[Math] Is the $\sigma$-algebra generated by a product topology a subset of the product $\sigma$-algebra of the individual topologies

general-topologymeasure-theory

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

$$\sigma\left(\prod_{\theta\in\Theta}\mathcal{T}_\theta\right)\subseteq\bigotimes_{\theta\in\Theta}\sigma\left(\mathcal{T}_\theta\right)$$

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.

Best Answer

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.

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