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Dear StackExchange community

Given an arbitrary probability space ($\Omega, \Sigma, P$) (without a topological structure), is it true that for any countably generated sub sigma algebra there exists a real-valued random variable that generates it?

After researching this question, I came up with mixed results.

  • I know there is this post Every countably generated sigma algebra is generated by a random variable, but it seems to me that the proof makes implicit use of some topology on $\Omega$
  • This text https://arxiv.org/pdf/0809.3066.pdf (Prop. 3.2) states that a sigma algebra is countably generated iff it is generated by a mapping ($\Omega, \Sigma$) $\rightarrow$ ({$0,1$}$^\mathbb{N}$, Borel sigma algebra on {$0,1$}$^\mathbb{N}$), which is not the same as our usual real-valued random variable $(\Omega, \Sigma) \rightarrow (\mathbb{R},$ Borel sigma algebra on $\mathbb{R}$)

Why this matters: Several results on the existence of regular conditional probability make use of the condition "$\Sigma$ is countably generated". Life would have been much easier if we could replace this condition with "$\Sigma$ = $\sigma (X)$ for some real-valued random variable $X$" (because $\sigma (X)$ has some nice properties).

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1 Answer 1

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$\{0,1\}^{\mathbb N}$ is a Polish space with the same cardinality as $\mathbb R$. Hence, there is a bijection $f: \{0,1\}^{\mathbb N} \to \mathbb R$ such that $f$ and $f^{-1}$ are measurable (for the Borel sigma algebras on both sides). Composing a $\{0,1\}^{\mathbb N}$ valued measurable function with this $f$ gives the answer.

Reference for existence of $f$: Measure Theory by Cohn.

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