Product sigma-algebra: approximating elements arbitrary good using the generating sets

Question

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.

Answer 1

$\newcommand\Om\Omega\newcommand\A{\mathcal A}\newcommand\I{\mathbb I}$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. 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 or here.