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

lemmalem:coupling-graph-pushforward-euclidean-2026a
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· 2,829 chars · 4 deps · depth 19 Reason: First publication: on the graph, membership in a Borel set is a condition on the first coordinate, which identifies the coupling with the push-forward without any pi-system argument.

The two measures agree on the measurable rectangles, a generating pi-system, because the graph condition turns a rectangle into a preimage under the first projection.

Proof

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

1. The coupling ignores the complement of the graph. Since π\pi is a probability measure and π(ΓS)=1\pi(\Gamma_{S})=1, claim 3 of Basic Properties of a Measure gives π(Rd+dΓS)=11=0\pi(\mathbb{R}^{d+d}\setminus\Gamma_{S})=1-1=0. Hence for every BB(Rd+d)B\in\mathcal{B}(\mathbb{R}^{d+d}) the sets BΓSB\cap\Gamma_{S} and BΓSB\setminus\Gamma_{S} are disjoint members of B(Rd+d)\mathcal{B}(\mathbb{R}^{d+d}) with union BB, so claims 1 and 2 of that lemma give

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

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

2. On the graph, membership in BB is a condition on the first coordinate. Let zΓSz\in\Gamma_{S} and put x=pr1(z)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 §projections we have z=ι(pr1(z),pr2(z))z=\iota(\mathrm{pr}_{1}(z),\mathrm{pr}_{2}(z)), and pr2(z)=S(x)\mathrm{pr}_{2}(z)=S(x) by the definition of ΓS\Gamma_{S}, so z=ι(x,S(x))=(id,S)(x)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 §pairing. Consequently, for BB(Rd+d)B\in\mathcal{B}(\mathbb{R}^{d+d}) and zΓSz\in\Gamma_{S},

zB    (id,S)(pr1(z))B    pr1(z)E,E=(id,S)1(B),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 EB(Rd)E\in\mathcal{B}(\mathbb{R}^{d}) because (id,S)(\mathrm{id},S) is Borel. Therefore BΓS=pr11(E)ΓSB\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 pr11(E)\mathrm{pr}_{1}^{-1}(E),

π(B)=π(pr11(E)ΓS)=π(pr11(E))=μ(E)=μ((id,S)1(B))=(id,S)#μ(B),\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 (pr1)#π=μ(\mathrm{pr}_{1})_{\#}\pi=\mu by Couplings of Two Probability Measures on Euclidean Space and Their Quadratic Cost §coupling and the last by Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pushforward. As BB(Rd+d)B\in\mathcal{B}(\mathbb{R}^{d+d}) was arbitrary, π=(id,S)#μ\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 §projections and Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §pairing we have pr2((id,S)(x))=pr2(ι(x,S(x)))=S(x)\mathrm{pr}_{2}((\mathrm{id},S)(x))=\mathrm{pr}_{2}(\iota(x,S(x)))=S(x) for every xRdx\in\mathbb{R}^{d}, so (id,S)1(pr21(B))=S1(B)(\mathrm{id},S)^{-1}(\mathrm{pr}_{2}^{-1}(B))=S^{-1}(B) for every BB(Rd)B\in\mathcal{B}(\mathbb{R}^{d}). Hence, using step 3 and (pr2)#π=ν(\mathrm{pr}_{2})_{\#}\pi=\nu from Couplings of Two Probability Measures on Euclidean Space and Their Quadratic Cost §coupling,

ν(B)=π(pr21(B))=μ((id,S)1(pr21(B)))=μ(S1(B))=S#μ(B),\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#μ=νS_{\#}\mu=\nu. This proves claim 1 of the statement.

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