Extreme points of the set of two probability measures with same marginals I am interested in the extreme points of the set $S$ of pairs of probability measures on $[0, 1]^2$ having the same marginals. More specifically, $(\mu_1, \mu_2) \in S$, where $\mu_1$ and $\mu_2$ are two probability measures on $[0, 1]^2$, if:
$$\mu_1(A\times[0,1]) = \mu_2(A\times[0,1]), \forall A \in \mathcal{B}([0, 1]) \\
\mu_1([0,1] \times B) = \mu_2([0,1] \times B), \forall B \in \mathcal{B}([0, 1]).$$
This set is a convex set and I would like to characterize it's extreme points. My hypothesis is that the extreme points of this set are made of a finite number of Dirac measures.
It's clear that measures such as $(\delta_{a_1, b_1}, \delta_{a_1, b_1})$ for $(a_1, b_1) \in \mathbb{R}^2$, where $\delta_{a_1, b1}$ denotes the Dirac measure supported at $(a_1, b_1) \in [0, 1]^2$, are extreme points but you can also find more exotic ones such as:
$$\left( \frac{1}{2} (\delta_{a_1, b_1} + \delta_{a_2, b_2}), \frac{1}{2} (\delta_{a_1, b_2} + \delta_{a_2, b_1}) \right), ~\text{for}~a_1, a_2, b_1, b_2 \in [0, 1]^4.$$
You can construct extreme measures similar as this one with $n$ Dirac measures for all $n \in \mathbb{N}$. For a given $n$, they are equivalent to:
$$\left( \frac{1}{n} \sum_{k=1}^n \delta_{a_k, b_k}, \frac{1}{n}( \sum_{k=1}^{n-1} \delta_{a_k, b_{k+1}} + \delta_{a_n, b_1})\right), ~\text{for}~a_1, \ldots, a_n, b_1, \ldots, b_n \in [0, 1]^{2n}$$
Let's denote the set made of all extreme probability measures with a finite number of Dirac measures $A$. I've been able to prove that the convex hull of $A$ is dense in $S$. From Milman theorem it follows that the extreme points are included in the closure, but I'm unable to conclude from there.
Is there a way to conclude or to disprove my hypothesis ? I've looked at some elements of the closure but they are not extreme points.
 A: I could be misinterpreting your question/hypothesis, but this seems to be a counterexample:
Let $\mu_1$ be the "uniform" probability measure on the line $L_1 \subset [0, 1)^2$ given by $y = x$ and let $\mu_2$ be the "uniform" probability measure on the line $L_2 \subset [0, 1)^2$ given by $y = x + \alpha$ (mod $1$), where $\alpha$ is some irrational number. Specifically, for measurable $E \subset [0, 1]^2$ we have 
\begin{align*}
\mu_1(E) &= \mu(\{x \in [0, 1) \mid (x, x) \in E\}) \\
\mu_2(E) &= \mu(\{x \in [0, 1) \mid (x, T(x)) \in E\}),
\end{align*}
where $\mu$ is the Lebesgue measure on $[0, 1)$ and $T : [0, 1) \to [0, 1)$ is the irrational rotation given by $x \mapsto x + \alpha \text{ (mod $1$)}$. Note the marginals $\mu_{1x}, \mu_{1y}, \mu_{2x}, \mu_{2y}$ are all equal to $\mu$, so $(\mu_1, \mu_2) \in S$.
Suppose we had $(\mu_1, \mu_2) = \lambda(\mu_1', \mu_2') + (1-\lambda)(\mu_1'', \mu_2'')$ for some other $(\mu_1', \mu_2'), (\mu_1'', \mu_2'') \in S$ and $0 < \lambda < 1$. Then since $\mu_i([0, 1]^2 - L_i) = 0$, we must have $\mu_i'([0, 1]^2 - L_i) = \mu_i''([0, 1]^2 - L_i) = 0$ as well, hence $\mu_i', \mu_i''$ are supported on the line $L_i \subset [0, 1)^2$ for $i = 1, 2$. 
Now since $\mu_1'$ is supported on $L_1$, we must have $\mu_{1x}' = \mu_{1y}'$ because for any measurable $A \subset [0, 1)$, $\mu_{1x}'(A) = \mu_{1y}'(A) = \mu_1'(\{(a, a) \mid a \in A\})$. Similarly, since $\mu_2'$ is supported on $L_2$, we must have 
$$\mu_{2x}'(A) = \mu_2'(\{(a, T(a)) \mid a \in A\}) = \mu_{2y}'(T(A))$$
hence $\mu_{2x}' = \mu_{2y}' \circ T$. But because $(\mu_1', \mu_2') \in S$, we have $\mu_{1x}' = \mu_{2x}'$ and $\mu_{1y}' = \mu_{2y}'$, hence 
$$\mu_{1x}' = \mu_{1y}' = \mu_{2y}' = \mu_{2x}' \circ T^{-1} = \mu_{1x}' \circ T^{-1}$$
so $\mu_{1x}'$ is invariant under $T$. It's a standard result that the unique invariant measure under $T$ is the Lebesgue measure $\mu$ (see e.g. Example 4.11 of Einsiedler and Ward's Ergodic Theory), hence $\mu_{1x}' = \mu$, and thus $\mu_{1y}', \mu_{2x}', \mu_{2y}' = \mu$ as well. But since they're supported on $L_1$ and $L_2$ respectively, each of $\mu_1', \mu_2'$ is completely determined by either of its marginals, so $(\mu_1', \mu_2') = (\mu_1, \mu_2)$ (and the same holds for $(\mu_1'', \mu_2'')$). This means $(\mu_1, \mu_2)$ is an extreme point.
