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Let $\mathcal{F}$ and $\mathcal{G}$ be any two famillies of subset of a space $X$ (neither $\mathcal{F}$, nor $\mathcal{G}$ is a sigma-field). $$\sigma( A\times B , A\in \mathcal{F}, B\in \mathcal{G}) \subset \sigma(\mathcal{F})\otimes \sigma(\mathcal{G})$$ is true. My question : is it true that conversely $$ \sigma(\mathcal{F})\otimes \sigma(\mathcal{G}) \subset \sigma( A\times B , A\in \mathcal{F}, B\in \mathcal{G})$$

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1 Answer 1

The converse does not hold assuming I interpreted your notation correctly.

Assume that $X$ is an uncountable set and $\mathcal{F}=\mathcal{G}=\{\{x\}|x\in X\}$. In other words, $\mathcal{F},\mathcal{G}$ are both the set of singletons in $X$. Then $\sigma(\mathcal{F})=\sigma(\mathcal{G})$ is the set of all countable and cocountable elements of $X$. Furthermore, $\{A\times B|A\in\mathcal{F},B\in\mathcal{G}\}$ is the set of all singletons in $X\times X$, so $\sigma(A\times B|A\in\mathcal{F},B\in\mathcal{G})$ is the collection of countable and cocountable sets in $X\times X$. On the other hand, $\{x\}\times X$ is in $\sigma(\mathcal{F})\otimes\sigma(\mathcal{G})$, but is neither countable nor cocountable, so $\{x\}\times X\not\in\sigma(A\times B|A\in\mathcal{F},B\in\mathcal{G})$. Therefore $\sigma(\mathcal{F})\otimes\sigma(\mathcal{G})\not\subseteq\sigma\{A\times B|A\in\mathcal{F},B\in\mathcal{G}\}$.`

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Do you know some large sufficient conditions on the underlying space and on the probability so that the converse holds ? –  Polite Apr 19 '13 at 9:33

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