TITLE: Proof that the preimage of generated $\sigma$-algebra is the same as the generated $\sigma$-algebra of preimage.
QUESTION [5 upvotes]: This question has also been asked here, but the answer there didn't help me.
I am trying to prove that, given some measurable space $(X, \Sigma)$, if $G$ is a collection of subset of $X$ such that $\sigma(G) = \Sigma$, then
$$f^{-1}(\sigma(G)) = \sigma(f^{-1}(G)).$$
So far, I have been able to show that $\sigma(f^{-1}(G)) \subseteq f^{-1}(\sigma(G))$, but I am having trouble with the opposite inclusion. I have shown that $f^{-1}(G)$ is a $\sigma$-algebra, from which it follows that $f^{-1}(G) \subseteq \sigma(f^{-1}(G))$, but that's about all I have been able to show. How do I proceed from there?
The answer in the other question suggests the following approach:
Prove that

$f^{-1}(G) \in \sigma(f^{-1}(G))$,
if $f^{-1}(A_i) \in \sigma(f^{-1}(G))$ for $i \in \mathbb{N}$ (i.e. a countable collection), then $f^{-1}(\bigcup_i A_i) \in \sigma(f^{-1}(G))$, and
if $f^{-1}(A) \in \sigma(f^{-1}(G))$ then $f^{-1}(Y \setminus A) \in \sigma(f^{-1}(G))$.

However, I am not sure how to actually prove this, and even then, I'm not sure how it follows from this that $f^{-1}(\sigma(G)) \subseteq \sigma(f^{-1}(G))$.

REPLY [8 votes]: After working on it some more, I have come up with a proof. It follows the ideas of the question I linked, and uses the properties of preimage that saz listed in the comments. I also found out that the proof does not actually need the premise that $\sigma(\mathcal{G}) = \Sigma$, and it is in fact not even necessary that $X$ is a measurable space.
The proof is as follows. We wish to prove that $f^{-1}(\sigma(\mathcal{G})) \subseteq \sigma(f^{-1}(\mathcal{G}))$.
  First we define
  $$D = \{G \subseteq Y \mid f^{-1}(G) \in \sigma(f^{-1}(\mathcal{G}))\}.$$
  Observe that what we want to show will follow if we prove that
  $D$ is a $\sigma$-algebra and that $\mathcal{G} \subseteq D$.
  This is because it will imply that $\sigma(\mathcal{G}) \subseteq D$,
  which, by the definition of $D$, will imply that if $G \in \sigma(\mathcal{G})$,
  then $f^{-1}(G) \in \sigma(f^{-1}(\mathcal{G}))$.
  This further implies that
  $$f^{-1}(\sigma(\mathcal{G})) = \{f^{-1}(G) \mid G \in \sigma(\mathcal{G})\} \subseteq \sigma(f^{-1}(\mathcal{G})).$$
  Hence, we now show that $D$ is a $\sigma$-algebra and that $\mathcal{G} \subseteq D$.
  First observe that $f^{-1}(\mathcal{G}) \subseteq \sigma(f^{-1}(\mathcal{G}))$.
  This implies that if $G \in \mathcal{G}$ then $f^{-1}(G) \in \sigma(f^{-1}(\mathcal{G}))$.
  By the definition of $D$, this further implies that $\mathcal{G} \subseteq D$.
  To show that $D$ is a $\sigma$-algebra,
  we verify properties of a $\sigma$-algebra.

Since $\emptyset = f^{-1}(\emptyset)$ and
                         $\emptyset \in \sigma(f^{-1}(\mathcal{G}))$,
                         we have $\emptyset \in D$.
Assume $G \in D$. Then $f^{-1}(G) \in \sigma(f^{-1}(\mathcal{G}))$.
                         Since $\sigma(f^{-1}(\mathcal{G}))$ is closed under complement,
                         we get
                         $$X \setminus f^{-1}(G) = f^{-1}(Y) \setminus f^{-1}(G) = f^{-1}(Y \setminus G) = f^{-1}(G^c) \in \sigma(f^{-1}(\mathcal{G})),$$
                         which implies that $G^c \in D$.
Assume $G_1,G_2,\dots \in D$. Then for all $G_i$
                         we have $f^{-1}(G_1) \in \sigma(f^{-1}(\mathcal{G}))$.
                         Since $\sigma(f^{-1}(\mathcal{G}))$ is closed under
                         countable union, we get
                         $$\bigcup_i f^{-1}(G_i) = f^{-1}(\bigcup_i G_i) \in \sigma(f^{-1}(\mathcal{G})),$$
                         which implies that $\bigcup_i G_i \in D$.