Take $([0,1],\mathscr B,\lambda)$ and the corresponding product measure $\lambda\times \lambda$ on the unit square. $\lambda$ is, of course. Lebesgue measure. If we assume CH, we can find a bijection $j:[0,1]\to \omega_1.$ Then, $j(x)$ has countably many predecessors in $\omega_1$ so if we define $Q=\left \{ (x,y):j(x)\le j(y) \right \},$ it's easy to see that $Q_x$ contains all but countably many points of $[0, 1 ]$ and $Q^y$ contains at most countably many points of $[0, 1].$ Now, define $f=\chi_Q$ , and note that $f_x$ and $f^y$ are Borel measurable (because their inverse images by any open set in $[0,1]$ is either countable, the complement of a countable set or a countable union of these), and that $\int_{0}^{1}f_xdy=1$ and $\int_{0}^{1}f^ydy=0,$ from which a simple calculation shows that Fubini's Theorem fails, which means that $f$ is not measureable with respect to $\mathscr B_{[0,1]}\otimes \mathscr B_{[0,1]}.$
My question is how can we see this without appealing to Fubini? And what specific property or properties of $Q$ cause $f$ to be non-measurable?