Let $(\Omega , F)$ and $(\Omega', F'$) be measurable spaces. Let $f: \Omega \rightarrow \Omega'$ be a measurable function. I am trying to prove that $f^{-1} (F') $ is a sigma-algebra.
It is trivial that $\Omega \in f^{-1} (F') $. However, I am unable to show that it is closed under taking complements. Here is how I started: Let $E \in f^{-1} (F') $. Then $f(E)\in F'$. As $F'$ is a sigma-algebra, $\Omega' \setminus f(E) \in F'$. Then $$ f^{-1}(F') \ni f^{-1}(\Omega' \setminus f(E)) = f^{-1}(\Omega') \setminus f^{-1}(f(E)) = \Omega \setminus f^{-1} (f(E)) \subset \Omega \setminus E$$ as $E \subset f^{-1}(f(E))$, but this is not getting me anywhere. How should I proceed?
Suppose that $A\in f^{-1}(F')$. We need to prove that $A^{c}\in f^{-1}(F')$. Indeed, this is the case. If we assume that $A\in f^{-1}(F')$, there is some $B\in F'$ such that $A = f^{-1}(B)$. Consequently, it results that:
\begin{align*} A = f^{-1}(B) \Rightarrow A^{c} = [f^{-1}(B)]^{c} = f^{-1}(B^{c})\in f^{-1}(F') \end{align*} that is because $B\in F'$ and $B^{c}\in F'$ ($F'$ is a $\sigma$-algebra).
Hopefully this helps!