Wasserstein metric and continuity-equation foundations

The product law $\mu \otimes \nu$ is a Borel probability on ${\mathbb{R}}^d\times {\mathbb{R}}^d$ with marginals $\mu ,\nu$ by [F1], so it belongs to $\mathrm{\Gamma }(\mu ,\nu )$. Thus the coupling set in the definition is nonempty.

By the inequality in [F2] and the finite second moments of $\mu ,\nu$, Tonelli gives $\int \mathrm{\mid }x-y{\mathrm{\mid }}^{2}d(\mu \otimes \nu )\le 2\int \mathrm{\mid }x{\mathrm{\mid }}^{2}d\mu +2\int \mathrm{\mid }y{\mathrm{\mid }}^{2}d\nu <\mathrm{\infty }\mathrm{.}$ All coupling costs are nonnegative, so their infimum is finite and nonnegative; its square root in the displayed definition is therefore a finite real number.

Let $G\subseteq {\mathbb{R}}^d$ be open and let $t_n\to t$. Narrow continuity gives ${\mu }_{t_n}\Rightarrow {\mu }_t$, so [F2] gives ${\mu }_t(G)\le {\mathrm{lim\hspace{0.17em}inf}}_n{\mu }_{t_n}(G)$. Thus $t\mapsto {\mu }_t(G)$ is lower semicontinuous and hence Borel measurable.

Let $\mathcal{D}$ be the Borel sets $A\subseteq {\mathbb{R}}^d$ for which $t\mapsto {\mu }_t(A)$ is measurable. It contains the whole space because each section is a probability. If $A\in \mathcal{D}$, then ${\mu }_t(A^c)=1-{\mu }_t(A)$, so $A^c\in \mathcal{D}$. For pairwise disjoint $A_j\in \mathcal{D}$, countable additivity gives ${\mu }_t({\bigcup }_j{A}_j)={\lim }_{N\to \mathrm{\infty }}{\sum }_{j=0}^N{\mu }_t(A_j)$, a measurable increasing limit. Hence $\mathcal{D}$ is a lambda-system. The open sets form a pi-system generating the Borel sigma-algebra and belong to $\mathcal{D}$ by step 1.1, so [F3] and [F9] give $\mathcal{D}=\mathcal{B}({\mathbb{R}}^d)$. Together with [F4], this proves that $t\mapsto {\mu }_t$ is a probability kernel.

By [F5], a product-Borel set $E$ is measurable for $\mathcal{B}([0,T])\otimes \mathcal{B}({\mathbb{R}}^d)$. Its sections $E_t$ are Borel by [F6]. Since $t\mapsto {\mu }_t$ is a probability kernel by step 2.1, [F7] applied to ${\mathbf{1}}_E$ shows that $t\mapsto {\mu }_t(E_t)$ is measurable. Thus $M(E)={\int }_0^T{\mu }_t(E_t)dt$ is defined; $M(\mathrm{\varnothing })=0$ and $M([0,T]\times {\mathbb{R}}^d)={\int }_0^{T}1dt=T$.

If $E_j$ are disjoint product-Borel sets, their sections are disjoint and ${\mu }_t(({\bigcup }_j{E}_j{)}_t)={\sum }_j{\mu }_t((E_j{)}_t)$. Integrating the increasing partial sums and using [F8] gives $M({\bigcup }_j{E}_j)={\sum }_{j}M(E_j)$; with step 2.1 this makes $M$ a finite Borel measure. For an indicator, the iterated-integral formula is the definition of $M$; finite nonnegative simple combinations give it for simple functions. For arbitrary nonnegative product-Borel $f$, use [F9] to choose $s_n\uparrow f$ and apply [F8] to both $M$ and the time-section integrals. This proves $\int fdM={\int }_0^T\int f(t,x)d{\mu }_t(x)dt$.

Apply [F2] to $\alpha$ with its second coordinate y as conditioning variable, obtaining a probability kernel $K_y(dx)$ with $\alpha (A\times B)={\int }_B{K}_y(A)d\rho (y)$. The supplier conditions on the second marginal, so for $\beta$ first define the coordinate swap $S(y,z)=(z,y)$ and $\widehat{\beta }=S_{\mathrm{\#}}\beta$ on ${\mathbb{R}}^{d_3}\times {\mathbb{R}}^{d_2}$. By [F7], this is a Borel probability whose second marginal is the common law $\rho$. Apply [F2] to $\widehat{\beta }$ to obtain a probability kernel $L_y(dz)$ with $\widehat{\beta }(C\times B)={\int }_B{L}_y(C)d\rho (y)$, equivalently $\beta (B\times C)={\int }_B{L}_y(C)d\rho (y)$. Thus for every Borel $A\subseteq {\mathbb{R}}^{d_1}$, $B\subseteq {\mathbb{R}}^{d_2}$, and $C\subseteq {\mathbb{R}}^{d_3}$ both joint laws are disintegrated over the same conditioning law.

For each $y$, let $Q_y=K_y\otimes L_y$, the product probability on ${\mathbb{R}}^{d_1}\times {\mathbb{R}}^{d_3}$. For a rectangle $A\times C$, $Q_y(A\times C)=K_y(A)L_y(C)$ is measurable in $y$. The sets $D$ for which $y\mapsto Q_y(D)$ is measurable form a lambda-system: the whole space has value one, complements have value $1-Q_y(D)$, and disjoint unions have value the increasing limit of finite sums. Rectangles form a pi-system generating the product Borel sigma-algebra by [F1] and [F4]. Thus [F4] proves measurability for every Borel $D$. The same lambda-system argument applied to rectangles $B\times A\times C$ in the variables $(y,x,z)$ shows that $y\mapsto Q_y(E_y)$ is measurable for each Borel $E\subseteq {\mathbb{R}}^{d_2}\times {\mathbb{R}}^{d_1}\times {\mathbb{R}}^{d_3}$, where $E_y=\{(x,z):(y,x,z)\in E\}$.

Define $\tilde{\gamma }(E)=\int Q_y(E_y)d\rho (y)$ on the product Borel sigma-algebra in the coordinate order $(y,x,z)$. It is nonnegative and has total mass one. For disjoint $E_j$, their sections are disjoint, so countable additivity of $Q_y$ followed by monotone convergence [F6] gives $\tilde{\gamma }({\bigcup }_j{E}_j)={\sum }_j\tilde{\gamma }(E_j)$. Hence $\tilde{\gamma }$ is a Borel probability.

For Borel rectangles $B\times A\times {\mathbb{R}}^{d_3}$ and $B\times {\mathbb{R}}^{d_1}\times C$, the definition and the two identities in step 1.1 give respectively $\tilde{\gamma }(B\times A\times {\mathbb{R}}^{d_3})={\int }_B{K}_y(A)L_y({\mathbb{R}}^{d_3})d\rho (y)=\alpha (A\times B)$ and $\tilde{\gamma }(B\times {\mathbb{R}}^{d_1}\times C)={\int }_B{L}_y(C)K_y({\mathbb{R}}^{d_1})d\rho (y)=\beta (B\times C)$. Rectangles determine finite Borel measures by [F4], so these identities identify the two pair marginals in coordinate order $(y,x,z)$. Define the Borel coordinate permutation $P(y,x,z)=(x,y,z)$ and set $\gamma =P_{\mathrm{\#}}\tilde{\gamma }$, meaning $\gamma (H)=\tilde{\gamma }(P^{-1}(H))$ for each Borel $H$; by [F7] this is a Borel probability. The coordinate projections satisfy ${\mathrm{pr}}_{x,y}\circ P(y,x,z)=(x,y)$ and ${\mathrm{pr}}_{y,z}\circ P(y,x,z)=(y,z)$, hence their pushforwards are exactly $\alpha$ and $\beta$.

The product probability $\mu \otimes \nu$ is a coupling by [F1]. The inequality in [F5] gives its cost at most $2\int \mathrm{\mid }x{\mathrm{\mid }}^{2}d\mu +2\int \mathrm{\mid }y{\mathrm{\mid }}^{2}d\nu <\mathrm{\infty }$. Thus the coupling set is nonempty and its cost infimum $c$ is finite and nonnegative.

For each integer $n\ge 1$, by AC choose a coupling ${\pi }_n$ with $\int \mathrm{\mid }x-y{\mathrm{\mid }}^{2}d{\pi }_n\le c+1\mathrm{/}n$. Their marginals are the fixed probabilities $\mu ,\nu$. For any $\epsilon >0$, tightness of each marginal supplies compact sets $K_1,K_2$ with $\mu (K_1^c)<\epsilon \mathrm{/}2$ and $\nu (K_2^c)<\epsilon \mathrm{/}2$. Then ${\pi }_n((K_1\times K_2{)}^c)\le \mu (K_1^c)+\nu (K_2^c)<\epsilon$ for every $n$. Hence $({\pi }_n)$ is tight on the Polish space ${\mathbb{R}}^{2d}$; [F2] gives a subsequence ${\pi }_{n_j}\Rightarrow \pi$. By [F3] its marginals are $\mu ,\nu$.

For $N\ge 1$, the function $c_N(x,y)=\min \{\mathrm{\mid }x-y{\mathrm{\mid }}^2,N\}$ is bounded and continuous. Weak convergence therefore gives $\int c_{N}d\pi ={\lim }_j\int c_{N}d{\pi }_{n_j}\le {\mathrm{lim\hspace{0.17em}inf}}_j\int \mathrm{\mid }x-y{\mathrm{\mid }}^{2}d{\pi }_{n_j}\le c$. Letting $N\uparrow \mathrm{\infty }$ and using [F4] yields $\int \mathrm{\mid }x-y{\mathrm{\mid }}^{2}d\pi \le c$. Since $\pi$ is a coupling and $c$ is the infimum, equality holds; the infimum is attained.

Put $L={\mathrm{lim\hspace{0.17em}inf}}_n{W}_2({\mu }_n,{\nu }_n{)}^2$. If $L=+\mathrm{\infty }$, the asserted inequality is immediate. Otherwise choose a subsequence indexed by integers $j\ge 1$, denoted $n_j$, along which the costs converge to $L$, and choose by AC couplings ${\sigma }_j$ with costs at most $W_2({\mu }_{n_j},{\nu }_{n_j}{)}^2+1\mathrm{/}j$. The marginal sequences are tight by [F2]; choosing compact sets for each coordinate and applying the same union bound as in step 2.1 proves $({\sigma }_j)$ is tight on ${\mathbb{R}}^{2d}$. A further subsequence converges weakly to a probability $\sigma$, whose marginals are $\mu ,\nu$ by [F3]. For every $N$, bounded continuity of $c_N$ gives $\int c_{N}d\sigma \le L$. Monotone convergence [F4] then gives $\int \mathrm{\mid }x-y{\mathrm{\mid }}^{2}d\sigma \le L$. The definition of $W_2$ and the coupling property imply $W_2(\mu ,\nu {)}^2\le L$.

The product coupling has finite cost by [F1], so $W_2$ is finite and nonnegative. The diagonal map $\mathrm{\Delta }(x)=(x,x)$ is Borel, and its pushforward ${\mathrm{\Delta }}_{\mathrm{\#}}\mu$ is a Borel probability by [F4]. Its two coordinate marginals are $({\mathrm{pr}}_i\circ \mathrm{\Delta }{)}_{\mathrm{\#}}\mu =\mu$, and its cost is $\int \mathrm{\mid }\mathrm{\Delta }(x{)}_1-\mathrm{\Delta }(x{)}_2{\mathrm{\mid }}^{2}d\mu (x)=0$ by [F4], hence $W_2(\mu ,\mu )=0$. The coordinate swap $S(x,y)=(y,x)$ is Borel; for every coupling $\pi$ of $\mu ,\nu$, [F4] shows $S_{\mathrm{\#}}\pi$ is a probability with first marginal $\nu$ and second marginal $\mu$, while its cost is $\int \mathrm{\mid }S(x,y{)}_1-S(x,y{)}_2{\mathrm{\mid }}^{2}d\pi =\int \mathrm{\mid }x-y{\mathrm{\mid }}^{2}d\pi$ by [F4]. Since $S$ is its own inverse, this correspondence is onto and preserves costs, proving equality of the two infima.

If $W_2(\mu ,\nu )=0$, choose an optimal coupling $\pi$ using [F1]. Its integral of $\mathrm{\mid }x-y{\mathrm{\mid }}^2$ is zero. For each positive integer $k$, the set $\{\mathrm{\mid }x-y\mathrm{\mid }\ge 1\mathrm{/}k\}$ has $\pi$-measure zero, since its contribution to the integral is at least $k^{-2}$ times its measure. Their union is the complement of the diagonal, so $\pi$ is concentrated on the diagonal. For every Borel set $A$, both coordinate indicators agree $\pi$-almost everywhere, giving $\mu (A)=\pi (A\times {\mathbb{R}}^d)=\pi ({\mathbb{R}}^d\times A)=\nu (A)$. Conversely, if $\mu =\nu$, step 1.1 gives distance zero.

Choose optimal couplings ${\pi }_{12}$ of $(\mu ,\nu )$ and ${\pi }_{23}$ of $(\nu ,\lambda )$ using [F1]. Glue them by [F2] to obtain a Borel probability $\gamma$ on ${\mathbb{R}}^{3d}$. The Borel projection $Q(x,y,z)=(x,z)$ has pushforward $Q_{\mathrm{\#}}\gamma$ a Borel probability by [F4]; its coordinate marginals are $\mu$ and $\lambda$, so it is a coupling. By [F4] its cost is $\int \mathrm{\mid }u-v{\mathrm{\mid }}^{2}d(Q_{\mathrm{\#}}\gamma )(u,v)=\int \mathrm{\mid }x-z{\mathrm{\mid }}^{2}d\gamma (x,y,z)$. Pointwise, $\mathrm{\mid }x-z\mathrm{\mid }\le \mathrm{\mid }x-y\mathrm{\mid }+\mathrm{\mid }y-z\mathrm{\mid }$; applying [F3] on $({\mathbb{R}}^{3d},\gamma )$ gives $W_2(\mu ,\lambda )\le {(\int \mathrm{\mid }x-z{\mathrm{\mid }}^{2}d\gamma )}^{1\mathrm{/}2}\le W_2(\mu ,\nu )+W_2(\nu ,\lambda )\mathrm{.}$ This proves the triangle inequality and all metric axioms.

For each integer $k\ge 1$, the Cauchy property gives a threshold N_k such that $W_2({\mu }_n,{\mu }_m)<2^{-k}$ whenever $n,m\ge N_k$. Set $n_1=N_1$ and recursively $n_{k+1}=\max \{N_{k+1},n_k+1\}$. Then $n_k<n_{k+1}$ and $W_2({\mu }_{n_k},{\mu }_{n_{k+1}})\le 2^{-k}$. By AC choose an optimal coupling ${\pi }_k$ of these consecutive laws for each k using [F1].

Put $S=({\mathbb{R}}^d{)}^{\{1,2,\dots \}}$, the set of sequences of Euclidean vectors indexed by positive integers, and equip $S$ with $D(x,y)={\sum }_{j\ge 1}{2}^{-j}\min (1,\mathrm{\mid }{x}_j-y_j\mathrm{\mid })$. This metric induces the product topology: each coordinate projection is continuous from the bound on its summand, and a product neighborhood restricting finitely many coordinates contains a sufficiently small D-ball. Conversely, given a D-ball radius, make the geometric tail less than half that radius and restrict the finitely many remaining coordinates so their summands total less than the other half; this gives a product neighborhood inside the D-ball. A D-Cauchy sequence is Cauchy in each coordinate, so coordinate completeness gives a limit in S; bounding the first finitely many coordinate terms and then the geometric tail proves convergence in D. Sequences of rational vectors which are zero after finitely many coordinates form a countable dense subset, since the tail is arbitrarily small and each of finitely many coordinates can be approximated by rationals. Thus S is complete and separable, hence Polish by definition.

Set $Q_1={\pi }_1$, the probability on the first two coordinates. For $N\ge 2$, assume $Q_{N-1}$ is a probability on coordinates $1,\dots ,N$ with adjacent pair marginals ${\pi }_1,\dots ,{\pi }_{N-1}$ and last marginal ${\mu }_{n_N}$. Regard it as a law on ${\mathbb{R}}^{d(N-1)}\times {\mathbb{R}}^d$, and regard ${\pi }_N$ as a law on ${\mathbb{R}}^d\times {\mathbb{R}}^d$; they have the common second/first marginal ${\mu }_{n_N}$. The gluing statement [F1] gives $Q_N$ on coordinates $1,\dots ,N+1$ preserving all earlier adjacent marginals and adding ${\pi }_N$. Let $I_N:({\mathbb{R}}^d{)}^{N+1}\to S$ send $(x_1,\dots ,x_{N+1})$ to $(x_1,\dots ,x_{N+1},0,0,\dots )$; this map is continuous for the product topology. Define ${\eta }_N=(I_N{)}_{\mathrm{\#}}{Q}_N$, that is, ${\eta }_N(B)=Q_N(I_N^{-1}(B))$ for Borel $B\subset S$. By [F7] it is a Borel probability, and its first $N+1$ coordinates have law $Q_N$ while every later coordinate equals zero.

The $j$th marginal of any ${\eta }_N$ is either ${\mu }_{n_j}$ (when $N+1\ge j$) or ${\delta }_0$. Each fixed probability ${\mu }_{n_j}$ is tight by [F2]; AC chooses a compact $K_j\subset {\mathbb{R}}^d$ containing zero with ${\mu }_{n_j}(K_j^c)<\epsilon 2^{-j-1}$. Countable subadditivity gives ${\eta }_N(({\prod }_j{K}_j{)}^c)\le {\sum }_{j\ge 1}\epsilon 2^{-j-1}<\epsilon$ for every $N$. The product $K={\prod }_j{K}_j$ is compact in $S$: it is closed and complete, and it is totally bounded because for any radius one first makes the metric tail small and then takes finite nets in the finitely many compact coordinates; [F5] applies. Hence $({\eta }_N)$ is tight. By [F2] and Prokhorov, a subsequence converges weakly to a probability $\eta$ on $S$.

For fixed $k$, all sufficiently late ${\eta }_N$ have their $(k,k+1)$ coordinate marginal equal to ${\pi }_k$. The coordinate projection $S\to {\mathbb{R}}^{2d}$ is continuous, so the projected laws of ${\eta }_N$ converge weakly to the corresponding marginal of $\eta$; because they are eventually exactly ${\pi }_k$, uniqueness of weak limits [F6] shows that this marginal is ${\pi }_k$. Let $X_k(x)=x_k$ be the coordinate maps on $(S,\eta )$. By [F4] and the pair marginal identity, ${\int }_S\mathrm{\mid }{X}_{k+1}-X_k{\mathrm{\mid }}^{2}d\eta =\int \mathrm{\mid }x-y{\mathrm{\mid }}^{2}d{\pi }_k=W_2({\mu }_{n_k},{\mu }_{n_{k+1}}{)}^2\le 4^{-k}\mathrm{.}$ Therefore $\int \mathrm{\mid }{X}_{k+1}-X_k\mathrm{\mid }d\eta \le 2^{-k}$ by Cauchy–Schwarz. Monotone convergence [F4] shows $\int {\sum }_k\mathrm{\mid }{X}_{k+1}-X_k\mathrm{\mid }d\eta \le {\sum }_k{2}^{-k}<\mathrm{\infty }$, so this sum is finite almost everywhere. The convergence set is Borel; its pointwise limit there, extended by zero on the complement, is a Borel map $X_{\mathrm{\infty }}:S\to {\mathbb{R}}^d$.

Minkowski [F3] gives, for $m>n$, $\mathrm{\parallel }{X}_m-X_n{\mathrm{\parallel }}_{L^2(\eta )}\le {\sum }_{k=n}^{m-1}\mathrm{\parallel }{X}_{k+1}-X_k{\mathrm{\parallel }}_2\le {\sum }_{k=n}^{m-1}{2}^{-k}$. Fatou's lemma [F3] as $m\to \mathrm{\infty }$ yields $\mathrm{\parallel }{X}_{\mathrm{\infty }}-X_n{\mathrm{\parallel }}_2\le {\sum }_{k=n}^{\mathrm{\infty }}{2}^{-k}$. Since $X_1$ has law ${\mu }_{n_1}\in {\mathcal{P}}_2$, the same inequality at $n=1$ and Minkowski imply $\int \mathrm{\mid }{X}_{\mathrm{\infty }}{\mathrm{\mid }}^{2}d\eta <\mathrm{\infty }$. Let ${\mu }_{\mathrm{\infty }}=(X_{\mathrm{\infty }}{)}_{\mathrm{\#}}\eta$, a Borel probability by [F7]. The map $x\mapsto (X_n(x),X_{\mathrm{\infty }}(x))$ is Borel, so its pushforward is a Borel probability by [F7]; its marginals are the laws of $X_n$ and $X_{\mathrm{\infty }}$, namely ${\mu }_{n_n}$ and ${\mu }_{\mathrm{\infty }}$. Thus ${\mu }_{\mathrm{\infty }}\in {\mathcal{P}}_2$, and this joint law is a coupling, so $W_2({\mu }_{n_n},{\mu }_{\mathrm{\infty }})\le \mathrm{\parallel }{X}_n-X_{\mathrm{\infty }}{\mathrm{\parallel }}_2\le {\sum }_{k=n}^{\mathrm{\infty }}{2}^{-k}⟶0.$ The original Cauchy sequence converges to the same limit: for any $ϵ>0$, choose n large enough that the displayed tail bound is below $ϵ\mathrm{/}2$ and $n_n$ exceeds the Cauchy threshold for $ϵ\mathrm{/}2$; then $W_2({\mu }_m,{\mu }_{n_n})<ϵ\mathrm{/}2$ for all sufficiently large m and apply the triangle inequality [F1].

Suppose $W_2({\mu }_n,\mu )\to 0$. For each $n$ choose an optimal coupling ${\pi }_n$ and regard its coordinates as $X_n,Y$ with laws ${\mu }_n,\mu$. The cost gives $\int \mathrm{\mid }{X}_n-Y{\mathrm{\mid }}^{2}d{\pi }_n\to 0$. For bounded continuous $\phi$, choose a ball $B(0,K)$ with $\mu (B(0,K{)}^c)$ small. On the event $\mathrm{\mid }Y\mathrm{\mid }\le K$ and $\mathrm{\mid }{X}_n-Y\mathrm{\mid }\le \delta <1$, both points lie in the compact ball $\bar{B}(0,K+1)$, where $\phi$ is uniformly continuous. The complement has probability at most $\mu (B(0,K{)}^c)+{\delta }^{-2}\int \mathrm{\mid }{X}_n-Y{\mathrm{\mid }}^{2}d{\pi }_n$. First choose $K$, then $\delta$, and then $n$; the boundedness of $\phi$ shows $\int \phi d{\mu }_n-\int \phi d\mu \to 0$. Thus ${\mu }_n\Rightarrow \mu$.

Conversely suppose narrow convergence and the stated uniform tail condition. Fix $\epsilon >0$. Choose $R$ so that the second-moment tails outside $Q=[-R,R{]}^d$ of all ${\mu }_n$ and of $\mu$ are less than $\epsilon$; the tail of $\mu$ vanishes because $\mu \in {\mathcal{P}}_2$. Enlarge R slightly if necessary so its coordinate faces have zero $\mu$-mass. Partition each coordinate interval $[-R,R]$ by finitely many levels that are not atoms of the corresponding one-dimensional marginal of $\mu$, with mesh small enough that each resulting rectangular cell has diameter at most $\epsilon$. Such levels exist because each probability has at most countably many atoms. Together with $Q^c$, these cells form a finite Borel partition $(C_i{)}_{i=0}^N$ whose boundaries have $\mu$-measure zero. By [F2], $a_{n,i}:={\mu }_n(C_i)\to a_i:=\mu (C_i)$ for each $i$.

Write $Z_n=X_n-Y$ and $a_n=\int \mathrm{\mid }{Z}_n{\mathrm{\mid }}^{2}d{\pi }_n\to 0$. For $R>K$, on $\{\mathrm{\mid }{X}_n\mathrm{\mid }>R\}$ either $\mathrm{\mid }Y\mathrm{\mid }>K$ or $\mathrm{\mid }{Z}_n\mathrm{\mid }>R-K$. Since $\mathrm{\mid }{X}_n{\mathrm{\mid }}^2\le 2\mathrm{\mid }Y{\mathrm{\mid }}^2+2\mathrm{\mid }{Z}_n{\mathrm{\mid }}^2$, for every $n$, ${\int }_{\mathrm{\mid }{X}_n\mathrm{\mid }>R}\mathrm{\mid }{X}_n{\mathrm{\mid }}^{2}d{\pi }_n\le 2{\int }_{\mathrm{\mid }Y\mathrm{\mid }>K}\mathrm{\mid }Y{\mathrm{\mid }}^{2}d{\pi }_n+2K^2\frac{a_n}{(R-K{)}^2}+2a_n\mathrm{.}$ Given a target $\eta >0$, first choose K so the first term is below $\eta \mathrm{/}4$; then choose N so $a_n<\eta \mathrm{/}8$ for $n\ge N$; then choose R>K so the middle term is below $\eta \mathrm{/}8$ for all $n\ge N$. The last term is below $\eta \mathrm{/}4$. Finally enlarge R so the finitely many tails for $n<N$ are below $\eta$. This proves the required uniform tail limit.

For each bounded cell $C_i\subset Q$, match mass $r_{n,i}=\min (a_{n,i},a_i)$. If $r_{n,i}>0$, take $r_{n,i}$ times the product of the probability measures ${\mu }_n{\mathrm{\mid }}_{C_i}\mathrm{/}{a}_{n,i}$ and $\mu {\mathrm{\mid }}_{C_i}\mathrm{/}{a}_i$; if $r_{n,i}=0$, use the zero measure. This gives a coupling piece supported on $C_i\times C_i$ with cost at most ${\epsilon }^2{r}_{n,i}$. Subtract its two marginals from ${\mu }_n$ and $\mu$. The residual measures have equal mass. If that mass is positive, their normalized product, multiplied by that mass, couples them; if it is zero there is no residual piece. The sum is a coupling ${\pi }_n$ of ${\mu }_n,\mu$. Its matched cost is at most ${\epsilon }^2$. For the residual coupling, $\mathrm{\mid }x-y{\mathrm{\mid }}^2\le 2\mathrm{\mid }x{\mathrm{\mid }}^2+2\mathrm{\mid }y{\mathrm{\mid }}^2$. Its residual second moments are bounded by the two tails outside $Q$ plus $d{R}^2{\sum }_{i=1}^N\mathrm{\mid }{a}_{n,i}-a_i\mathrm{\mid }$ on each side: inside a bounded cell, unmatched mass is exactly the positive part of the difference of its two cell masses. Therefore $W_2({\mu }_n,\mu {)}^2\le \int \mathrm{\mid }x-y{\mathrm{\mid }}^{2}d{\pi }_n\le {\epsilon }^2+4\epsilon +4d{R}^2{\sum }_{i=1}^N\mathrm{\mid }{a}_{n,i}-a_i\mathrm{\mid }\mathrm{.}$ For this fixed finite partition the sum tends to zero, so ${\mathrm{lim\hspace{0.17em}sup}}_n{W}_2({\mu }_n,\mu {)}^2\le {\epsilon }^2+4\epsilon$. Letting $\epsilon \downarrow 0$ proves $W_2({\mu }_n,\mu )\to 0$.