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Let $(E,\mathcal E)$ be a measurable space. Under which assumption on $(E,\mathcal E)$ can we show that $\Delta:=\left\{(x,x):x\in E\right\}\in\mathcal E\otimes\mathcal E$? Note that this doesn't hold in general. Is it correct, for example, if $E$ is a Polish space and $\mathcal E=\mathcal B(E)$? A reference with a proof would be enough for me.

0xbadf00d
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  • Looking at the accepted answer to the question you linked, my guess is that $\Delta\in\mathcal{E}\otimes\mathcal{E}$ iff $|E|\leq2^{\aleph_0}$ and ${x}\in\mathcal{E}$ for all $x\in E$. – SmileyCraft Jan 02 '19 at 00:29

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True for the Borel sigma algebra of any second countable space (in particular separable metric space).

ViktorStein
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Kavi Rama Murthy
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  • Do you have a reference at hand? – 0xbadf00d Jan 02 '19 at 10:15
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    It is easy to show that the product sigma field is nothing but the Borel sigma field of the product space under the assumption of second countability. Since the Diagonal is a closed set its complement is open, hence Borel in the product, hence belongs to the product sigma filed. – Kavi Rama Murthy Jan 02 '19 at 11:40