The Kantorovich relaxation is useful because minimizers survive limiting operations. The proof is a clean instance of the direct method: compactness of feasible plans supplies a limit, closed marginal constraints preserve feasibility, and lower semicontinuity prevents the cost from dropping in the limit.
Lower semicontinuity of integral costs
Portmanteau lemma for costs. Let \(Z\) be Polish, \(\eta_n\Rightarrow\eta\), and let \(f:Z\to(-\infty,\infty]\) be lower semicontinuous and bounded below. Then \[ \int_Z f\,\mathrm{d}\eta \le\liminf_{n\to\infty}\int_Z f\,\mathrm{d}\eta_n. \]
Proof. After adding a constant, assume \(f\ge0\). There exist bounded continuous functions \(f_k\uparrow f\); for example one may use the infimal regularizations \(f_k(z)=\min\{k,\inf_w(f(w)+k\,d(z,w))\}\). For fixed \(k\), \[ \int f_k\,\mathrm{d}\eta=\lim_n\int f_k\,\mathrm{d}\eta_n \le\liminf_n\int f\,\mathrm{d}\eta_n. \] Letting \(k\to\infty\) and applying monotone convergence proves the claim.
Existence by the direct method
Existence theorem. Let \(X,Y\) be Polish, \(\mu\in\mathcal P(X)\), \(\nu\in\mathcal P(Y)\), and \(c:X\times Y\to(-\infty,\infty]\) be lower semicontinuous and bounded below. If at least one \(\pi_0\in\Pi(\mu,\nu)\) has finite cost, then the Kantorovich problem admits an optimal plan.
Proof. Choose a minimizing sequence \(\pi_n\in\Pi(\mu,\nu)\). Page 1 showed that \(\Pi(\mu,\nu)\) is weakly compact, so a subsequence converges weakly to some \(\pi\in\Pi(\mu,\nu)\). The preceding lemma gives \[ \int c\,\mathrm{d}\pi \le\liminf_n\int c\,\mathrm{d}\pi_n =K_c(\mu,\nu). \] Since \(\pi\) is feasible, the reverse inequality follows from the definition of the infimum. Thus equality holds and \(\pi\) is optimal.
Nothing in this argument implies uniqueness or determinism. Those are geometric properties of the cost and the marginals, not consequences of compactness.
Stability when the marginals vary
Stability theorem. Suppose \(\mu_n\Rightarrow\mu\), \(\nu_n\Rightarrow\nu\), and \(\pi_n\in\Pi(\mu_n,\nu_n)\). Then \((\pi_n)\) is tight. Every weak limit point \(\pi\) belongs to \(\Pi(\mu,\nu)\). If \(c\) is lower semicontinuous and bounded below, then \[ \int c\,\mathrm{d}\pi\le\liminf_n\int c\,\mathrm{d}\pi_n. \]
Proof. Weakly convergent sequences of probability measures are uniformly tight. Choose compact \(K_X,K_Y\) carrying mass at least \(1-\varepsilon/2\) for all sufficiently large \(n\); enlarge them to handle the finitely many remaining indices. Then \[ \pi_n(K_X\times K_Y)\ge1-\mu_n(K_X^c)-\nu_n(K_Y^c)\ge1-\varepsilon. \] Prokhorov gives subsequential compactness. If \(\pi_{n_k}\Rightarrow\pi\), pushing forward by the continuous projections identifies the marginals of \(\pi\) as \(\mu,\nu\). The cost inequality is the Portmanteau lemma.
Convergence of optimal values
The preceding theorem gives the lower-bound half. To identify the limit as optimal, one also needs recovery plans whose costs approach that of a prescribed limiting plan.
Continuity for bounded continuous costs. If \(c\in C_b(X\times Y)\), \(\mu_n\Rightarrow\mu\), and \(\nu_n\Rightarrow\nu\), then \[ K_c(\mu_n,\nu_n)\longrightarrow K_c(\mu,\nu). \] Moreover, every weak limit of optimal plans for \((\mu_n,\nu_n)\) is optimal for \((\mu,\nu)\).
Proof. The lower bound follows from stability. For the upper bound, fix an optimal \(\pi\in\Pi(\mu,\nu)\). Because weak convergence on Polish spaces can be metrized by couplings, choose \(\alpha_n\in\Pi(\mu,\mu_n)\) and \(\beta_n\in\Pi(\nu,\nu_n)\) that concentrate near the diagonals. Glue \(\alpha_n\), \(\pi\), and \(\beta_n\), then take the \((\mu_n,\nu_n)\)-marginal \(\widetilde\pi_n\). The pair \((x_n,y_n)\) under \(\widetilde\pi_n\) converges in probability to \((x,y)\) under \(\pi\). Bounded continuity of \(c\) gives \[ \int c\,\mathrm{d}\widetilde\pi_n\to\int c\,\mathrm{d}\pi. \] Hence \(\limsup_nK_c(\mu_n,\nu_n)\le K_c(\mu,\nu)\). Combining the two inequalities proves convergence. If \(\pi_n\) are optimal and \(\pi_{n_k}\Rightarrow\bar\pi\), lower semicontinuity and convergence of the values force \(\bar\pi\) to attain \(K_c(\mu,\nu)\).
Unbounded costs and moment control
Proposition. Let \(X=Y=\mathbb R^d\), \(p\ge1\), and suppose \(\mu_n\to\mu\), \(\nu_n\to\nu\) in \(W_p\). If \(c\) is continuous and \[ |c(x,y)|\le C(1+|x|^p+|y|^p), \] then \(K_c(\mu_n,\nu_n)\to K_c(\mu,\nu)\), and limit points of optimal plans are optimal.
Proof. Truncate \(c\) by bounded continuous functions and apply the preceding theorem. The \(W_p\) convergence makes \((|x|^p)\) uniformly integrable under \(\mu_n\) and \((|y|^p)\) uniformly integrable under \(\nu_n\). Since every coupling has these marginals, the same tail bound holds uniformly over all admissible plans. The truncation error therefore tends to zero uniformly, which passes the bounded-cost conclusion to \(c\).
Failure without the hypotheses
Moment escape. Let \(\mu_n=(1-n^{-1})\delta_0+n^{-1}\delta_n\) and \(\mu=\delta_0\). Then \(\mu_n\Rightarrow\mu\), but \[ W_1(\mu_n,\mu)=\int|x|\,\mathrm{d}\mu_n(x)=1. \] Weak convergence alone does not control an unbounded transport cost: a vanishing amount of mass may travel an unbounded distance.
Loss of lower semicontinuity. On \(\mathbb R^2\), define \(c(0,0)=1\) and \(c(x,y)=0\) elsewhere. For \(\pi_n=\delta_{(1/n,0)}\Rightarrow\delta_{(0,0)}=\pi\), the costs satisfy \(\int c\,\mathrm{d}\pi_n=0\) but \(\int c\,\mathrm{d}\pi=1\). A discontinuous upward jump can destroy existence by allowing minimizing sequences to approach a more expensive limit.