{d}$, that $B$ and $B'$ are positive definite, and $\det B\cdot\det B'=1$, which are the remaining assertions.