# Countably generated sigma algebra vs. generated by a real-valued random variable

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).

$$\{0,1\}^{\mathbb N}$$ 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.