Is the generated $\sigma$-algebra of a product topology a subset of the product $\sigma$-algebra of the individual topologies?

Formally, let $\Theta$ be some non-empty set (to serve as a set of indices). To each $\theta\in\Theta$ let $\left(\Omega_\theta,\mathcal{T}_\theta\right)$ be a topological space ($\Omega_\theta\neq\emptyset$). Is it true that


where $\prod_{\theta\in\Theta}\mathcal{T}_\theta$ is the product topology of the $\mathcal{T}_\theta$'s, $\sigma\left(\prod_{\theta\in\Theta}\mathcal{T}_\theta\right)$ is the $\sigma$-algebra generated by the product topology, $\sigma\left(\mathcal{T}_\theta\right)$ is the Borel field generated by the topology $\mathcal{T}_\theta$ and $\otimes_{\theta\in\Theta}\sigma\left(\mathcal{T}_\theta\right)$ is the product $\sigma$-algebra of the Borel fields?

The context for this question is my trying to grapple with a proof given in Schervish's Theory of Statistics (1995) (Lemma B. 40, p. 621)) to the claim that a Polish space is a Borel Space.

The proof proceeds by showing that a Polish space is bi-measurable with $\mathbb{R}^\infty$, which was proved previously to be a Borel space. The bi-measurable function is called $f$ and is defined thus

$$f:\mathcal{X}\rightarrow\mathbb{R}^\infty,\space\space f(x):=\left(d\left(x,y_1\right),d\left(x,y_2\right),\dots\right)$$ where $\left\{y_1, y_2,\dots\right\}\subset\mathcal{X}$ is dense in $\mathcal{X}$ and $d$ is the metric on $\mathcal{X}$.

To show that $f$ is measurable, Schervish points to the fact that $f$ is continuous, since each of its coordinates is continuous.

However the implication "$f$ is continuous $\implies$ $f$ is measurable" is only valid if the product topology is contained in the $\sigma$-algebra on $\mathbb{R}^\infty$, hence my question.


For a countable collection of separable metric spaces (like $\mathbb{R}^\infty$) these two $\sigma$-algebras are actually equal; see, for example, Kallenberg's Foundations of Modern Probability, second edition, page 3.

In general the inclusion need not hold. See Michael Greinecker's example in the comment.

  • 4
    $\begingroup$ That is not correct. While it is true that both are generated from open cylinder sets, in a topology you are allowed to form arbitrary unions, which can give you some sets you do not get in the $\sigma$-algebra. Let $\kappa$ be uncountable and $\{0,1\}$ be endowed witht the discrete topology. Then every singleton is closed in $\{0,1\}^\kappa$ since the product of Hausdorff spaces is a Hausdorff space, but the singleton is not in the product $\sigma$-algebra, which is determined by countably many coordiantes. $\endgroup$ – Michael Greinecker Mar 9 '13 at 15:26
  • $\begingroup$ @MichaelGreinecker You are right. I have changed my answer. Thank you. $\endgroup$ – Gibarian Mar 9 '13 at 15:56
  • $\begingroup$ Thank you very much, Gibarian and Michael. $\endgroup$ – Evan Aad Mar 9 '13 at 16:08

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