I was hoping someone would check my proof. The question is Let $f: \Omega \to \Omega'$ and let C be a class of subsets of $\Omega'$ Show that $\sigma(f^{-1}(C))=f^{-1}(\sigma(C))$
My proof is below.
We have that $C\subseteq \sigma(C)$, so $f^{-1}(C)\subseteq f^{-1}(\sigma(C))$. $f^{-1}(\sigma(C))$ is a sigma algebra, so $\sigma(f^{-1}(C)) \subseteq f^{-1}(\sigma(C))$
Now let $D= \{B\in \sigma(C): f^{-1}(B)\in \sigma(f^{-1}(C))\}$ It is not difficult to show that D is a sigma algebra.
If $A\in C$ then $f^{-1}(A)\in f^{-1}(C)$, so $C\subseteq D$, and since $D$ is a sigma-algebra, we have that $\sigma(C)\subseteq D$, but of course $D\subseteq \sigma(C)$ so $D=\sigma(C)$. But this means that if $A\in \sigma(C)$, then $A\in D$, hence $f^{-1}(A) \subseteq \sigma(f^{-1}(C))$ and we have that $f^{-1}(\sigma(C))\subseteq\sigma(f^{-1}(C))$ and hence $\sigma(f^{-1}(C))=f^{-1}(\sigma(C))$