Proof that the preimage of generated $\sigma$-algebra is the same as the generated $\sigma$-algebra of preimage.

Solution 1:

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 $\bigcup_i G_i \in D$.

Solution 2:

This is based on the following properties of $f^{-1}$: $$ f^{-1}(\bigcup_{i=1}^{\infty}G_i)=\bigcup_{i=1}^{\infty}f^{-1}(G_i)\tag1 $$ $$ f^{-1}(\bigcap_{i=1}^{\infty}G_i)=\bigcap_{i=1}^{\infty}f^{-1}(G_i)\tag2 $$ $$ f^{-1}(G_i^c)=f^{-1}(G_i)^c\tag3 $$ Suppose $A\in \sigma(G)$ and $\:A=\bigcup_{i=1}^{\infty}G_i,\:G_i\in G$, i.e. $A$ is formed by the countable union of sets in $G$. Then $$ f^{-1}(A)=f^{-1}(\bigcup_{i=1}^{\infty}G_i)=\bigcup_{i=1}^{\infty}f^{-1}(G_i)\in\sigma(f^{-1}(G)) $$ For $A\in \sigma(G)$ and $\:A=\bigcap_{i=1}^{\infty}G_i,\:G_i\in G$, i.e. $A$ is formed by the countable intersection of sets in $G$. Then $$ f^{-1}(A)=f^{-1}(\bigcap_{i=1}^{\infty}G_i)=\bigcap_{i=1}^{\infty}f^{-1}(G_i)\in\sigma(f^{-1}(G)) $$ For $A\in \sigma(G)$ and $\:A=G_i^c,\:G_i\in G$, i.e. $A$ is formed by the complement of set in $G$. Then $$ f^{-1}(A)=f^{-1}(G_i^c)=f^{-1}(G_i)^c\in \sigma(f^{-1}(G)) $$ $G_i\in \sigma(G)$ is again the countable union, intersection and complement of sets and so $f^{-1}(G_i)\in \sigma(f^{-1}(G))$. So we have $$f^{-1}(\sigma(G)) \subseteq \sigma(f^{-1}(G))$$