Let $X$ be a topological space equipped with the Borel $\sigma$-algebra. Inspired by this question whether a measure $\pi$ on the product space $X \times X$ having marginals $\mu = \delta_x$ and $\nu$ implies that $\pi = \mu \otimes \nu$, which is true, I wondered
If $\pi$ has marginals $\mu = \frac{1}{2} \delta_x + \frac{1}{2} \delta_y$ for $x \ne y \in X$ and $\nu$, must $\pi = \mu \otimes \nu$?
(The measure $\pi$ on $X \times X$ having marginals $\mu$ and $\nu$ means that $\pi(A \times X) = \mu(A)$ and $\pi(X \times B) = \nu(B)$ for all measurable subsets $A, B \subset X$.)
Let $A, B \subset X$ be measurable.
I have been able to adapt the proof from the question mentioned about for the case that $x, y \in A$ and that $x, y \not\in A$, but I am struggling to show the statement for the case that $x \not\in A$ but $y \in A$ (or the other way around).
My attempts so far. As the Borel-$\sigma$-Algebra is generated by measure rectangles of the form $A \times B \subset X \times X$, it suffices to show that $\mu(A) \nu(B) = \pi(A \times B)$.
If $x \not\in A$ but $y \in A$, then $\mu(A)\nu(B) = \frac{1}{2} \nu(B)$. We have \begin{equation*} 0 \le \pi\big( (A \setminus \{ y \}) \times B\big) \le \pi\big( (A \setminus \{ y \}) \times X\big) = \mu(A \setminus \{ y \}) = 0, \end{equation*} so $\pi\big( (A \setminus \{ y \}) \times B\big) = 0$. Hence $\pi(\{ y \} \times B) = \pi(A \times B)$. Analogously, we have \begin{align*} 0 \le \pi\big( \big((X \setminus A) \setminus \{ x \}\big) \times B\big) \le \pi\big( \big((X \setminus A) \setminus \{ x \}\big) \times X\big) = \mu\big((X \setminus A) \setminus \{ x \}\big) = 0, \end{align*} implying $\pi\big( \big((X \setminus A) \setminus \{ x \}\big) \times B\big) = 0$ or, equivalently, $ \pi\big( (X \setminus A) \times B\big) = \pi(\{ x \} \times B)$. These two equations imply \begin{equation*} \pi(\{ x, y \} \times B) = \pi\big( (X \setminus A) \times B) + \pi(A \times B) = \pi(X \times B) = \nu(B). \end{equation*} Hence it remains to show \begin{equation*} \frac{1}{2} \nu(B) = \frac{1}{2} \pi(\{ x, y \} \times B) \overset{!}{=} \pi(\{ y \} \times B), \end{equation*} which we can simplify to \begin{equation*} \pi(\{ x \} \times B) \overset{!}{=} \pi(\{ y \} \times B). \end{equation*} We have \begin{align*} 1 - \nu(B) & = \nu(X \setminus B) = \pi\big(X \times (X \setminus B)\big) \ge \pi\big(\{ x \} \times (X \setminus B)\big) \\ & = \pi(\{ x \} \times X) - \pi(\{ x \} \times B) = \frac{1}{2} - \pi(\{ x \} \times B), \end{align*} that is $\frac{1}{2} \ge \pi(\{ y \} \times B) - \pi(\{ x \} \times B)$.
The result is false. Take $X=\{0,1\}$ (with the discrete topology) and let $\nu=\frac12\delta_0 + \frac12 \delta_1$. Then $\pi=\frac12\delta_{(0,1)} + \frac12\delta_{(1,0)}$ has both marginals equal to $\nu$, but is not equal to $\nu\otimes \nu$.