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?
Best Answer
$\{B \subset Y: f^{-1}(B)\in \mathcal M(f^{-1}(\mathcal E)\}$ 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.