Wasserstein metric

Remarks

The condition on $\mu$ does not depend on the choice of $x_0$, by the triangle inequality. The product measure $\mu \otimes \nu$ is always a coupling, so $\Gamma(\mu, \nu)$ is not empty and $W_p(\mu, \nu) < \infty$. The function $W_p$ is a metric on $\mathcal{P}_p(M)$, and the infimum is attained.

Example: let $\delta_x$ be the Dirac measure at $x$. The only coupling of $\delta_x$ and $\delta_y$ is $\delta_{(x, y)}$, so $W_p(\delta_x, \delta_y) = d(x, y)$.

Non-example: on $M = \R$, the standard Cauchy distribution has $\int \abs{x} \, d\mu = \infty$. So it does not lie in $\mathcal{P}_1(\R)$, and $W_1$ is not defined for it.

On $\R$, with distribution functions $F_\mu$ and $F_\nu$, one has $W_1(\mu, \nu) = \int_{-\infty}^{\infty} \abs{F_\mu(t) - F_\nu(t)} \, dt$. In $\mathcal{P}_p(M)$, $W_p(\mu_n, \mu) \to 0$ if and only if $\mu_n \to \mu$ weakly and the $p$-th moments converge.