Approximation of $f(x, y)$ by $\sum_{n=1}^N c_n \chi_{A_n}(x) \chi_{B_n}(y)$

Question

I believe the following statement is true, but I cannot prove it. I would be grateful for your advice.

Let $(X, \mathcal{A}, \mu)$ and $(Y, \mathcal{B}, \nu)$ be $\sigma$-finite measure spaces, and let $(X \times Y, \mathcal{A} \times \mathcal{B}, \mu \times \nu)$ be product measure space. Define \begin{equation} D := \left \{\sum_{n=1}^N c_n \chi_{A_n \times B_n} \mid N \in \mathbb{Z}_+ , c_n \in \mathbb{C}, A_n \in \mathcal{A} , B_n \in \mathcal{B}\right \}. \end{equation} Then, $D$ is dense in $L^p(X \times Y; \mathbb{C})$ for all $p \in [1, \infty)$.

Answer 1

Here's a sketch. You already know that the set $D' := \big\{ \sum c_n \chi_{E_n} : E_n \text{ is $(\mathcal{A} \times \mathcal{B})$-measurable} \big\}$ is dense in $L^p(X \times Y)$. Then just use the fact that the product $\sigma$-algebra $\mathcal{A} \times \mathcal{B}$ is generated by sets of the form $A \times B$ where $A \in \mathcal{A}$ and $B \in \mathcal{B}$. This allows you to conclude that any $(\mathcal{A} \times \mathcal{B})$-measurable set can be approximated up to an arbitrarily small error by a finite union of sets of the form $A \times B$. This then implies that $D$ is dense in $D'$.