Kantorovich–Rubinstein duality on compact second-countable metric spaces

Use [F1] at scale $1$ to obtain a finite net $z_1,\dots ,z_s$. Let $M:=2+{\max }_{k,\mathrm{\ell }}d(z_k,z_{\mathrm{\ell }})$, whose maximum exists by [F7]. For any $x,y\in X$, choose net points within $1$ of $x$ and $y$; [F8] gives $d(x,y)<M$. Hence an anchored feasible $f$ satisfies $\mathrm{\mid }f(x)\mathrm{\mid }\le d(x,x_0)<M$. For every $x$ and $r>0$, $d(x,y)<r$ implies $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }<r$, so the metric-topology definition makes $f$ continuous; [F13] then makes it Borel. Also $0\le f^{+},f^{-}\le M$. By [F9] and [F10], each nonnegative integral is at most $M\mu (X)=M$, so [F12] makes $f$ integrable and both dual integrals are finite. The zero function is feasible, so the dual objective set is nonempty and its supremum $D_X$ is a finite real number.

Fix $\epsilon >0$ and use [F1] to obtain distinct net points $x_1,\dots ,x_m$, Borel nearest-center cells $C_1,\dots ,C_m$, and a Borel map $Q$ with $Q(x)=x_i$ on $C_i$ and $d(x,Q(x))<\epsilon$. Set $p_i=\mu (C_i)$ and $q_i=\nu (C_i)$. Countable additivity and the finite Borel partition imply $p_i,q_i\ge 0$ and ${\sum }_i{p}_i={\sum }_i{q}_i=1$. Let $c_{\epsilon }$ be the minimum finite transport cost between these vectors on $F=\{x_1,\dots ,x_m\}$; [F4] ensures that this finite minimum exists and is attained.

By [F4], choose a minimizing finite coupling ${\pi }^{\epsilon }$ and a maximizing normalized finite 1-Lipschitz potential; their common finite primal and dual value is $c_{\epsilon }$.

For any $\gamma \in {\mathrm{\Pi }}_X(\mu ,\nu )$, set ${\pi }_{ij}:=\gamma (C_i\times C_j)$. Its row and column sums are $p_i,q_j$ by the marginal identities and finite additivity, so it is a finite coupling. On $C_i\times C_j$, [F8] gives $d(x_i,x_j)\le d(x_i,x)+d(x,y)+d(y,x_j)<d(x,y)+2\epsilon$. The cells $C_i\times C_j$ form a finite Borel partition of $X\times X$. Integrating the pointwise bounds on these cells and using finite additivity of the nonnegative integral [F11] gives ${\sum }_{i,j}{\pi }_{ij}d(x_i,x_j)\le \int d(x,y)d\gamma +2\epsilon$, since $\gamma (X\times X)=1$. Taking the infimum over $\gamma$ gives $c_{\epsilon }\le W_1^X(\mu ,\nu )+2\epsilon$.

Let $D_F$ be the maximum finite dual value. By [F4], $D_F=c_{\epsilon }$. For any feasible finite potential $(f_i)$, [F3] gives an extension $\tilde{f}$; put $h:=\tilde{f}-\tilde{f}(x_0)$. This anchors $h$ without changing either probability-difference objective. The function $h\circ Q$ is finite-valued Borel simple, so its integrals are ${\sum }_i{p}_{i}h(x_i)$ and ${\sum }_i{q}_{i}h(x_i)$ by [F10]. Pointwise, $\mathrm{\mid }h(x)-h(Q(x))\mathrm{\mid }<\epsilon$. For either probability measure $\rho \in \{\mu ,\nu \}$, let $r=h-h\circ Q$. It is integrable because both terms are bounded Borel functions. Monotonicity [F9] gives $0\le \int (r+\epsilon )d\rho$ and $0\le \int (\epsilon -r)d\rho$; [F10] gives $\int \epsilon d\rho =\epsilon$, and linearity [F12] yields $\mathrm{\mid }\int rd\rho \mathrm{\mid }\le \epsilon$. Hence each finite objective differs from the corresponding compact objective by at most $2\epsilon$, so $D_F\le D_X+2\epsilon$.

Conversely, let $g$ be any feasible compact potential and restrict it to $F$, subtracting $g(x_1)$ to normalize the first coordinate. The finite objective is unchanged by this shift because both probability vectors sum to one. The functions $g\circ Q$ are Borel simple and have the finite sums as their integrals, by [F10]. For each $\rho \in \{\mu ,\nu \}$, the integrable difference $r=g-g\circ Q$ satisfies $-\epsilon \le r\le \epsilon$; applying the two monotonicity and linearity inequalities detailed in step 2.3 gives $\mathrm{\mid }\int rd\rho \mathrm{\mid }\le \epsilon$. Thus each compact integral differs from its finite sum by at most $\epsilon$. By the definition of $D_F$ in step 2.3, every restricted finite objective is at most $D_F$, and $D_X\le D_F+2\epsilon$.

For any $\gamma \in {\mathrm{\Pi }}_X(\mu ,\nu )$ and feasible compact $f$, the Lipschitz constraint gives $f(x)-f(y)\le d(x,y)$. By step 1.1 and [F13], the functions $f\circ {\mathrm{p}\mathrm{r}}_1$ and $f\circ {\mathrm{p}\mathrm{r}}_2$ are bounded Borel, and the metric cost is bounded and Borel; hence they are integrable. The first-coordinate pushforward of $\gamma$ agrees with $\mu$ on every Borel set by the marginal condition, and the second-coordinate pushforward agrees with $\nu$. Equal measures integrate indicators, then simple functions, then all nonnegative Borel functions equally by [F10]; positive-negative decomposition and signed-integral linearity [F12] give equality for bounded signed Borel functions. Thus $\int (f(x)-f(y))d\gamma =\int fd\mu -\int fd\nu$. Since $d(x,y)-(f(x)-f(y))\ge 0$, [F9]–[F10] give a nonnegative integral for this difference, so [F12] implies the dual value is at most the cost of $\gamma$. Taking the infimum gives $D_X\le W_1^X(\mu ,\nu )$.

Apply [F2] to the minimizing plan ${\pi }^{\epsilon }$ of step 2.1. It gives a Borel coupling ${\gamma }^{\epsilon }\in {\mathrm{\Pi }}_X(\mu ,\nu )$ with cost at most $c_{\epsilon }+2\epsilon$. Hence the coupling set is nonempty and $W_1^X(\mu ,\nu )\le c_{\epsilon }+2\epsilon$.

Steps 2.2 and 3.3 give $\mathrm{\mid }{W}_1^X-c_{\epsilon }\mathrm{\mid }\le 2\epsilon$. Steps 2.3 and 3.1 give $\mathrm{\mid }{D}_X-D_F\mathrm{\mid }\le 2\epsilon$, and step 2.3 gives $D_F=c_{\epsilon }$. Combining with step 3.2 yields $D_X\le W_1^X\le c_{\epsilon }+2\epsilon =D_F+2\epsilon \le D_X+4\epsilon$. Since this holds for every $\epsilon >0$, $W_1^X(\mu ,\nu )=D_X$.