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I've been trying to understand various questions to do with sigma algebras on uncountable product spaces.

Let $T$ be an uncountable set and for each $t \in T$, let $\Omega_t$ be a topological space. Let $\Omega := \prod_{t \in T}\Omega_t$ be the product space equipped with the product topology in the usual way. The Borel sigma-algebra $\mathcal{Borel}(\Omega)$ is the sigma-algebra generated by the open sets in this topology. The Baire sigma-algebra is the sigma-algebra generated by sets of the form {$\omega \in \Omega : f(\omega) > 0$} for continuous functions $f : \Omega \to \mathbf{R}$. (Equivalently use zero sets if you prefer but this makes it look a little more on par with open sets)

EDT: The Baire sigma algebra is contained in the Borel sigma algebra but presumably(??) it isn't the case that they are equal in general? i.e. Is it the case that the Baire is strictly smaller and if so how can we show it?

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    $\begingroup$ Trivially, the Baire $\sigma$-algebra is contained in the Borel one, since $\{\omega \in \Omega: f(\omega) > 0\}$ is open. Are you asking whether it must be strictly smaller? $\endgroup$
    – Nate Eldredge
    Commented Aug 22, 2023 at 15:52
  • $\begingroup$ sorry yes. Very tired; typing on autopilot. Edited. $\endgroup$
    – SBK
    Commented Aug 22, 2023 at 15:56
  • $\begingroup$ Oh hang on maybe the answer is shown by the sources in this question mathoverflow.net/questions/364062/… that you also commented on @NateEldredge ? $\endgroup$
    – SBK
    Commented Aug 22, 2023 at 16:00
  • $\begingroup$ As you suspected, in general, the $\sigma$-algebras don't agree (clear by common sense, or else why would anyone talk about Baire sets). See here for example: math.stackexchange.com/questions/3792682/… $\endgroup$
    – Christian Remling
    Commented Aug 22, 2023 at 17:12
  • $\begingroup$ The product structure has nothing to do with it, you can just take $\Omega_0$ as your desired space and the other $\Omega_t$'s as one-point spaces. $\endgroup$
    – Christian Remling
    Commented Aug 22, 2023 at 17:13

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Let all factor spaces be nontrivial compact Hausdorff spaces. Then every continuous function is determined by countably many coordinates, and so is, consequently, every Baire measurable set.

It follows that singletons are not Baire sets. However, they are closed and, consequently, Borel.

The same argument works if all factor spaces are nontrivial separable metrizable spaces.

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