Example for a non-Borel set $A$ in $\mathbb{R}^2$ such that all $x$- and $y$- sections are measurable

Question

It is well known that the for two sigma algebras $\mathfrak{S}_1$ and $\mathfrak{S}_2$ that for set $A \in \mathfrak{S}_1 \times \mathfrak{S}_2$ all $x$- and $y$-sections are measurable: $$A(.,y)=\{x:(x,y\in A)\} \in \mathfrak{S}_1$$ $$A(x,.)=\{y:(x,y\in A)\} \in \mathfrak{S}_2.$$

I have read that the converse is not true and I want to gain some intuition about it. I want to find an example where $A \not\in \mathfrak{B}(\mathbb{R}^2)$ such that all sections are measurable, i.e. are in $\mathfrak{B}(\mathbb{R})$.

Can someone help how two approach this? I have tried using known non measurable sets in $\mathbb{R}$ building products of them, but I ended up with non measurable sections.

Answer 1

Consider the diagonal $\Delta=\{(x,x): x\in \mathbb R\}$. Not every subset of this is a Borel set. (Why?). For any subset of $\Delta$ every section is measurable. (Why?).