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Let

  • $(\Omega_i,\mathcal A_i)$ be a measurable space
  • $\emptyset\ne A_i\subseteq\Omega_i$

Are we able to show that $\left.\mathcal A_1\otimes\mathcal A_2\right|_{A_1\times A_2}=\left.\mathcal A_1\right|_{A_1}\otimes\left.\mathcal A_2\right|_{A_2}$?

Above $\left.\mathcal B\right|_B:=\left\{B\cap B':B'\in\mathcal B\right\}$ denotes the trace $\sigma$-algebra of $B$ in $\mathcal B$ and $\mathcal B_1\otimes\mathcal B_2$ the product $\sigma$-algebra of $\mathcal B_1$ and $\mathcal B_2$.

I'm pretty sure that the statement is correct, but I didn't find an approach to show it.

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Well, $\mathcal A_1\otimes\mathcal A_2|_{A_1\times A_2}$ is a $\sigma$-algebra that contains all rectangles made from trace sets and therefore a subset of $\left.\mathcal A_1\right|_{A_1}\otimes\left.\mathcal A_2\right|_{A_2}$.

For the other inclusion, it suffices to show that $B_1\times A_2\in \mathcal A_1\otimes\mathcal A_2|_{A_1\times A_2}$ for all $B_1\in\left.\mathcal A_1\right|_{A_1}$ and $A_1\times B_2\in\mathcal A_1\otimes\mathcal A_2|_{A_1\times A_2}$ for all $B_2\in\left.\mathcal A_2\right|_{A_2}$. Let's do the first case. If $B_1\in\left.\mathcal A_1\right|_{A_1}$, then there is some $B_1'\in\mathcal A_1$ with $B_1=B_1'\cap A_1$. But then we have $B_1\times A_2=(B_1'\times\Omega_2) \cap (A_1\times A_2)\in\mathcal A_1\otimes\mathcal A_2|_{A_1\times A_2}$.

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