I have a question regarding the following proof (Theorem 8.1). In the proof, I don't quite understand the part (the last sentences in the proof of Theorem 8.1 where it says "Therefore .... "):
$X^{-1}(F) \subset \sigma(X^{-1}(C)) $
that is the part I don't understand. How is the $X^{-1}(F)$ being a subset of $\sigma (X^{-1}(C))$ arrive?
Could someone give a comment or explanation?
thank you
It is worth pointing out that this is not a necessary step in the proof. Once we have shown that $\mathcal{B} := \{ A \in \mathcal{F} : X^{-1} (A) \in \mathcal{E} \} = \mathcal{F}$, we know that for any set $A \in \mathcal{F}$ it is the case that $X^{-1}(A) \in \mathcal{E}$, which is precisely what it means for $X$ to be measurable. In other words, $X^{-1} (\mathcal{F}) \subseteq \mathcal{E}$ is all that was needed to finish the proof.
To answer your question, let $\mathcal{D} = \{ A \in \mathcal{F} : X^{-1} (A) \in \sigma (X^{-1} (\mathcal{C} )) \}$. Note that $\mathcal{C} \subseteq \mathcal{D}$, because $X^{-1} (\mathcal{C}) \subseteq \sigma (X^{-1} (\mathcal{C}))$, and that $\mathcal{D}$ is a $\sigma$-algebra because of the three properties of preimages with unions/intersections/complements. Therefore, $\sigma (\mathcal{C}) \subseteq \mathcal{D}$, because $\sigma (\mathcal{C})$ is the smallest $\sigma$-algebra containing $\mathcal{C}$. But by the construction of $\mathcal{D}$, any set in $\mathcal{D}$ must belong to $\sigma (X^{-1} (\mathcal{C}))$. Thus $X^{-1} (\sigma (\mathcal{C})) \subseteq \sigma (X^{-1} (\mathcal{C}))$.