Optimal coupling of two measures Let,$\mu $ and $\nu$ be two probability measures on $\Omega$ such that $|\Omega| < \infty$. let $(X,Y)$ be an optimal coupling. How to prove that the optimal coupling is not unique , by finding a counter example?
 A: Here's a counterexample under my interpretation of the problem.
If this isn't what you want, you should add more detail to your question
so that we know precisely what you are asking.  
Let $\Omega=\{0,1,2,3\}$ and let $\mu,\nu$ be the uniform distribution over
$\{0,1\}$ and $\{0,1,2,3\}$, respectively. A coupling of these measures is 
a measure $m$ on the product space $\Omega\times\Omega$ whose marginals are 
$\mu$ and $\nu$. An optimal coupling is such a measure $m$ that minimizes
$$\inf\{\varepsilon>0: m((x,y): |x-y|\geq \varepsilon) \leq \varepsilon\}.$$
This infimum is achieved and the minimum value is $d(\mu,\nu)$, the 
distance between the two measures under the Lévy-Prohorov metric 
(see Theorem 1.2 in Chapter 3 (page 96) of Markov Processes: Characterization 
and Convergence by Ethier and Kurtz).
In our case, we want a probability measure $m$ on $\Omega\times\Omega$ 
that maximizes the probability of the diagonal, subject to 
$m(0,0)+m(1,0)+m(2,0)+m(3,0)=1/2$, $m(0,0)+m(0,1)=1/4$,
$m(1,0)+m(1,1)=1/4$,  and $m(2,0)+m(2,1)=1/4$.   
Here are two distinct solutions:
$$ m_1(0,0)=m_1(1,1)=m_1(2,0)=m_1(3,1)=1/4, \text{ otherwise zero} $$
$$ m_2(0,0)=m_2(1,1)=m_2(2,1)=m_2(3,0)=1/4, \text{ otherwise zero} .$$
In other words, outside the diagonal we can move 
the mass around in any way, as long as we respect the 
marginal distributions. 
