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Let $(X, \mathcal{J})$ and $(Y, \mathcal{F})$ be two measure spaces. Let us assume that $J$ is a collection of subsets of $X$ which generates $\mathcal{J}$, i.e. $\sigma(J)=\mathcal{J}$. Similarly, assume $\sigma(F)=\mathcal{F}$. Is it always true that $$ \sigma(J)\times\sigma(F)=\sigma(J\times F)? $$ Here, $J\times F$ is the set of all cartesian products of sets in $J$ with sets in $F$.

Please note: in the above expression the two product signs mean different things. On the left I am considering the product sigma algebra $\mathcal{J}\times \mathcal{F}$, while on the right I am considering the $\sigma$-algebra generated by the cartesian products of my ``elementary sets.''

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$\sigma(J) \times \sigma(F)$ might not even be a $\sigma$-algebra. – Clive Newstead Sep 14 '12 at 18:51
It's defined to be the sigma algebra generated by all measurable rectangles. – Alex Lapanowski Sep 14 '12 at 18:55
Ah right, I'm used to the notation $\sigma(J)\otimes\sigma(F)$ for that purpose. – Clive Newstead Sep 14 '12 at 19:39
up vote 0 down vote accepted

We have to show that if $(X_1,\mathcal F_1)$ and $(X_2,\mathcal F_2)$ are two measurable spaces, and $\mathcal C_j$ are collections of subsets generating $\mathcal F_j$ then $$\mathcal F_1\otimes\mathcal F_2=\sigma(\mathcal C_1\times\mathcal C_2),$$ where $\mathcal C_1\times\mathcal C_2=\{C_1\times C_2,C_j\in\mathcal C_j\}$.

Since $\mathcal C_1\times\mathcal C_2\subset \mathcal F_1\otimes\mathcal F_2$, we have $\supset$. To see the other inclusion, define $$\mathcal B:=\{A\subset X_1,A\times A_2\in\sigma(\mathcal C_1\times\mathcal C_2)\:\forall A_2\in\mathcal F_2\}.$$ It's a $\sigma$-algebra which contains $\mathcal C_1$.

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