I am struggling to find a reference for the following statement, which I still believe to be true.
"Let $(\Omega_1, \mathcal{A}_1, \mu_1), (\Omega_2, \mathcal{A}_2, \mu_2)$ be finite measure spaces. Furthermore, let $(\Omega_1\times\Omega_2, \mathcal{A}_1\otimes\mathcal{A}_2, \mu_1\otimes\mu_2)$ the usual product measure space. Then, for every $A\in \mathcal{A}_1\otimes\mathcal{A}_2$ there are sequences $(B_1^i)_{i\in\mathbb{N}}\subset\mathcal{A}_1, (B_2^i)_{i\in\mathbb{N}}\subset\mathcal{A}_2$ such that $\bigcup_{i=1}^n B_1^i\times B_2^i\subset A$ and $(\mu_1\times\mu_2)(A\backslash \bigcup_{i=1}^n B_1^i\times B_2^i)\to 0$$(\mu_1\otimes\mu_2)(A\backslash \bigcup_{i=1}^n B_1^i\times B_2^i)\to 0$ for $n\to\infty$."
So, in $\mathbb{R}^2$ with the Borel-$\sigma$-algebra that is the classical picture that you can approach measurable sets by unions of rectangluar sets. I would also be fine with a statement that approximates $A$ with bigger sets, so $A\subset \cup_{i=1}^{n_j} B_1^{i,j}\times B_2^{i,j}$ for any $j\in\mathbb{N}$ and for $j\to\infty$ I have that $(\mu_1\times\mu_2)((\cup_{i=1}^{n_j} B_1^{i,j}\times B_2^{i,j})\backslash A)$$(\mu_1\otimes\mu_2)((\cup_{i=1}^{n_j} B_1^{i,j}\times B_2^{i,j})\backslash A)$.
Thank you very much for every answer in advance.
EDIT: Sorry, I forgot one property, which I added. Also, if it is easier I would also be happy with an approximation with "bigger" sets.