Proof of McCann's Jacobian Equation Along an Optimal Map Between Absolutely Continuous Measures

Each result cited is universally quantified over the data in its own statement. Elementary order and field arithmetic in $\mathbb{R}$ is used without mention (Elementary Order Arithmetic in an Ordered Field, Elementary Arithmetic in an Ordered Field). Integrals against $\lambda_{d}$ are written $\int\cdots d\lambda_{d}$; a Borel set is $\mu$-full if its complement is $\mu$-null, and finitely many $\mu$-full sets have a $\mu$-full intersection (claim 4 of Basic Properties of a Measure); likewise for $\nu$. The densities $\rho,\rho_{1}$ are Borel, real and nonnegative, $\mu$ is the measure with density $\rho$ with respect to $\lambda_{d}$, and so, by claim 3 of Image Measures, Measures with Densities, and Change of Variables, $\int u\,d\mu=\int u\rho\,d\lambda_{d}$ for every Borel $u\ge0$; likewise for $\nu$ and $\rho_{1}$. In particular $\int\rho\,d\lambda_{d}=\mu(\mathbb{R}^{d})=1$ and $\int\rho_{1}\,d\lambda_{d}=1$. By Optimal Transport Maps and Uniquely Mapped Pairs of Probability Measures §map, $T$ is Borel and $T_{\#}\mu=\nu$.

Step 1: the Hessians. By Along an Optimal Map between Absolutely Continuous Measures the Hessians of the Two Convex Potentials are Inverse Matrices §hessians, applied to $\mu,\nu,T,S,G,G',\varphi,\psi,D,D'$, there is a $\mu$-full Borel $X_{0}\subseteq D$ every point of which has the properties listed there; in particular every $x\in X_{0}$ satisfies: $T(x)\in D'$; $\varphi$ is twice differentiable at $x$ with first-order coefficient $T(x)$; $\psi$ is twice differentiable at $T(x)$ with first-order coefficient $x$; $D^{2}\varphi(x)$ and $D^{2}\psi(T(x))$ are positive definite; and $\det D^{2}\varphi(x)\cdot\det D^{2}\psi(T(x))=1$. By Determinants of Positive Definite Matrices: Positivity, the Bound $\log\det A\le\mathrm{tr}\,A-d$, Bounds under Pinching, and the Expansion of $\det(I+tB)$ §positive, $0<\det D^{2}\varphi(x)$ for $x\in X_{0}$.

Step 2: two area inequalities. (a) Let $k_{1}$ be the function of The Area Inequality for the Gradient of a Convex Function §area for $f=\varphi$, $U=G$, $A=X_{0}$ (a Borel subset of $G$ at whose points $\varphi$ is twice differentiable) and $h=\rho_{1}$: by Step 1, $k_{1}(x)=\rho_{1}(T(x))\det D^{2}\varphi(x)$ for $x\in X_{0}$ and $k_{1}(x)=0$ otherwise. That theorem gives that $k_{1}$ is Borel and nonnegative with

(b) 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 for $\psi$ on $G'$ there is a Borel $A_{\psi}\subseteq G'$ with $\lambda_{d}(G'\setminus A_{\psi})=0$ at whose points $\psi$ is twice differentiable; put $A'=D'\cap A_{\psi}$, which is $\nu$-full ($\nu(G'\setminus A_{\psi})=0$ by absolute continuity). Let $k$ be the function of The Area Inequality for the Gradient of a Convex Function §area for $f=\psi$, $U=G'$, $A=A'$ and $h=\rho$: $k(y)=\rho(D\psi(y))\det D^{2}\psi(y)$ for $y\in A'$ and $k(y)=0$ otherwise; that theorem gives $\int k\,d\lambda_{d}\le\int\rho\,d\lambda_{d}=1$. The sets $P=\{\rho>0\}$ and $P_{1}=\{\rho_{1}>0\}$ are Borel, and $\mu(\mathbb{R}^{d}\setminus P)=\int\mathbf{1}_{\{\rho=0\}}\rho\,d\lambda_{d}=0$, so $P$ is $\mu$-full; likewise $P_{1}$ is $\nu$-full. Let $u:\mathbb{R}^{d}\to\mathbb{R}$ be $u=k\rho_{1}^{-1}$ on $P_{1}$ and $u=0$ elsewhere, a nonnegative Borel function. Then $\int u\,d\nu=\int\mathbf{1}_{P_{1}}k\,d\lambda_{d}\le\int k\,d\lambda_{d}\le1$ (monotonicity, claim 1 of Linearity and Monotonicity of the Lebesgue Integral), and since $T_{\#}\mu=\nu$, Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pushforward gives

Step 3: the ratio and its reciprocal. Let $X_{1}=X_{0}\cap P\cap T^{-1}(P_{1}\cap A')$, a Borel set which is $\mu$-full because $\mu(T^{-1}(P_{1}\cap A'))=\nu(P_{1}\cap A')=1$. Let $x\in X_{1}$ and $y=T(x)$. Then $y\in A'\cap P_{1}$ and $D\psi(y)=x$ by Step 1, so $k(y)=\rho(x)\det D^{2}\psi(y)=\rho(x)\,(\det D^{2}\varphi(x))^{-1}$ and, writing $w(x)=u(T(x))$,

using $\rho(x)>0$, $\rho_{1}(y)>0$, $\det D^{2}\varphi(x)>0$ and Step 2(a). Let $v:\mathbb{R}^{d}\to\mathbb{R}$ be $v=k_{1}\rho^{-1}$ on $X_{1}$ and $v=0$ elsewhere, a nonnegative Borel function. Then $v\rho=\mathbf{1}_{X_{1}}k_{1}$ everywhere, so by monotonicity and (1)

Step 4: the ratio equals $1$. Let $F:\mathbb{R}^{d}\to\mathbb{R}$ be $F(x)=u(T(x))+v(x)-2$ for $x\in X_{1}$ and $F(x)=0$ otherwise; it is Borel. For $x\in X_{1}$, (3) gives $F(x)=w+w^{-1}-2=(w-1)^{2}w^{-1}\ge0$ with $w=w(x)>0$; so $0\le F$ everywhere, and

everywhere. By claim 1 of Linearity and Monotonicity of the Lebesgue Integral (additivity and monotonicity of the integral of nonnegative functions), $\mu(X_{1})=1$, (2) and (4),

the integrals lying in $[0,\infty]$; hence $\int F\,d\mu=0$, and by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §vanishing the Borel set $Z=\{x\in\mathbb{R}^{d}:F(x)=0\}$ is $\mu$-full. Let $X=X_{1}\cap Z$, a $\mu$-full Borel set with $X\subseteq X_{0}\subseteq D$; so every $x\in X$ has all the properties listed in Along an Optimal Map between Absolutely Continuous Measures the Hessians of the Two Convex Potentials are Inverse Matrices §hessians, since $x\in X_{0}$. For $x\in X$, $(w(x)-1)^{2}w(x)^{-1}=F(x)=0$ gives $w(x)=1$, which by (3) is

and $\rho(x)>0$ since $x\in P$. This is the Jacobian equation.