= Solution
Let $\Delta=\{(x,y):x=y\}$. For any <transport plan> $\pi$ and any <Borel set> $A$,
$$
\mu(A)-\nu(A)=\pi(A\times A^c)-\pi(A^c\times A).
$$
Both rectangles are contained in $\Delta^c$, so
$$
|\mu(A)-\nu(A)|\leq\pi(\Delta^c)=\mathbb K(\pi).
$$
Thus $\inf\mathbb K\geq\sup_A|\mu(A)-\nu(A)|$.
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 by
$$
d\alpha=\min(f,g)\,d\rho,\qquad r=1-\alpha(\mathbb R^d).
$$
Since $\int(f-g)\,d\rho=0$, the positive and negative parts have equal mass. Taking $A=\{f>g\}$ gives
$$
r=\int(f-g)_+\,d\rho=\frac12\int|f-g|\,d\rho
=\sup_A|\mu(A)-\nu(A)|.
$$
If $r=0$, then $\mu=\nu$ and the diagonal <transport plan> has zero cost. If $r>0$, put $\beta=\mu-\alpha$, $\gamma=\nu-\alpha$ and define
$$
\pi^\dagger=(\operatorname{Id},\operatorname{Id})_\#\alpha+\frac1r\,\beta\otimes\gamma.
$$
The residual measures each have mass $r$, so the <marginal distributions> of $\pi^\dagger$ 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 $\mathbb K(\pi^\dagger)=r$. This is a <maximal coupling>.
In the paper's convention for the zero-mass signed measure $\mu-\nu$, the answer is
$$
\boxed{\inf_{\pi\in\Pi(\mu,\nu)}\mathbb K(\pi)
=\sup_A|\mu(A)-\nu(A)|=\frac12\|\mu-\nu\|_{\mathrm{TV}}.}
$$
The 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>.
Back to article page
