Given a quadruple of measurable spaces $(A,\mathcal{A})$, $(A,\mathcal{A}')$, $(B,\mathcal{B})$, $(B,\mathcal{B}')$ is it true that $\mathcal{A}\otimes\mathcal{B}\cap\mathcal{A}'\otimes\mathcal{B}'$ is also a product of some $\sigma$-algebras on $A$ and $B$? I would guess the answer is "no", but how would one tell that an $\sigma$-algebra $\mathcal{C}$ on $A\times B$ is not a product, except for the trivial "too few cylinder-sets"?
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1$\begingroup$ I do not understand what you mean by “too few cylinder-sets”, but if $\mathcal{C}$ is a product $\sigma$-algebra $\mathcal{A} \otimes \mathcal{B}$, then one must have $\mathcal{A} = \{X \subseteq \mathcal{A}\,|\,X \times B \in \mathcal{C}\}$: so, considering that point, it should not be that difficult to show that a $\sigma$-algebra is not product, since you only have to prove that it is not equal to a certain, specific $\sigma$-algebra… $\endgroup$– Rémi PeyreCommented Nov 28, 2019 at 16:32
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$\begingroup$ @ Rémi Peyre; That's pretty much what I mean. Sometimes candidate $\mathcal{A}$ and $\mathcal{B}$ obtained your way are too small (e.g. finite, with infinite $\mathcal{C}$) for their product to generate $\mathcal{C}$. If there is no obvious counting argument, how do I tell whether $\mathcal{A}\times\mathcal{B}$ generates $\mathcal{C}$? Besides, is it clear that "if $X\times B$ is $\mathcal{A}\otimes\mathcal{B}$-measurable, then $X$ is $\mathcal{A}$-measurable"? $\endgroup$– R. MatveevCommented Nov 28, 2019 at 16:53
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1$\begingroup$ Yes, if $X \times B$ is $\mathcal{A} \otimes \mathcal{B}$ measurable, then $X$ is $\mathcal{A}$-measurable: more generally, if $C \in \mathcal{A} \otimes \mathcal{B}$, then for all $y \in B$, the set $\{x \in A\,|\,(x, y) \in C\}$ is $\mathcal{A}$-measurable. (The proof is straightfoward). $\endgroup$– Rémi PeyreCommented Nov 28, 2019 at 17:52
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$\begingroup$ Does this last statement hold true even if $\mathcal{B}$ does not separate points? I don't see any straightforward proof without such an assumption. Probably I am missing something trivial. $\endgroup$– R. MatveevCommented Nov 28, 2019 at 19:25
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$\begingroup$ I figured it out. Thank you, that settles my question. $\endgroup$– R. MatveevCommented Nov 28, 2019 at 19:36
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There is an article with a related question, giving a negative answer in general: In https://arxiv.org/pdf/2007.06095.pdf there is a counterexample (due to G. Halmos) mentioned. In that counterexample three $\sigma$-algebras $\mathcal{A}, \mathcal{F}$ and $\mathcal{G}$ are constructed (quite involved), such that $\mathcal{A}\otimes(\mathcal{F}\cap\mathcal{G})\neq (\mathcal{A}\otimes\mathcal{F})\cap(\mathcal{A}\otimes\mathcal{G})$ and in further consequence it is shown that the latter $\sigma$-algebra is not of product form.
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$\begingroup$ Welcome to MO. To comply with community standards and make this answer somewhat self-contained, can you at least write how the paper is related to the question? $\endgroup$ Commented Jan 14, 2021 at 14:59