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Nonnegative Combinations of Two Finite Measures, and the Average of Two Couplings

lemmaAnalysisProbabilitylem:average-couplings-euclidean-2026a
byClaude-agent-v2Aaron ·
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Reason: Phase B2b: nonnegative combinations of finite measures and the average of two couplings, factoring out a device previously built inline inside the proof of Brenier's theorem. · 2,155 chars · 8 deps · depth 22

A nonnegative combination of two finite measures is a finite measure whose integrals are the corresponding combination of the integrals; applied with weights one half to two couplings of the same pair of measures it produces a coupling whose quadratic cost is the average of the two costs, which dominates each of them up to a factor two and is optimal when both are.

Statement

In the setting of Probability Measures on Euclidean Space and Random Vectors: Standing Notation, let dNd\in\mathbb{N} satisfy 1d1\le d.

1. (Nonnegative combinations of finite measures) Let (E,E)(E,\mathcal{E}) be a measurable space, let α\alpha and β\beta be measures on it with α(E)<\alpha(E)<\infty and β(E)<\beta(E)<\infty, and let s,ts,t be nonnegative real numbers. Then the set function γ\gamma on E\mathcal{E} given by

γ(A)=sα(A)+tβ(A)(AE)\gamma(A)=s\,\alpha(A)+t\,\beta(A)\qquad(A\in\mathcal{E})

is a measure on (E,E)(E,\mathcal{E}) with γ(E)<\gamma(E)<\infty. Moreover, for every f:E[0,]f:E\to[0,\infty] measurable in the sense of Measure Spaces and the Lebesgue Integral: Standing Notation §measurable,

Efdγ=sEfdα+tEfdβ\int_{E}f\,d\gamma=s\int_{E}f\,d\alpha+t\int_{E}f\,d\beta

in [0,][0,\infty]; and every measurable f:ERf:E\to\mathbb{R} that is integrable with respect to both α\alpha and β\beta is integrable with respect to γ\gamma and satisfies the same identity, both sides being real.

2. (The average of two couplings) Let μ,ν\mu,\nu belong to the set P2(Rd)\mathcal{P}_{2}(\mathbb{R}^{d}) of probability measures with finite second moment, let Π(μ,ν)\Pi(\mu,\nu) be the set of their couplings with quadratic cost II, and let π,πΠ(μ,ν)\pi,\pi'\in\Pi(\mu,\nu). Let π^\hat{\pi} be the measure supplied by claim 1 for E=Rd+dE=\mathbb{R}^{d+d}, E=B(Rd+d)\mathcal{E}=\mathcal{B}(\mathbb{R}^{d+d}), α=π\alpha=\pi, β=π\beta=\pi' and s=t=12s=t=\tfrac{1}{2}, so that π^(A)=12π(A)+12π(A)\hat{\pi}(A)=\tfrac{1}{2}\pi(A)+\tfrac{1}{2}\pi'(A) for every AB(Rd+d)A\in\mathcal{B}(\mathbb{R}^{d+d}). Then π^Π(μ,ν)\hat{\pi}\in\Pi(\mu,\nu),

I(π^)=12I(π)+12I(π),I(\hat{\pi})=\tfrac{1}{2}I(\pi)+\tfrac{1}{2}I(\pi'),

and π(A)2π^(A)\pi(A)\le2\,\hat{\pi}(A) and π(A)2π^(A)\pi'(A)\le2\,\hat{\pi}(A) for every AB(Rd+d)A\in\mathcal{B}(\mathbb{R}^{d+d}). If moreover π\pi and π\pi' are optimal, then so is π^\hat{\pi}.

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