Context
Suppose I have two measurable spaces $(\mathsf{X}, \mathcal{X})$ and $(\mathsf{Y}, \mathcal{Y})$ on which we have two probability measures $\pi$ and $\nu$ respectively. I form a joint distribution $\mu = \pi\otimes \nu$ on the joint space $(\mathsf{Z}, \mathcal{Z}) := (\mathsf{X}\times\mathsf{Y}, \mathcal{X}\otimes \mathcal{Y})$.
Question
Let $\mathsf{C}\in\mathcal{Z}$. Is it true that there always exist $\mathsf{A}\in\mathcal{X}$ and $\mathsf{B}\in\mathcal{Y}$ such that $$ \mu(\mathsf{C}) = \pi(\mathsf{A})\nu(\mathsf{B}) $$
Attempted Solution
Given any two subsets $\mathsf{A}\subset\mathsf{X}$ and $\mathsf{B}\in\mathsf{Y}$ (not necessarily measurable subsets), the Cartesian product of these subsets is the set of all pairs as follows $$ \mathsf{A}\times\mathsf{B} := \left\{(x, y)\,:\, x\in\mathsf{A}, y\in\mathsf{B}\right\}. $$ Now, if these two sets are measurable, meaning $\mathsf{A}\in\mathcal{X}$ and $\mathsf{B}\in\mathcal{Y}$ then we call their Cartesian product $\mathsf{A}\times\mathsf{B}$ a measurable rectangle. Now, the sigma algebra $\mathcal{X}\otimes\mathcal{Y}$ is the sigma algebra generated by the collection of measurable rectangles $$ \mathcal{X}\otimes\mathcal{Y} := \sigma\left(\{\mathsf{A}\times\mathsf{B}\,:\, \mathsf{A}\in\mathcal{X}, \mathsf{B}\in\mathcal{Y}\}\right) $$ This means that we have taken all the sigma algebras containing the collection of all measurable rectangles and then we have taken the intersection of all these sigma algebras. The remaining sigma algebra, $\mathcal{X}\otimes\mathcal{Y}$, is the smallest sigma algebra containing this collection. Now, the fact that it contains the collection does not mean that every set $\mathsf{C}\in\mathcal{Z}$ is necessarily a measurable rectangle, or is it? I want to try to understand this better since I constantly see $$ \mu(\mathsf{C}) = \pi(\mathsf{A})\nu(\mathsf{B}) $$ in papers or books but it is not motivated.
If it is not always possible, then can you give a counter example? I am particularly interested in examples where $(\mathsf{X}, \mathcal{X}) = (\mathsf{Y}, \mathcal{Y}) = (\mathbb{R}^d, \mathcal{B}(\mathbb{R}^d))$.
If one of the measures, say $\nu$, is diffuse, the answer is Yes. Given $C$, take $A=\mathsf{X}$ and $B$ any element of $\mathcal Y$ with $\nu(B)=\nu(C)$. The fact that $\nu$ is diffuse guarantees that $\nu(\mathcal Y) = [0,1]$.
Example: Take $\mathsf{X}=\mathsf{Y}=\{0,1\}$, $\pi={1\over 2}\delta_0+{1\over 2}\delta_1$ and $\nu={1\over 3}\delta_0+{2\over 3}\delta_1$. Then $\pi$ takes only the values $0,{1\over 2}, 1$, while $\nu$ takes only the values $0,{1\over 3}, {2\over 3}, 1$. Therefore $\mu\left(\{(0,1), (1,0),(1,1)\}\right) ={5\over 6}$ cannot be of the form $\pi(A)\nu(B)$.