Brenier's Theorem: Optimal Couplings out of an Absolutely Continuous Measure are Induced by a Unique Map

In the setting of Probability Measures on Euclidean Space and Random Vectors: Standing Notation, let $d\in\mathbb{N}$ satisfy $1\le d$ and let $\mu,\nu$ belong to the set $\mathcal{P}_{2}(\mathbb{R}^{d})$ of probability measures with finite second moment, whose second moments are written $M_{2}(\mu)$ and $M_{2}(\nu)$. Let $\Pi(\mu,\nu)$ be the set of their couplings, and let optimality of a coupling be as defined there.

Write $\mathrm{id}$ for the identity map of $\mathbb{R}^{d}$, which is Borel, being continuous. For a Borel map $T:\mathbb{R}^{d}\to\mathbb{R}^{d}$ the pairing $(\mathrm{id},T)$ is Borel by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets (ii), so the push-forward $(\mathrm{id},T)_{\#}\mu$ of Probability Measures on Euclidean Space and Random Vectors: Standing Notation (iv) is defined, and it belongs to $\Pi(\mu,T_{\#}\mu)$ by Couplings on Euclidean Space: Product Coupling, Swap, Finiteness of the Cost, Push-Forward Couplings, Modifying One Marginal, Quantisation, Gluing over a Finitely Supported Measure, and the Lipschitz Bound (iv). For Borel maps $S,S':\mathbb{R}^{d}\to\mathbb{R}^{d}$ the function $x\mapsto\lVert S(x)-S'(x)\rVert^{2}$ is Borel by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets (iv) and vanishes exactly where $S$ and $S'$ agree, by claim 2 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square together with the metric axioms of $\mathbb{R}^n$ with the Euclidean Distance Is a Complete Metric Space with Coordinatewise Convergence and claims 1 and 3 of Zero Products and Elementary Identities in a Field; hence $\{x\in\mathbb{R}^{d}:S(x)\ne S'(x)\}$, the preimage of the Borel set $\mathbb{R}\setminus\{0\}$ under that function, belongs to $\mathcal{B}(\mathbb{R}^{d})$.

Assume that $\mu$ is absolutely continuous. Then the following hold.

1. (Every optimal coupling is induced by a map)(i) Let $\pi\in\Pi(\mu,\nu)$ be optimal. Then there is a Borel map $T:\mathbb{R}^{d}\to\mathbb{R}^{d}$ with $\pi=(\mathrm{id},T)_{\#}\mu$, and every Borel map $T$ with $\pi=(\mathrm{id},T)_{\#}\mu$ satisfies $T_{\#}\mu=\nu$ and

$$\int_{\mathbb{R}^{d}}\lVert T(x)\rVert^{2}\,\mu(dx)=M_{2}(\nu)<\infty .$$

2. (The map is the subgradient of a convex potential)(ii) Let $\pi\in\Pi(\mu,\nu)$ be optimal. Then there is a Borel map $T:\mathbb{R}^{d}\to\mathbb{R}^{d}$ with $\pi=(\mathrm{id},T)_{\#}\mu$ for which there are an open convex set $G\subseteq\mathbb{R}^{d}$, which belongs to $\mathcal{B}(\mathbb{R}^{d})$ by Product Measures and Lebesgue Measure on ℝⁿ: Standing Notation and Background (iv) and satisfies $\mu(G)=1$, a function $\phi:G\to\mathbb{R}$ that is convex on $G$, and a set $D\in\mathcal{B}(\mathbb{R}^{d})$ with $D\subseteq G$ and $\mu(D)=1$, such that $\phi$ is differentiable at every point of $D$ and its subdifferential relative to $G$ satisfies

$$\partial_{G}\phi(x)=\{T(x)\}\qquad\text{for every }x\in D .$$

3. (Uniqueness of the optimal coupling)(iii) If $\pi$ and $\pi'$ are optimal couplings of $\mu$ and $\nu$, then $\pi=\pi'$.

4. (Uniqueness of the map)(iv) If $T$ and $T'$ are Borel maps from $\mathbb{R}^{d}$ to $\mathbb{R}^{d}$ such that $(\mathrm{id},T)_{\#}\mu$ and $(\mathrm{id},T')_{\#}\mu$ are optimal couplings of $\mu$ and $\nu$, then

$$\mu\bigl(\{x\in\mathbb{R}^{d}:\ T(x)\ne T'(x)\}\bigr)=0 .$$