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

lemmalem:average-couplings-euclidean-2026a
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· 7,673 chars · 21 deps · depth 22 Reason: Phase B2b: proof by termwise addition of convergent series for countable additivity, and by indicators, simple functions and monotone convergence for the integral identity.

Countable additivity of the combination follows from termwise addition of convergent series; the integral identity is checked on indicators, extended to simple functions by linearity and to nonnegative functions by monotone convergence, and the coupling statements are read off from it.

Proof

Throughout, each result cited is universally quantified over the data appearing in its own statement and is applied to the data named here.

Step 1 (The combination is a finite measure). Let (E,E)(E,\mathcal{E}), α\alpha, β\beta, ss, tt and γ\gamma be as in claim 1. Every value α(A)\alpha(A), β(A)\beta(A) is a real number in [0,α(E)][0,\alpha(E)], respectively [0,β(E)][0,\beta(E)], by claim 2 of Basic Properties of a Measure, so γ(A)\gamma(A) is a nonnegative real number and γ(E)=sα(E)+tβ(E)<\gamma(E)=s\alpha(E)+t\beta(E)<\infty. Also γ()=0\gamma(\varnothing)=0, since α()=β()=0\alpha(\varnothing)=\beta(\varnothing)=0 by Measure, Measure Space, and Probability Measure. Let (Am)mN(A_{m})_{m\in\mathbb{N}} be a sequence of pairwise disjoint members of E\mathcal{E} and A=mAmA=\bigcup_{m}A_{m}. By Measure, Measure Space, and Probability Measure the series mα(Am)\sum_{m}\alpha(A_{m}) and mβ(Am)\sum_{m}\beta(A_{m}) converge with sums α(A)\alpha(A) and β(A)\beta(A). For nNn\in\mathbb{N}, claims 2 and 3 of Properties of Finite Sums give

m=1n(sα(Am)+tβ(Am))=sm=1nα(Am)+tm=1nβ(Am),\sum_{m=1}^{n}\bigl(s\,\alpha(A_{m})+t\,\beta(A_{m})\bigr)=s\sum_{m=1}^{n}\alpha(A_{m})+t\sum_{m=1}^{n}\beta(A_{m}),

and by claims 1 and 3 of Arithmetic of Limits of Real Sequences the right-hand side converges to sα(A)+tβ(A)=γ(A)s\,\alpha(A)+t\,\beta(A)=\gamma(A) as nn\to\infty. Hence mγ(Am)\sum_{m}\gamma(A_{m}) converges with sum γ(A)\gamma(A), and γ\gamma is a measure on (E,E)(E,\mathcal{E}) with γ(E)<\gamma(E)<\infty.

Step 2 (The integral identity). Let f:E[0,]f:E\to[0,\infty] be measurable in the sense of Measure Spaces and the Lebesgue Integral: Standing Notation §measurable. If f=1Af=\mathbf{1}_{A} for some AEA\in\mathcal{E}, then The Integral of an Indicator Function is the Measure of the Set, applied to each of γ\gamma, α\alpha and β\beta, turns the identity into the definition of γ(A)\gamma(A). If ff is a nonnegative simple function, write f=j=1rcj1Ajf=\sum_{j=1}^{r}c_{j}\mathbf{1}_{A_{j}} with cjc_{j} the distinct values of ff, all nonnegative, and Aj=f1({cj})EA_{j}=f^{-1}(\{c_{j}\})\in\mathcal{E}; then claim 1 of Linearity and Monotonicity of the Lebesgue Integral, applied to each of the three measures, together with claims 2 and 3 of Properties of Finite Sums, gives the identity for ff from the case of indicators.

For general ff, Approximation of Measurable Functions by Simple Functions §nonnegative supplies a nondecreasing sequence (um)mN(u_{m})_{m\in\mathbb{N}} of nonnegative simple functions with pointwise supremum ff. Applying Monotone Convergence Theorem on (E,E,γ)(E,\mathcal{E},\gamma), on (E,E,α)(E,\mathcal{E},\alpha) and on (E,E,β)(E,\mathcal{E},\beta), the three sequences (umdγ)m\bigl(\int u_{m}\,d\gamma\bigr)_{m}, (umdα)m\bigl(\int u_{m}\,d\alpha\bigr)_{m} and (umdβ)m\bigl(\int u_{m}\,d\beta\bigr)_{m} have suprema fdγ\int f\,d\gamma, fdα\int f\,d\alpha and fdβ\int f\,d\beta in [0,][0,\infty]. Write A=EfdαA=\int_{E}f\,d\alpha, B=EfdβB=\int_{E}f\,d\beta and Γ=Efdγ\Gamma=\int_{E}f\,d\gamma, and am=umdαa_{m}=\int u_{m}\,d\alpha, bm=umdβb_{m}=\int u_{m}\,d\beta, cm=umdγc_{m}=\int u_{m}\,d\gamma, so that cm=sam+tbmc_{m}=s\,a_{m}+t\,b_{m} for every mm by the simple case, and AA, BB, Γ\Gamma are the suprema in [0,][0,\infty] of the nondecreasing sequences (am)(a_{m}), (bm)(b_{m}), (cm)(c_{m}), the sequences being nondecreasing by claim 1 of Linearity and Monotonicity of the Lebesgue Integral. We must show Γ=sA+tB\Gamma=sA+tB, the right-hand side formed with the conventions of Measure, Measure Space, and Probability Measure.

Suppose first that AA and BB are both real. Then (am)(a_{m}) and (bm)(b_{m}) are nondecreasing and bounded above, so they converge to AA and BB by claim 1 of A Bounded Monotone Sequence of Real Numbers Converges, and (cm)(c_{m}) converges to sA+tBsA+tB by claims 1 and 3 of Arithmetic of Limits of Real Sequences; being nondecreasing, (cm)(c_{m}) has supremum its limit by the same claim of A Bounded Monotone Sequence of Real Numbers Converges, so Γ=sA+tB\Gamma=sA+tB.

Otherwise at least one of AA, BB is \infty; say A=A=\infty, the case B=B=\infty being symmetric. If s=0s=0 and t=0t=0 then cm=0c_{m}=0 for every mm and Γ=0=sA+tB\Gamma=0=sA+tB by the convention 0=00\cdot\infty=0. If s=0s=0 and 0<t0<t, then cm=tbmc_{m}=t\,b_{m} and the assertion is the case just treated with the single sequence (bm)(b_{m}), whose supremum is BB; explicitly, Γ=supmtbm=tB\Gamma=\sup_{m}t\,b_{m}=tB whether BB is real or \infty, since multiplication by the positive tt preserves upper bounds by claim 5 of Elementary Arithmetic in an Ordered Field applied in both directions with the factors tt and t1t^{-1}. If 0<s0<s, then samcms\,a_{m}\le c_{m} for every mm, because tbm0t\,b_{m}\ge0, and supmsam=\sup_{m}s\,a_{m}=\infty by the same argument, so Γ==sA+tB\Gamma=\infty=sA+tB.

In every case

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

Now let f:ERf:E\to\mathbb{R} be measurable and integrable with respect to both α\alpha and β\beta. Applying the nonnegative case to f|f| shows fdγ=sfdα+tfdβ<\int|f|\,d\gamma=s\int|f|\,d\alpha+t\int|f|\,d\beta<\infty, so ff is integrable with respect to γ\gamma. Writing f=max(f,0)max(f,0)f=\max(f,0)-\max(-f,0), both parts are nonnegative, measurable by claim 4 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, and have finite integrals against all three measures, so the nonnegative case applies to each; subtracting the two identities and using claim 2 of Linearity and Monotonicity of the Lebesgue Integral on each of the three measures gives the identity for ff, both sides being real. This proves claim 1.

Step 3 (Claim 2). Let μ,ν,π,π\mu,\nu,\pi,\pi' and π^\hat{\pi} be as in claim 2; by claim 1, π^\hat{\pi} is a measure on (Rd+d,B(Rd+d))(\mathbb{R}^{d+d},\mathcal{B}(\mathbb{R}^{d+d})) with π^(Rd+d)=12+12=1\hat{\pi}(\mathbb{R}^{d+d})=\tfrac{1}{2}+\tfrac{1}{2}=1, so π^P(Rd+d)\hat{\pi}\in\mathcal{P}(\mathbb{R}^{d+d}).

Marginals. For BB(Rd)B\in\mathcal{B}(\mathbb{R}^{d}), the set (pr1)1(B)(\mathrm{pr}_{1})^{-1}(B) belongs to B(Rd+d)\mathcal{B}(\mathbb{R}^{d+d}), the projections being Borel, and by Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pushforward together with π,πΠ(μ,ν)\pi,\pi'\in\Pi(\mu,\nu),

(pr1)#π^(B)=π^((pr1)1(B))=12μ(B)+12μ(B)=μ(B),(\mathrm{pr}_{1})_{\#}\hat{\pi}(B)=\hat{\pi}\bigl((\mathrm{pr}_{1})^{-1}(B)\bigr)=\tfrac{1}{2}\mu(B)+\tfrac{1}{2}\mu(B)=\mu(B),

and likewise (pr2)#π^=ν(\mathrm{pr}_{2})_{\#}\hat{\pi}=\nu. Hence π^Π(μ,ν)\hat{\pi}\in\Pi(\mu,\nu) by Couplings of Two Probability Measures on Euclidean Space and Their Quadratic Cost §coupling.

Cost. The function zpr1(z)pr2(z)2z\mapsto\lVert\mathrm{pr}_{1}(z)-\mathrm{pr}_{2}(z)\rVert^{2} is Borel and nonnegative by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §functions, so claim 1, applied to it with s=t=12s=t=\tfrac{1}{2}, gives

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

by Couplings of Two Probability Measures on Euclidean Space and Their Quadratic Cost §cost; the two costs are real numbers 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 §cost-finite, μ\mu and ν\nu having finite second moment, so the identity holds in R\mathbb{R}.

Domination. For AB(Rd+d)A\in\mathcal{B}(\mathbb{R}^{d+d}) one has 12π(A)0\tfrac{1}{2}\pi'(A)\ge0, hence 12π(A)π^(A)\tfrac{1}{2}\pi(A)\le\hat{\pi}(A), and multiplying by the nonnegative number 22 gives π(A)2π^(A)\pi(A)\le2\hat{\pi}(A) by claim 5 of Elementary Arithmetic in an Ordered Field; symmetrically π(A)2π^(A)\pi'(A)\le2\hat{\pi}(A).

Optimality. If π\pi and π\pi' are optimal, then I(π)=I(π)=W2(μ,ν)2I(\pi)=I(\pi')=W_{2}(\mu,\nu)^{2}, so the cost identity gives I(π^)=12W2(μ,ν)2+12W2(μ,ν)2=W2(μ,ν)2I(\hat{\pi})=\tfrac{1}{2}W_{2}(\mu,\nu)^{2}+\tfrac{1}{2}W_{2}(\mu,\nu)^{2}=W_{2}(\mu,\nu)^{2}, and π^\hat{\pi} is optimal.

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