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I'm working on proving the equality between the preimage of generated $\sigma$-algebra and generated $\sigma$-algebra of preimage.

I noticed that there's a duplicate: Proof that the preimage of generated $\sigma$-algebra is the same as the generated $\sigma$-algebra of preimage., but I'm failing to understand the proof in in the post, so I tried my own take and wanted some feedback:

I have proved two properties to help with this problem:

Let $f:X \to Y$ be a mapping. If $\mathcal B$ is a $\sigma$-algebra on $Y$, then $f^{-1}(\mathcal B)$={$f^{-1}(B):B \in \mathcal B$} is a $\sigma$-algebra on X, and that if $\mathcal A$ is a $\sigma$-algebra on X, then $f_*$$\mathcal A$={$E\subset Y$:$f^{-1}(E)\in \mathcal A$} is a $\sigma$-algebra on Y.

Now I used the first property to show that if $\mathcal E$ is a collection of subsets of $Y$, $\sigma(f^{-1}$($\mathcal E$))$\subset f^{-1}(\sigma(\mathcal E))$ since $f^{-1}$($\mathcal E$)$\subset f^{-1}(\sigma(\mathcal E))$, and $f^{-1}(\sigma(\mathcal E))$ is a $\sigma$-algebra.

Now I want to show the other direction, and I think I just have to pick a 'clever' $\mathcal A$. I tried both $\sigma(f^{-1}(\mathcal E))$, and $f^{-1}(\sigma(\mathcal E))$ but didn't arrive at the conclusion I want.

How can I show $f^{-1}(\sigma(\mathcal E)) \subset \sigma(f^{-1}(\mathcal E))$ using the second property above?

Sank
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1 Answers1

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$\{B \subset Y: f^{-1}(B)\in \mathcal M(f^{-1}(\mathcal E)\}$ is a sigma algebra by the second property. It contains $\mathcal E$, hence it contains $\mathcal M(\mathcal E)$. This is exactly what we are trying to prove.

Kavi Rama Murthy
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  • I got to $\mathcal M(\mathcal E)$ being contained in the first set, but I'm not sure how that implies the conclusion that I am trying to get to. – Sank Sep 19 '18 at 23:39
  • What you have to show is $f^{-1}(D) \in \mathcal M( \mathcal f^{-1}( \mathcal E))$ if $D \in \mathcal M (\mathcal E)$ and that is what we just proved. – Kavi Rama Murthy Sep 19 '18 at 23:43
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    So, by definition of the set, $\mathcal E$ is contained in it, consequently $\mathcal M(\mathcal E)$. Then for any $D \in \mathcal M(\mathcal E)$, $f^{-1}(D)\in \mathcal M(f^{-1}(\mathcal E))$. Thus $f^{-1}(\mathcal M(\mathcal E)) \subset \mathcal M(f^{-1}(\mathcal E))$. Is that correct? – Sank Sep 19 '18 at 23:52
  • Yes, that is the argument. – Kavi Rama Murthy Sep 19 '18 at 23:57
  • thank you, I understand now! Is my logic behind for the other direction of inclusion ok? – Sank Sep 19 '18 at 23:58
  • Yes, that part is fine. – Kavi Rama Murthy Sep 19 '18 at 23:59