The quadratic Wasserstein distance is a metric

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.