Proof of A Coupling Concentrated on the Graph of a Borel Map is the Push-Forward by That Map

Each result cited below is universally quantified over the data in its own statement.

1. The coupling ignores the complement of the graph. Since $\pi$ is a probability measure and $\pi(\Gamma_{S})=1$, claim 3 of Basic Properties of a Measure gives $\pi(\mathbb{R}^{d+d}\setminus\Gamma_{S})=1-1=0$. Hence for every $B\in\mathcal{B}(\mathbb{R}^{d+d})$ the sets $B\cap\Gamma_{S}$ and $B\setminus\Gamma_{S}$ are disjoint members of $\mathcal{B}(\mathbb{R}^{d+d})$ with union $B$, so claims 1 and 2 of that lemma give

$$\pi(B)=\pi(B\cap\Gamma_{S})+\pi(B\setminus\Gamma_{S})=\pi(B\cap\Gamma_{S}),$$

because $\pi(B\setminus\Gamma_{S})\le\pi(\mathbb{R}^{d+d}\setminus\Gamma_{S})=0$.

2. On the graph, membership in $B$ is a condition on the first coordinate. Let $z\in\Gamma_{S}$ and put $x=\mathrm{pr}_{1}(z)$. By Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets (i) we have $z=\iota(\mathrm{pr}_{1}(z),\mathrm{pr}_{2}(z))$, and $\mathrm{pr}_{2}(z)=S(x)$ by the definition of $\Gamma_{S}$, so $z=\iota(x,S(x))=(\mathrm{id},S)(x)$, the pairing being as in Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets (ii). Consequently, for $B\in\mathcal{B}(\mathbb{R}^{d+d})$ and $z\in\Gamma_{S}$,

$$z\in B\iff(\mathrm{id},S)\bigl(\mathrm{pr}_{1}(z)\bigr)\in B\iff\mathrm{pr}_{1}(z)\in E,\qquad E=(\mathrm{id},S)^{-1}(B),$$

and $E\in\mathcal{B}(\mathbb{R}^{d})$ because $(\mathrm{id},S)$ is Borel. Therefore $B\cap\Gamma_{S}=\mathrm{pr}_{1}^{-1}(E)\cap\Gamma_{S}$.

3. Identification of the coupling. Combining steps 1 and 2, and applying step 1 once more to the Borel set $\mathrm{pr}_{1}^{-1}(E)$,

$$\pi(B)=\pi\bigl(\mathrm{pr}_{1}^{-1}(E)\cap\Gamma_{S}\bigr)=\pi\bigl(\mathrm{pr}_{1}^{-1}(E)\bigr)=\mu(E)=\mu\bigl((\mathrm{id},S)^{-1}(B)\bigr)=(\mathrm{id},S)_{\#}\mu\,(B),$$

the third equality because $(\mathrm{pr}_{1})_{\#}\pi=\mu$ by Couplings of Two Probability Measures on Euclidean Space and Their Quadratic Cost (i) and the last by Probability Measures on Euclidean Space and Random Vectors: Standing Notation (iv). As $B\in\mathcal{B}(\mathbb{R}^{d+d})$ was arbitrary, $\pi=(\mathrm{id},S)_{\#}\mu$.

4. The map transports the first marginal to the second. By Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets (i) and Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets (ii) we have $\mathrm{pr}_{2}((\mathrm{id},S)(x))=\mathrm{pr}_{2}(\iota(x,S(x)))=S(x)$ for every $x\in\mathbb{R}^{d}$, so $(\mathrm{id},S)^{-1}(\mathrm{pr}_{2}^{-1}(B))=S^{-1}(B)$ for every $B\in\mathcal{B}(\mathbb{R}^{d})$. Hence, using step 3 and $(\mathrm{pr}_{2})_{\#}\pi=\nu$ from Couplings of Two Probability Measures on Euclidean Space and Their Quadratic Cost (i),

$$\nu(B)=\pi\bigl(\mathrm{pr}_{2}^{-1}(B)\bigr)=\mu\bigl((\mathrm{id},S)^{-1}(\mathrm{pr}_{2}^{-1}(B))\bigr)=\mu\bigl(S^{-1}(B)\bigr)=S_{\#}\mu\,(B),$$

so $S_{\#}\mu=\nu$. This proves claim 1 of the statement.