given polish space $(X,d)$, consider weak* topology of probability. optimal transport of probability $u,v$ is defined by $\pi(u,v)$ such that $\pi(u,v)$ minimizes:
$\{\int d(x,y) d \pi(dx,dy): \pi \text{ is the couple of } u,v\}$
if $u_k \to u, v_k \to v$, can we show $\pi(u_k, v_k) \to \pi(u,v)$?
and how is the measurability w.r.t. the $\sigma$-alg generated by weak* top?
Of course not! This is hopeless if $X$ has unbounded diameter.
As a counterexample, consider $X=\mathbb Z$ with the standard metric; $\mu_k=(1-\frac 1k)\delta_0+\frac 1k\delta_k$; and $\nu_k=\nu=\mu=\delta_0$. Then $\nu_k\to\nu$, $\mu_k\to \mu$, but $\pi(\mu_k,\nu_k)=1$ for each $k$.
Anthony's example is based on the well-known fact that in the non-compact case the transportation distance does not actually metrize the weak topology unless one imposes additional moment conditions on the considered measures. However the actual point of this question is, in my opinion, that, to begin with, "the" optimal transportation plan ($\pi$ in the OP notation) is not well-defined, since the transportation functional can achieve its minimum on many distributions. Assume, for instance, that there are 4 points $\alpha_1,\alpha_2,\omega_1,\omega_2\in X$ with the property that all distances $d(\alpha_i,\omega_j)$ are equal to 1, and let $u$ and $v$ be two probability measures supported by the sets $\{\alpha_1,\alpha_2\}$ and $\{\omega_1,\omega_2\}$, respectively. Then the transportation functional has the same value 1 for any measure $\pi$ with the marginals $u,v$.
By using this observation one can easily construct an example, in which there are two sequences of measures $u_n',u_n''$ and a measure $v$ on a compact metric space with the following properties:
1) Both sequences weakly converge to a measure $u$;
2) The optimal transportation plans $\pi_n',\pi_n''$ for the couples $(u_n',v)$ and $(u_n'',v)$, respectively, are unique;
3) Both sequences $\pi_n'$ and $\pi_n''$ converge, but their limits are different.
