If the sections of a function are $\mathscr{L}^2$, would the function be jointly $\mathscr{L}^2$?

Question

Let $(\varOmega,\mathcal{A},\mu)$ and $(\varXi,\mathcal{B},\nu)$ be $\sigma$-finite measure spaces, and $f\colon\varOmega\times\varXi\to\mathbb{R}$ be $\big(\mathcal{A}\otimes\mathcal{B}\big)$-$\mathcal{B}(\mathbb{R})$-measurable. Suppose also that the sections of $f$ are $\mathscr{L}^2$; i.e., $$f(\cdot,\xi)\in\mathscr{L}^2(\varOmega,\mu),~\text{for $\nu$-a.e.}~\xi,~\text{and}~f(\omega,\cdot)\in\mathscr{L}^2(\varXi,\nu),~\text{for $\mu$-a.e.}~\omega.$$ Then, is it true that $f\in\mathscr{L}^2(\varOmega\times\varXi,\mu\otimes\nu)$? If not, what restrictions do we need?

Answer 1

The answer is no. You cannot conclude that $f\in\mathscr{L}^2(\varOmega\times\varXi,\mu\otimes\nu)$, see this essentially the same question.

One sufficient condition for the claim to hold though is that $f$ be a tensor product: $f(\omega, \xi) = f_1(\omega)f_2(\xi)$. Since the $\mathscr{L}^p$ play nice with tensor products: $$ \|f_1 \otimes f_2\|_{\mathscr{L}^p(\varOmega\times\varXi,\mu\otimes\nu)} = \|f_1\|_{\mathscr{L}^p(\varOmega,\mu)} \|f_2\|_{\mathscr{L}^p(\varXi,\nu)}.$$