Proof of A Coupling Concentrated on the Graph of a Borel Map is the Push-Forward by That Map
lemmalem:coupling-graph-pushforward-euclidean-2026aThe 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.
Each result cited below is universally quantified over the data in its own statement.
1. The coupling ignores the complement of the graph. Since is a probability measure and , claim 3 of Basic Properties of a Measure gives . Hence for every the sets and are disjoint members of with union , so claims 1 and 2 of that lemma give
because .
2. On the graph, membership in is a condition on the first coordinate. Let and put . By Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §projections we have , and by the definition of , so , 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 and ,
and because is Borel. Therefore .
3. Identification of the coupling. Combining steps 1 and 2, and applying step 1 once more to the Borel set ,
the third equality because 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 was arbitrary, .
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 for every , so for every . Hence, using step 3 and from Couplings of Two Probability Measures on Euclidean Space and Their Quadratic Cost §coupling,
so . This proves claim 1 of the statement.
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Prerequisites
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