# The Borel sigma-algebra of a product of two topological spaces

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 $$\Sigma_1\boxtimes\Sigma_2=\Sigma_{12}$$?

Remarks.

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

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. $$\square$$

• 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... Sep 7 at 21:33
• 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? Sep 16 at 22:33
• @YemonChoi You are totally right, that's a mistake. The argument works only if $\tau_1$ and $\tau_2$ are second-countable. Sep 17 at 7:57