The following problem arose while trying to justify some "known results" in abstract harmonic analysis on noncommutative groups, for which I couldn't find explicit statements in the literature. It is much more general than what I require, but I would like to know the "correct level of generality" for what I have been able to prove.

To establish terminology: if $\Omega$ is a set and $\Sigma$ a $\sigma$-algebra on it, we say that $(\Omega,\Sigma)$ is standard if there is a Polish topology on $\Omega$ that generates $\Sigma$.

Let $(\Omega_1,\tau_1)$ and $(\Omega_2,\tau_2)$ be topological spaces, and let $\Sigma_1$ and $\Sigma_2$ be the $\sigma$-algebras generated by $\tau_1$ and $\tau_2$. Equip $\Omega_1\times\Omega_2$ with the product topology, which I denote by $\tau_{12}$, and let $\Sigma_{12}$ be the $\sigma$-algebra generated by $\tau_{12}$.

It is easy to check that the product $\sigma$-algebra $\Sigma_1\boxtimes\Sigma_2$ is coarser than (i.e. contained in) $\Sigma_{12}$, and in general this can be strict, i.e. the two $\sigma$-algebras might be different.

Question. Suppose $(\Omega_1,\Sigma_1)$ and $(\Omega_2,\Sigma_2)$ are both standard. (This implies, since the product of Polish spaces is Polish, that $(\Omega,\Sigma_1\boxtimes\Sigma_2)$ is standard.) Does it follow that $\Sigma_1\boxtimes\Sigma_2=\Sigma_{12}$?


  1. The place where I am having difficulties is that the topologies $\tau_1$ and $\tau_2$ need not be Polish (indeed, need not be Hausdorff), even though the $\sigma$-algebras they generate are standard. (For the intended applications one can relate $\tau_1$ and $\tau_2$ to certain Polish topologies, but this requires "opening up a black box" and I am hoping to avoid this.)

  2. I can prove the desired result if I add the extra assumption that $(\Omega_1\times\Omega_2,\Sigma_{12})$ is standard (and this assumption holds in the intended applications). For then the identity map $(\Omega_1\times\Omega_2,\Sigma_{12})\to (\Omega_1\times\Omega_2,\Sigma_1\boxtimes\Sigma_2)$ is a measurable bijection between standard spaces, hence has measurable inverse by (a consequence of) Souslin's theorem. But this feels like a sledgehammer, and also I suspect that my extra assumption is not necessary.


1 Answer 1


Edit [2022-09-17]: The proof of Proposition 2 below is flawed as pointed out in the comments. However, the answer remains positive if $\tau_1$ and $\tau_2$ are second-countable. (Assuming $\tau'_1$ and $\tau'_2$ come from separable metrics, they are automatically second-countable.)

The answer is positive. This follows from the following two facts:

Proposition 1. Let $(\Omega_1,\tau_1)$ and $(\Omega_2,\tau_2)$ be topological spaces. Then, $\Sigma(\tau_1)\otimes\Sigma(\tau_2)\subseteq\Sigma(\tau_1\otimes\tau_2)$ (as you pointed out). The two $\sigma$-algebras coincide if $\tau_1$ and $\tau_2$ are second-countabe.

This is standard. See, for instance, Proposition 4.1.7 of Dudley's Real Analysis and Probability.

Proposition 2. Let $\tau_1$ and $\tau'_1$ be topologies on $\Omega_1$, and $\tau_2$ and $\tau'_2$ be topologies on $\Omega_2$. If $\Sigma(\tau_1)=\Sigma(\tau'_1)$ and $\Sigma(\tau_2)=\Sigma(\tau'_2)$, then $\Sigma(\tau_1\otimes\tau_2)=\Sigma(\tau'_1\otimes\tau'_2)$.

Proof. Let $A\in\tau_1$ and $B\in\tau_2$. Then, $A\in\Sigma(\tau_1)=\Sigma(\tau'_1)$ and $B\in\Sigma(\tau_2)=\Sigma(\tau'_2)$, hence $A\times B\in\Sigma(\tau'_1)\otimes\Sigma(\tau'_2)\subseteq\Sigma(\tau'_1\otimes\tau'_2)$, by Proposition 1. Therefore, $\tau_1\otimes\tau_2\subseteq\Sigma(\tau'_1\otimes\tau'_2)$, which implies $\Sigma(\tau_1\otimes\tau_2)\subseteq\Sigma(\tau'_1\otimes\tau'_2)$. The opposite inclusion holds by symmetry. $\square$

  • $\begingroup$ Thank you! That is much better than what I had come up with using Suslin's result. In fact, I had been trying to prove something like Proposition 2 but got myself confused/miscalculated at 3 in the morning... $\endgroup$
    – Yemon Choi
    Sep 7 at 21:33
  • $\begingroup$ Actually, while I was writing up the intended application of (a special case of) this result I realised I don't follow a step in the proof of Proposition 2. I agree that every $(\tau_1\times\tau_2)$-open rectangle belongs to $\Sigma(\tau_1'\otimes\tau_2')$, but why does this ensure that the open sets for $\tau_1\times\tau_2$ belong to this sigma-algebra, since we could be taking uncountable unions? $\endgroup$
    – Yemon Choi
    Sep 16 at 22:33
  • $\begingroup$ @YemonChoi You are totally right, that's a mistake. The argument works only if $\tau_1$ and $\tau_2$ are second-countable. $\endgroup$
    – Algernon
    Sep 17 at 7:57

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