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Let $(\mathcal{X}, \Sigma)$ be a measurable space, and let $\Delta = \{(x,x) \mid x \in \mathcal{X}\} \subseteq \mathcal{X}\times \mathcal{X}$ be the diagonal. What might I call a space $(\mathcal{X}, \Sigma)$ such that $\Delta \in \Sigma \otimes \Sigma$? Can anybody give me a necessary and sufficient condition on $(\mathcal{X}, \Sigma)$ for this to be true? A reference would be great as well.

AJ LaMotta
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    Here is one! Is that the type of thing you're interested in? – Izaak van Dongen May 13 '25 at 13:45
  • sufficient condition: if there are measurable set ${A_n}{n\in \mathbb{N}}$ on $\mathcal{X}$ that separate points, that is, for every $x \neq y$ in $\mathcal{X}$ there exists some $A{n_0}$ such that either $x \in A_{n_0}$ and $y \notin A_{n_0}$, or $y \in A_{n_0}$ and $x \notin A_{n_0}$. Then
    $$ \mathcal{X} \times \mathcal{X} \setminus \Delta = \left( \bigcup_n A_n \times A_n^c \right) \cup \left( \bigcup_n A_n^c \times A_n \right), $$ so $\mathcal{X} \times \mathcal{X} \setminus \Delta$ is measurable, and consequently, $\Delta$ is measurable.     – Daniel Smania May 13 '25 at 13:58
  • Most commonly used sufficient condition in probability is being Polish – Evangelopoulos Foivos May 13 '25 at 14:01
  • See this paper, Proposition 2.1. – Mittens May 13 '25 at 14:06
  • @IzaakvanDongen Perfect, thank you! – AJ LaMotta May 13 '25 at 21:50

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