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

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 $(E,\mathcal{E})$, $\alpha$, $\beta$, $s$, $t$ and $\gamma$ be as in claim 1. Every value $\alpha(A)$, $\beta(A)$ is a real number in $[0,\alpha(E)]$, respectively $[0,\beta(E)]$, by claim 2 of Basic Properties of a Measure, so $\gamma(A)$ is a nonnegative real number and $\gamma(E)=s\alpha(E)+t\beta(E)<\infty$. Also $\gamma(\varnothing)=0$, since $\alpha(\varnothing)=\beta(\varnothing)=0$ by Measure, Measure Space, and Probability Measure. Let $(A_{m})_{m\in\mathbb{N}}$ be a sequence of pairwise disjoint members of $\mathcal{E}$ and $A=\bigcup_{m}A_{m}$. By Measure, Measure Space, and Probability Measure the series $\sum_{m}\alpha(A_{m})$ and $\sum_{m}\beta(A_{m})$ converge with sums $\alpha(A)$ and $\beta(A)$. For $n\in\mathbb{N}$, claims 2 and 3 of Properties of Finite Sums give

$$\sum_{m=1}^{n}\bigl(s\,\alpha(A_{m})+t\,\beta(A_{m})\bigr)=s\sum_{m=1}^{n}\alpha(A_{m})+t\sum_{m=1}^{n}\beta(A_{m}),$$

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

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

For general $f$, Approximation of Measurable Functions by Dyadic Simple Functions (i) supplies a nondecreasing sequence $(u_{m})_{m\in\mathbb{N}}$ of nonnegative simple functions with pointwise supremum $f$. Applying Monotone Convergence Theorem on $(E,\mathcal{E},\gamma)$, on $(E,\mathcal{E},\alpha)$ and on $(E,\mathcal{E},\beta)$, the three sequences $\bigl(\int u_{m}\,d\gamma\bigr)_{m}$, $\bigl(\int u_{m}\,d\alpha\bigr)_{m}$ and $\bigl(\int u_{m}\,d\beta\bigr)_{m}$ have suprema $\int f\,d\gamma$, $\int f\,d\alpha$ and $\int f\,d\beta$ in $[0,\infty]$. Write $A=\int_{E}f\,d\alpha$, $B=\int_{E}f\,d\beta$ and $\Gamma=\int_{E}f\,d\gamma$, and $a_{m}=\int u_{m}\,d\alpha$, $b_{m}=\int u_{m}\,d\beta$, $c_{m}=\int u_{m}\,d\gamma$, so that $c_{m}=s\,a_{m}+t\,b_{m}$ for every $m$ by the simple case, and $A$, $B$, $\Gamma$ are the suprema in $[0,\infty]$ of the nondecreasing sequences $(a_{m})$, $(b_{m})$, $(c_{m})$, the sequences being nondecreasing by claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion. We must show $\Gamma=sA+tB$, the right-hand side formed with the conventions of Measure, Measure Space, and Probability Measure.

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

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

In every case

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

Now let $f:E\to\mathbb{R}$ be measurable and integrable with respect to both $\alpha$ and $\beta$. Applying the nonnegative case to $|f|$ shows $\int|f|\,d\gamma=s\int|f|\,d\alpha+t\int|f|\,d\beta<\infty$, so $f$ is integrable with respect to $\gamma$. Writing $f=\max(f,0)-\max(-f,0)$, both parts are nonnegative, measurable by claim 4 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps, 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, and the Integrability Criterion on each of the three measures gives the identity for $f$, both sides being real. This proves claim 1.

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

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

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

and likewise $(\mathrm{pr}_{2})_{\#}\hat{\pi}=\nu$. Hence $\hat{\pi}\in\Pi(\mu,\nu)$ by Couplings of Two Probability Measures on Euclidean Space and Their Quadratic Cost (i).

Cost. The function $z\mapsto\lVert\mathrm{pr}_{1}(z)-\mathrm{pr}_{2}(z)\rVert^{2}$ 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 (iv), so claim 1, applied to it with $s=t=\tfrac{1}{2}$, gives

$$I(\hat{\pi})=\tfrac{1}{2}I(\pi)+\tfrac{1}{2}I(\pi')$$

by Couplings of Two Probability Measures on Euclidean Space and Their Quadratic Cost (ii); 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 (iii), $\mu$ and $\nu$ having finite second moment, so the identity holds in $\mathbb{R}$.

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

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