Introduction:
When we prove the following result:
Let $f$ from $X$ to $Y$ ($X$ and $Y$ being endowed with $\sigma$-algebras). If $\mathscr G$ generates $\Sigma_Y$, and $f^{-1}(\mathscr G)\subseteq \Sigma_X$, then $f$ is measurable.
we usually consider the set $S_1=\{ B\subseteq Y \mid f^{-1}(B) \in \Sigma_X \}$. Then, using order theory considerations, we get to prove $\Sigma_Y \subseteq S_1$ (the desired result) by showing that $S_1\in \{\Sigma \mid \Sigma \text{ is a $\sigma$-algebra}, \mathscr G \subseteq \Sigma \}$.
Same proof, with a different set $S_2$?
Now, let's redefine $S_1$ as follows (and let's call it $S_2$): $S_2=\{\mathscr C\subseteq \mathscr P (Y)\mid f^{-1}(\mathscr C) \subseteq \Sigma_X \}$ where $\mathscr P(Y)$ denotes the power set of $Y$.
- How could the same proof be readapted with considerations on $S_2$ ?
It looks we should check this time that $S_2 \subseteq \{A \subseteq \mathscr P(Y) \mid A \text{ is some mathematical object}, \mathscr G \in A\}$, the "mathematical object" being something like a set of $\sigma$-algebras.
Assuming the two requirements (that $S_2$ is "a set of $\sigma$-algebras" and that $\mathscr G \in S_2$) are met, the conclusion should ideally be: $\Sigma_Y \in S_2$. But, here I don't see how an order theory argument could fit ?
- Additional question: in case it is impossible to adapt the proof of the above theorem with $S_2$, is there a way to construct $S_1$ out of $S_2$ ?
