 Solution

The Kantorovich–Rubinstein theorem concerns a Polish space $(X,\rho )$ and probability measures $\mu ,\nu$ with finite first moments, meaning $\int \rho (x,x_0)d\mu (x)+\int \rho (x,x_0)d\nu (x)<\infty$ for one, hence every, $x_0\in X$. It statesHere $\mathrm{Lip}(f)\le 1$ means $|f(x)-f(y)|\le \rho (x,y)$ for every $x,y$. The Wasserstein distance with ground cost $\rho$ therefore equals a supremum over functions with Lipschitz constant at most one. One may normalize $f(x_0)=0$, since adding a constant does not change the difference of integrals. This normalization bounds $|f(x)|$ by $\rho (x,x_0)$, so finite first moments ensure integrability. The supremum is unchanged if an absolute value is placed around the difference, because $-f$ is admissible whenever $f$ is.