Below we use [Bochner measurability](https://www.wikiwand.com/en/Bochner_measurable_function) and [Bochner integral](https://www.wikiwand.com/en/Bochner_integral). Let


* $(X, \mathcal A, \mu)$ and $(Y, \mathcal B, \nu)$ be complete $\sigma$-finite measure spaces,
* $(E, | \cdot |)$ a Banach space,
* $S (X)$ the space of $\mu$-simple functions from $X$ to $E$,
* $L^0 (X)$ the space of $\mu$-measurable functions from $X$ to $E$,
* $L^1 (X)$ the space of $\mu$-integrable functions from $X$ to $E$,
* $\mathcal C :=\mathcal A \otimes \mathcal B$ the product $\sigma$-algebra of $\mathcal A$ and $\mathcal B$,
* $\lambda := \mu \otimes \nu$ the product measure of $\mu$ and $\nu$.


I'm trying to prove a claim in [this](https://mathoverflow.net/a/184921/99469) thread.



> 
> **Theorem** Let $f \in L^0 (X \times Y)$. There is a sequence $(f\_n) \subset S(X \times Y)$ such that $f\_n \to f$ $\lambda$-a.e. and that each $f\_n$ is of the form
> $$
> f\_n (x, y) = \sum\_{i,j=1}^{\varphi\_n} e\_{n, i,j} 1\_{A\_{n, i}} (x) 1\_{B\_{n, j}} (y),
> \tag{$\*$}
> $$
> where $e\_{n, k} \in E$ and $(A\_{n, i})\_{i=1}^{\varphi\_n}, (B\_{n, j})\_{j=1}^{\varphi\_n}$ are finite measurable partitions of $X,Y$ respectively.
> 
> 
> 


Such $f\_n$ has a very nice form that I could not construct. Could you elaborate on how to get it?




---


*My failed attempt* First, we assume $\mu (X) + \nu (Y) +\sup\_{(x, y) \in X \times Y} |f(x, y)| < \infty$. Then $\lambda(X \times Y) < \infty$ and $f \in L^1(X \times Y)$. There is a sequence $(f\_n) \subset S(X \times Y)$ such that $f\_n \to f$ $\lambda$-a.e. and in $L^1(X \times Y)$. Let
$$
f\_n (x, y) = \sum\_{k=1}^{\varphi\_n} e\_{n, k} 1\_{C\_{n, k}} (x, y),
$$
where $e\_{n, k} \in E$ and $C\_{n, k} \in \cal C$ such that $\lambda(C\_{n, k}) < \infty$.



How to construct this sequence that converges a.e. in product measure and that has a very particular form?
$\newcommand\C{\mathcal C}\newcommand\A{\mathcal A}\newcommand\B{\mathcal B}$Using the $\sigma$-finiteness condition, truncation of $f$, and your *attempt*, we see that without loss of generality $f=1\_C$ for some $C\in\C$.


But in this case the result follows by (say) Theorem 1.4 of [this paper](https://link.springer.com/article/10.1007/s11117-017-0507-8) or [its arXiv version](https://arxiv.org/abs/1702.01142). Indeed, by that theorem, every $C\in\C$ can be approximated by sets in the algebra generated by product sets $A\times B$ such that $(A,B)\in\A\times\B$ with respect to the pseudometric $d$ given by the formula $d(E,F):=(\mu\otimes\nu)(E+F)$ for $E$ and $F$ in $\C$, where $+$ denotes the symmetric difference.


