 Solution

Let $\Delta =\{(x,y):x=y\}$. For any transport plan $\pi$ and any Borel set $A$,Both rectangles are contained in ${\Delta }^c$, soThus $\inf \mathbb{K}\ge {\sup }_A|\mu (A)-\nu (A)|$.

To attain this bound, use the dominating measure $\rho =\mu +\nu$ and Radon-Nikodym derivatives $f=d\mu /d\rho$, $g=d\nu /d\rho$. Define the common measure and residual mass bySince $\int (f-g)d\rho =0$, the positive and negative parts have equal mass. Taking $A=\{f>g\}$ givesIf $r=0$, then $\mu =\nu$ and the diagonal transport plan has zero cost. If $r>0$, put $\beta =\mu -\alpha$, $\gamma =\nu -\alpha$ and defineThe residual measures each have mass $r$, so the marginal distributions of ${\pi }^{†}$ are $\alpha +\beta =\mu$ and $\alpha +\gamma =\nu$, and its total mass is $1-r+r=1$. They are mutually singular measures, concentrated respectively on $\{f>g\}$ and $\{g>f\}$. Their product measure therefore gives no mass to $\Delta$, and $\mathbb{K}({\pi }^{†})=r$. This is a maximal coupling.

In the paper's convention for the zero-mass signed measure $\mu -\nu$, the answer isThe repository's total variation distance uses the supremum itself. The paper's formula $2{\sup }_A|\sigma (A)|$ agrees with the usual total variation norm when $\sigma$ has total mass zero, as here; it is not the usual norm for an arbitrary positive measure. All suprema above are over Borel sets.