Today during a seminar I heard that a measure on the unit square with uniform marginals that gives no weight above the main diagonal is the uniform measure on the diagonal.
I have no precise idea how to prove the statement. Any hint? Moreover, what does it exactly means to have uniform marginals?
Define $f:[0,1] \to [0,1] \times [0,1]$ by $f(x)=(x,x)$. Let $m$ be Lebesgue measure in $[0,1]$ and define $\mu$ on $[0,1] \times [0,1]$ by $\mu (E)=m(f^{-1}(E))$. This is the uniform measure on the diagonal. The first marginal of $\mu$ is the measure $\nu$ given by $\nu (A)= \mu(A \times [0,1])=m(f^{-1} (A \times [0,1])=m(A)$ so the first marginal is the uniform measure on $[0,1]$. Similarly, the second marginal is also the uniform measure on $[0,1]$. Of course, $\mu$ gives no weight above (or below) the diagonal. Let us show that $\mu$ is the only measure on the square which has the given properties: suppose $X ,Y$ are random variables with joint distribution $\mu$. Then $Y \leq X$ almost surely and $E(X-Y)=EX-EY=0$ so $X=Y$ almost surely. Hence $\mu$ is concentrated on the diagonal. Can you now use the marginals to verify that it is the uniform measure on the diagonal?