Proof of Tightness from Bounded Second Moments, and Tightness of the Couplings of Two Measures with Finite Second Moment

Each result cited below is universally quantified over the data in its own statement, and is applied to the data named where it is used.

Claim 1. Let $\varepsilon\in\mathbb{R}$ satisfy $0<\varepsilon$. Note first that, for $x\in\mathbb{R}^{m}$, $d_{E}(0,x)=\lVert 0-x\rVert=\lVert x\rVert$ by claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ and the vector identities of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space, together with claim 5 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$; so for every real $r\ge 0$ the set $\{x\in\mathbb{R}^{m}:\lVert x\rVert\le r\}$ is the closed ball of $(\mathbb{R}^{m},d_{E})$ with centre $0$ and radius $r$, which is bounded and closed by claims 2 and 3 of Elementary Properties of the Closed Ball in a Metric Space, hence compact in $(\mathbb{R}^{m},d_{E})$ by Heine-Borel Theorem in $\mathbb{R}^n$.

If $\mathcal{M}$ is empty, the requirement of Tight Family of Borel Measures on a Metric Space §tight is satisfied vacuously by the closed ball with centre $0$ and radius $1$, and $\mathcal{M}$ is tight. So assume $\mathcal{M}$ is nonempty and fix $\mu_{*}\in\mathcal{M}$; since $M_{2}(\mu_{*})$ is the integral of a nonnegative function, $0\le M_{2}(\mu_{*})\le R$, so $0\le R$.

By The Archimedean Property of the Real Numbers there is a natural number $n$ with $R\varepsilon^{-1}<\iota(n)$, where $\iota$ is the embedding of The Canonical Map from the Natural Numbers to a Field; put $t=\iota(n)$, which satisfies $0<t$ by claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field. Multiplying $R\varepsilon^{-1}<t$ by the positive number $\varepsilon$, claim 10 of Elementary Order Arithmetic in an Ordered Field gives $R<t\varepsilon$. Let $r=\sqrt{t}$ be the nonnegative square root of $t$, so $r^{2}=t$ and $0\le r$, and put

a compact subset of $(\mathbb{R}^{m},d_{E})$ by the first paragraph.

Let $\mu\in\mathcal{M}$. The map $x\mapsto\lVert x\rVert^{2}$ on $\mathbb{R}^{m}$ is Borel and nonnegative by Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pairs, and is read as a map into $[0,\infty]$ as in Probability Measures on Euclidean Space and Random Vectors: Standing Notation §measures. Applying claim 6 of The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere on the measure space $(\mathbb{R}^{m},\mathcal{B}(\mathbb{R}^{m}),\mu)$ to this map and to the positive real number $t$, the set $A_{t}=\{x\in\mathbb{R}^{m}:t\le\lVert x\rVert^{2}\}$ belongs to $\mathcal{B}(\mathbb{R}^{m})$ and

the middle expression being $M_{2}(\mu)$ by The Second Moment of a Probability Measure on Euclidean Space and the Probability Measures with Finite Second Moment §moment.

If $x\in\mathbb{R}^{m}\setminus K$ then $r<\lVert x\rVert$, and both numbers being nonnegative, claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives $t=r^{2}<\lVert x\rVert^{2}$, so $x\in A_{t}$. Hence $\mathbb{R}^{m}\setminus K\subseteq A_{t}$ and $\mu(\mathbb{R}^{m}\setminus K)\le\mu(A_{t})$ by claim 2 of Basic Properties of a Measure. Moreover $\mu(A_{t})\le\mu(\mathbb{R}^{m})=1$ by the same claim, so $\mu(A_{t})$ is a real number, and multiplying $t\,\mu(A_{t})\le R<t\varepsilon$ by the positive number $t^{-1}$, again by claim 10 of Elementary Order Arithmetic in an Ordered Field, gives $\mu(A_{t})<\varepsilon$. Therefore $\mu(\mathbb{R}^{m}\setminus K)\le\varepsilon$ for every $\mu\in\mathcal{M}$, and $\mathcal{M}$ is tight in $(\mathbb{R}^{m},d_{E})$.

Claim 2. Let $\pi\in\Pi(\mu,\nu)$. By Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pairs the concatenation map $\iota^{m,m}$ is a bijection from $\mathbb{R}^{m}\times\mathbb{R}^{m}$ onto $\mathbb{R}^{m+m}$ and the coordinate projections of Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §projections satisfy $\iota^{m,m}(\mathrm{pr}_{1}(z),\mathrm{pr}_{2}(z))=z$ for every $z\in\mathbb{R}^{m+m}$; hence claim 3 of Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space gives

Each of the two maps $z\mapsto\lVert\mathrm{pr}_{i}(z)\rVert^{2}$ is Borel and nonnegative, being the composition of the Borel projection $\mathrm{pr}_{i}$ with the Borel map $x\mapsto\lVert x\rVert^{2}$ of Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pairs, compositions of Borel maps being Borel by Probability Measures on Euclidean Space and Random Vectors: Standing Notation §borel-maps. By claim 1 of Linearity and Monotonicity of the Lebesgue Integral, applied on the measure space $(\mathbb{R}^{m+m},\mathcal{B}(\mathbb{R}^{m+m}),\pi)$ to these two nonnegative maps,

By the change-of-variables formula of Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pushforward, applied with the Borel map $\mathrm{pr}_{1}$ and the nonnegative Borel map $x\mapsto\lVert x\rVert^{2}$, the first integral equals $\int_{\mathbb{R}^{m}}\lVert x\rVert^{2}\,((\mathrm{pr}_{1})_{\#}\pi)(dx)$, and $(\mathrm{pr}_{1})_{\#}\pi=\mu$ by Couplings of Two Probability Measures on Euclidean Space and Their Quadratic Cost §coupling; so it equals $M_{2}(\mu)$. In the same way the second integral equals $M_{2}(\nu)$, and $M_{2}(\pi)=M_{2}(\mu)+M_{2}(\nu)$ in $[0,\infty]$.

Assume now that $\mu$ and $\nu$ lie in $\mathcal{P}_{2}(\mathbb{R}^{m})$, so that $M_{2}(\mu)$ and $M_{2}(\nu)$ are real by The Second Moment of a Probability Measure on Euclidean Space and the Probability Measures with Finite Second Moment §space, and put $R=M_{2}(\mu)+M_{2}(\nu)$. By Couplings of Two Probability Measures on Euclidean Space and Their Quadratic Cost §coupling we have $\Pi(\mu,\nu)\subseteq\mathcal{P}(\mathbb{R}^{m+m})$, and by the identity just proved every $\pi\in\Pi(\mu,\nu)$ satisfies $M_{2}(\pi)=R\le R$. Since $1\le m\le m+m$, claim 1 of the present lemma may be read with $m+m$ in place of $m$, the lemma being stated for every natural number $m$ with $1\le m$; applied to the family $\Pi(\mu,\nu)$ and the bound $R$, it gives that $\Pi(\mu,\nu)$ is tight in $(\mathbb{R}^{m+m},d_{E})$.