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Here is a problem from probability with martingales. I want to a better way of writing this than my waffle:

Let $Y$ be a random variable and $\pi (\mathbb{R})$ is a $\pi$-system generating the Borel $\sigma$-algebra of $\mathbb{R}$, show that $Y^{-1}(\pi (\mathbb{R}))$ is $\pi$-system generating $\sigma(Y)$.

I want to say something like this:

Showing it is a $\pi$ system is easy. inverse operations are preserved under taking a finite unions. Take a $B\in \sigma(Y)$ then $B=f^{-1}(B')$, for some $B'\in\mathcal{B}(\mathbb{R})$ here comes the waffle: $B'$ is generated by some arbitrary unions/intersections of open sets and taking pre-image preserves taking unions/intersections. so $B$ is the unions and intersections of pre-image of sets in $\pi(\mathbb{R})$

what is a neat way of saying this?

Lost1
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  • Do you know already that preimages of measurable sets form a $\sigma$-algebra? – Michael Greinecker Jan 25 '14 at 17:45
  • @MichaelGreinecker yes but how is that related to this? – Lost1 Jan 25 '14 at 17:57
  • Well, one side is that the $\sigma$-algebra generated by the preimages of sets in the $\pi$-system is not larger than the $\sigma$-algebra generated by $Y$, and that follows directly from that fact. – Michael Greinecker Jan 25 '14 at 18:08
  • @MichaelGreinecker nope, i am trying to show the pre-image of the $\pi$-system is a $\pi$-system? i am interested in writing down the proof using rigorous language rather than the result itself. – Lost1 Jan 25 '14 at 18:15
  • You want to show that $\sigma{Y^{-1}(B)\in\pi(\mathbb{R})}=\sigma{Y^{-1}(B)\in\mathcal{B}(\mathbb{R})}$ in a clean way, right? – Michael Greinecker Jan 25 '14 at 18:22
  • @MichaelGreinecker $\sigma{Y^{-1}(B): B\in\pi(\mathbb{R}) }=\sigma{Y^{-1}(B): B\in\mathcal{B}(\mathbb{R}) }$, yes – Lost1 Jan 25 '14 at 18:41

1 Answers1

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A neat way is to use the following result:

Let $\cal C$ be some collection of subsets of a set $Y$, and let $f$ be a function from some set $X$ to $Y$. I want to prove: $$f^{-1}(\sigma(\mathcal C))=\sigma(f^{-1}(\mathcal C)).$$

Community
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Davide Giraudo
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  • so doing this rigorously require transfinite induction or not? – Lost1 Jan 26 '14 at 12:54
  • because what i wrote in bald is basically that. every borel set is 'generated' by repeatedly taking intersections and unions. – Lost1 Jan 26 '14 at 12:57