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 $(\mu\_1\otimes\mu\_2)(A\backslash \bigcup\_{i=1}^n B\_1^i\times B\_2^i)\to 0$ 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 $(\mu\_1\otimes\mu\_2)((\cup\_{i=1}^{n\_j} B\_1^{i,j}\times B\_2^{i,j})\backslash A)$.


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.



Product sigma-algebra: approximating elements arbitrary good using the generating sets
$\newcommand\Om\Omega\newcommand\A{\mathcal A}\newcommand\I{\mathbb I}$Your original statement is trivial: Take $B\_1^i=\Omega\_1$ and $B\_2^i=\Omega\_2$ for all $i$.


If you additionally require that $B\_1^i\times B\_2^i\subseteq A$ for all $i$, then the statement will become false in general. For instance, for $j=1,2$ let $(\Om\_j,\A\_j,\mu\_j)$ be the standard Lebesgue measure space over $[0,1]$. Let $A=\{(x,y)\in[0,1]^2\colon x-y\in\I\}$, where $\I$ is the set of all irrational real numbers. Then (by, say, the Tonelli theorem), $(\mu\_1\otimes\mu\_2)(A)=1$. However, if $B\_1\times B\_2 \subseteq A$, then $(\mu\_1\otimes\mu\_2)(B\_1\times B\_2)=0$, because otherwise the set $B\_1-B\_2=\{x-y\colon x\in B\_1,y\in B\_2\}$ would contain a nonzero-length interval and therefore would contain a rational point as an element -- cf. e.g [this](https://www.researchgate.net/profile/Iosif-Pinelis/publication/280132389_150600537v2/data/55abc6d608aea3d0868593f2/150600537v2.pdf?origin=ResearchDetailAlternativeSimilarResearch&_rtd=eyJjb250ZW50SW50ZW50Ijoic2ltaWxhciJ9). So, for any $B\_j^i\in\A\_j$ such that $B\_1^i\times B\_2^i\subseteq A$ for all $i$, we will have $(\mu\_1\otimes\mu\_2)(A\setminus\bigcup\_{i=1}^n(B\_1\times B\_2))=1\not\to0$.


Similarly, in general you cannot approximate a set $A\in\A\_1\otimes\A\_2$ by bigger sets (containing $A$) belonging to the algebra of all finite unions of the products of sets in $\A\_1$ and $\A\_2$ -- just consider the complement of the set $A$ in the example above. (Do not confuse the product $\sigma$-algebra $\A\_1\otimes\A\_2$, generated by the mentioned algebra, with the product $\A\_1\times\A\_2$ of the $\sigma$-algebras $\A\_1$ and $\A\_2$.)




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What is possible is for each natural $n$ to find a set $B\_n$ belonging to the mentioned algebra of all finite unions of of the products of sets in $\A\_1$ and $\A\_2$ such that $(\mu\_1\otimes\mu\_2)(A+B\_n)\to0$ as $n\to\infty$, where $+$ denotes the symmetric difference -- see e.g. Theorem 1.1 [here](https://link.springer.com/article/10.1007/s11117-017-0507-8) or [here](https://arxiv.org/abs/1702.01142).


