Let $\mathcal{M}$ be the family comprising the laws of such $(X,Y)$, that satisfy
$X=\mathbb{E}(Y|X),$
$Y\in [0,1].$
In one article, I have found the following claim
It is well known that $ \text{ext}(\mathcal{M}) $ is the set of $ 1 \times 2 $ laws of $ (p, Y_p) $ for two-valued $ Y_p \in [0, 1] $ with $ \mathbb{E}(Y_p) = p$.
Here $\text{ext}(\mathcal{M})$ stands for the set of extreme points of $\mathcal{M}$. In the article however, there is no reference given. Do You know how one proves this fact? I would be extremely grateful if You could point me to the article/book in which I could find the proof of this (and maybe luckily, some more context).
A probability measure $\gamma$ on $[0,1]^2$ admits an essentially unique disintegration $$ \gamma(dx,dy) =\mu(dx)\pi_x(dy), $$ in which $\mu$ is the "first marginal" of $\gamma$ ($\mu(A) =\gamma(A\times[0,1])$) and $x\mapsto\pi_x$ is a $\mu$-a.s. uniquely determined measurable map from $[0,1]$ to the class of probability measure on $[0,1]$. (If $\gamma$ is the law of $(X,Y)$ then $\mu$ is the law of $X$ and $\pi_x$ is the conditional distribution of $Y$ given $X=x$.)
What distinguishes an element of $\mathcal M$ is the a.s. equality $$ \int_{[0,1]} y\,\pi_x(dy) = x,\qquad \mu\hbox{-a.e} x\in[0,1]. $$ In particular, if $\gamma$ is to be extreme in $\mathcal M$ then $\mu$ must be extreme as well; that is, $\mu$ must be a point mass at some $p\in[0,1]$. So if $\gamma$ is extreme then $\mu=\delta_p$ (point mass) and there is really only $\pi=\pi_p$ to reckon with. The martingale property means that $\int_{[0,1]}y\pi(dy)=p$. If $\pi$ is a two-point distribution, at $y_1<y_2$, say, then the mean condition forces $\pi\{y_1\} =(y_2-p)/(y_2-y_1)$ and then $\pi\{y_2\}=1-\pi\{y_1\}$. The $\gamma$ associated with such a $\pi$ is clearly extremal.
Conversely, if $\gamma$ is an extremal element of $\mathcal M$ then $\gamma$ admits a disintegration as above with $\mu=\delta_p$ for some $p\in[0,1]$ and $\pi$ a probability measure on $[0,1]$ that is extremal in the class of probability measures on $[0,1]$ with mean value $p$. It is known that this implies that the support of $\pi$ contains at most two points. (By a famous theorem of R.G. Douglas, the extremality of $\pi$ is equivalent to the statement that $L^1(\pi)$ is the linear span of the two functions $\{1,x\}$. From here it's a short step to see that all the mass of $\pi$ is concentrated on one or two points.) A nice discussion of this can be found in the paper "Extreme Points of Certain Sets of Probability Measures, with Applications" by A.F. Karr (https://www.jstor.org/stable/3689412).