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I've been looking through some posts about how projections of sets from product $\sigma$-algebras of two Borel $\sigma$-algebras onto their component spaces are not necessarily measurable which apparently leads into the rabbit hole of descriptive set theory.

My problem, though, is a little different and I thought perhaps someone had an idea of how to tackle it seeing as how I am not really knowledgeable when it comes to descriptive set theory.

What I have is a kind of Polish space $\mathcal{X}$ (complete, separable metric space), a countable (or even finite) set $\mathcal{A}$ (being polish by virtue of its discrete metric if I am not mistaken) and their respective Borel $\sigma$-algebras $\mathcal{B}({\mathcal{X}})$ and $\mathcal{B}(\mathcal{A})=2^{\mathcal{A}}$ since $\mathcal{A}$ is countable.

What I am wondering now is, if I look at some set from their product $S \in \mathcal{B}({\mathcal{X}}) \otimes \mathcal{B}(\mathcal{A})$, does the setup now give us that $$ \pi_{\mathcal{X}}(S) := \{ x \in \mathcal{X} | \, \exists\, a \in \mathcal{A}: \, (x,a) \in S \} $$

What I have tried to use is that since $\mathcal{B}(\mathcal{A})=2^{\mathcal{A}}$ and countable, we can write $S$ as

$$ S = \bigcup_{a \in \mathcal{A}} S_a \times \{a \} $$ where $ S_a = \{ x \in \mathcal{X}|\, (x,a) \in S \} $ hoping I could somehow conclude that $ S_a \in \mathcal{B}(\mathcal{X}) $ as intuitively, pairing Borel sets with countably (or even finitely) many point sets and using that as a generator for $ \mathcal{B}({\mathcal{X}}) \otimes \mathcal{B}(\mathcal{A}) $ shouldn't all of a sudden provide us with sets whose projections are non-Borel.

However, I cannot seem to properly prove this statement and perhaps it is false to begin with. I thought someone here could give some insights.

Thanks!

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Your argument goes through. The set $S_a = \{ x \in \mathcal{X}|\, (x,a) \in S \}$ is known as the $a$-section, and it is generally measurable for the product $\sigma$-algebra (the smallest $\sigma$-algebra generated by rectangles).

The proof that sections are measurable is a crucial step in the proof of Fubini's theorem, and that is where you can probably find a proof in most textbooks that treat measure theory. You can also find a short proof here.

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  • $\begingroup$ Right, I totally overlooked that! Thanks so much for your help :) $\endgroup$ – Sunshower Jul 2 '17 at 11:07

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