I've come up with this proposition in a lecture note.
Consider $f: X \to Y$.
For $A \subseteq X$, we define $f[A] := \{f(x) \in X \mid x \in A\}$.
For $A \subseteq Y$, we define $f^{-1}[A] := \{x \in X \mid f(x) \in A\}$.
For $\mathcal C \subseteq \mathcal P(X)$, we define $f [\![ \mathcal C ]\!] := \{f[A] \mid A\in \mathcal C\}$.
For $\mathcal C \subseteq \mathcal P(Y)$, we define $f^{-1} [\![ \mathcal C ]\!] := \{f^{-1}[A] \mid A\in \mathcal C\}$.
The smallest $\sigma$-algebra containing $\mathcal C$ is denoted by $\sigma(\mathcal C)$.
Prove that $\mathcal C \subseteq \mathcal P(Y)$ implies $\sigma(f^{-1} [\![ \mathcal C ]\!]) = f^{-1} [\![ \sigma(\mathcal C) ]\!]$.
A proof can be found here. It's so interesting that I tried to give it a shot. Could you please confirm if my attempt is fine?
My attempt:
We notice that if $\mathcal A$ is a $\sigma$-algebra over $X$, then $\overrightarrow{\mathcal A} := \{B \subseteq Y \mid f^{-1}[B] \in \mathcal A\}$ is a $\sigma$-algebra over $Y$. Similarly, if $\mathcal B$ is a $\sigma$-algebra over $Y$, then $\overleftarrow{\mathcal B} := \{f^{-1}[B] \mid B \in \mathcal B\}$ is a $\sigma$-algebra over $X$.
It maybe the case that $f[f^{-1}[B]] \subsetneq B$, so it's not necessarily true that $\overrightarrow{\mathcal A} \subseteq f [\![ \mathcal A ]\!]$. For $A \in f^{-1} [\![ \overrightarrow{\mathcal A} ]\!]$, $A = f^{-1}[B]$ for some $B \in \overrightarrow{\mathcal A}$. By construction of $\overrightarrow{\mathcal A}$, $A \in \mathcal A$ and thus $f^{-1} [\![ \overrightarrow{\mathcal A} ]\!] \subseteq \mathcal A$. Let $$\begin{aligned} \mathfrak S_1 &= \{\mathcal A \text{ is a } \sigma \text{-algebra} \mid f^{-1} [\![ \mathcal C ]\!] \subseteq \mathcal A\} \\ \mathfrak S_2 &= \{ f^{-1} [\![ \mathcal B ]\!] \mid \mathcal B \text{ is a } \sigma \text{-algebra and } \mathcal C \subseteq \mathcal B\} \end{aligned}.$$
Then $\sigma( f^{-1} [\![ \mathcal C ]\!] ) = \bigcap \mathfrak S_1$ and $f^{-1} [\![ \sigma(\mathcal C) ]\!] =\bigcap \mathfrak S_2$.
Let $\mathcal A \in \mathfrak S_2$. Then $\mathcal A = f^{-1} [\![ \mathcal B ]\!]$ for some $\sigma$-algebra $\mathcal B$ such that $\mathcal C \subseteq \mathcal B$. Then $\mathcal A$ is also a $\sigma$-algebra such that $f^{-1} [\![ \mathcal C ]\!] \subseteq f^{-1} [\![ \mathcal B ]\!] = \mathcal A$. This means $\mathcal A \in \mathfrak S_1$ and thus $\mathfrak S_2 \subseteq \mathfrak S_1$. We then obtain $\bigcap \mathfrak S_1 \subseteq \bigcap \mathfrak S_2$.
For $B \in \mathcal C$, $f^{-1} [B] \in f^{-1} [\![ \mathcal C ]\!] \subseteq \mathcal A$. Then $B \in \overrightarrow{\mathcal A}$ by construction of $\overrightarrow{\mathcal A}$. Hence $\mathcal C \subseteq \overrightarrow{\mathcal A}$ and thus $f^{-1} [\![ \overrightarrow{\mathcal A} ]\!] \in \mathfrak S_2$. This means for every $\mathcal A \in \mathfrak S_1$, we can find a $\sigma$-algebra $f^{-1} [\![ \overrightarrow{\mathcal A} ]\!] \in \mathfrak S_2$ such that $f^{-1} [\![ \overrightarrow{\mathcal A} ]\!] \subseteq \mathcal A$ (this inclusion is proved above). Hence $\bigcap \mathfrak S_2 \subseteq \bigcap \mathfrak S_1$.
Finally, $\bigcap \mathfrak S_1 = \bigcap \mathfrak S_2$. This completes the proof.
Update: From this comment by Nate Eldredge, I would like to apply his enlightening idea.
We want to prove $f^{-1} [\![ \sigma(\mathcal C) ]\!] \subseteq \sigma(f^{-1} [\![ \mathcal C ]\!])$ which is the hard part of the theorem. Let $$\mathcal D = \{A \subseteq Y \mid f^{-1} [A] \in \sigma(f^{-1} [\![ \mathcal C ]\!]\}.$$
Then it's easy to show that $\mathcal D$ is indeed a $\sigma$-algebra over $Y$ and $\mathcal C \subseteq \mathcal D$. Hence $\sigma(\mathcal C) \subseteq \mathcal D$. The inclusion then follows.
