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Preimage of generated $\sigma$-algebra

I wish to prove the following:

"Let $X$ be a set and $\mathcal{A}$ a family of subsets of $X$, and $\Sigma_{\mathcal{A}}$ the $\sigma$-algebra of subsets of $X$ generated by $\mathcal{A}$. Suppose that $Y$ is another set and $f : Y \rightarrow X$ a function. Then $\left\{f^{-1} \left[E \right] : E \in \Sigma_{\mathcal{A}} \right\}$ is the $\sigma$-algebra of subsets of $Y$ generated by $\left\{ f^{-1} \left[A\right] : A \in \mathcal{A} \right\}$."

I understand that the $\sigma$-algebra of subsets of $Y$ generated by $\left\{ f^{-1} \left[A\right] : A \in \mathcal{A} \right\}$ is defined to be $$\bigcap \left\{ \Sigma : \Sigma \text{ is a } \sigma \text{-algebra of subsets of }Y, \left\{ f^{-1} \left[A\right] : A \in \mathcal{A} \right\} \subseteq \Sigma\right\}.$$

I'm not sure how this leads to the desired result, though. Any help much appreciated.

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marked as duplicate by Nate Eldredge, t.b., Willie Wong Mar 26 '12 at 12:36

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Don't you mean $f:Y\to X$, not $f:X\to Y$? – Zev Chonoles Jan 24 '12 at 1:14
I think the very last $X$ should be a $Y$ too. – cardinal Jan 24 '12 at 1:46
Your set-up does not make sense as given, since $f^{-1}[E]$ for $E\in\Sigma_A$ does not make sense; because $E\subseteq X$, but $f\colon X\to Y$; that's what Zev is pointing at. – Arturo Magidin Jan 24 '12 at 4:39
Thank you. My apologies. – Harry Williams Jan 25 '12 at 2:02
The answer to your question is here:… – Byron Schmuland Feb 18 '12 at 23:54

1 Answer 1

I think the best approach will be to just show inclusion of the two sigma algebras in both directions.

I am studying this kind of material as well at the moment, so my answer needs considered with this caveat in mind. I m sure better answers will be provided in due course !

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Something to consider: What do you mean by "$f$ is a measurable function"? (Indeed, in a sense, that's precisely what you're trying to create here.) :) – cardinal Jan 24 '12 at 2:23
I think what you are saying is that from the way we define the domain of $\quad f \quad$ it will necessarily be a measurable function right ? Bit new to this material, so just wondering whether I understand you correctly. (post amended accordingly...) – Beltrame Jan 24 '12 at 2:32
Yes, essentially. – cardinal Jan 24 '12 at 2:44
I understand $f:D \rightarrow \mathbb{R}$ to be a measurable function if any, or equivalently all, of the following are true: (i) ${x : f(x) < a} \in \Sigma_D$ for every $a \in \mathbb{R}$; (ii) ${x : f(x) \leq a} \in \Sigma_D$ for every $a \in \mathbb{R}$; (iii) (i) ${x : f(x) > a} \in \Sigma_D$ for every $a \in \mathbb{R}$; (iv) (i) ${x : f(x) \geq a} \in \Sigma_D$ for every $a \in \mathbb{R}$, where $D \subseteq X$ and $\Sigma_D$ is the subspace $\sigma$-algebra of subsets of $D$. – Harry Williams Feb 6 '12 at 22:12
At least, this is the technical definition in my lecture notes; I can see there are some pre-images here, but they seem more specific than those in the problem here, so I'm not sure how it helps? – Harry Williams Feb 6 '12 at 22:17

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