Let $(\Omega, {\cal F})$ and $(E, {\cal E})$ be measurable spaces and $X$ a r.v. from $\Omega$ into $E$ (i.e., $X$ is ${\cal F}/{\cal E}$-measurable). We assume that $\cal E$ is generated by ${\cal A}$, i.e., ${\cal E} = \sigma({\cal A})$. I wish to show the following important fact: $\sigma(X) = \sigma(X^{-1}{\cal A})$ where $\sigma(X) := \{X^{-1}A; A\in {\cal E}\}$ and $X^{-1}{\cal A} := \{X^{-1}A; A\in {\cal A}\}$.
Is the following proof correct? Or is there any other simpler proof of it?
Since $\sigma(X) \supset X^{-1}{\cal A}$, we have $\sigma(X) \supset \sigma(X^{-1}{\cal A})$.
To show the other inclusion, we note that for any $A\in {\cal A}$, $X^{-1}A \in X^{-1}{\cal A} \subset \sigma(X^{-1}{\cal A})$. Since $\cal A$ generates $\cal E$, this shows that the r.v. $X$ is $\sigma(X^{-1}{\cal A})$-measurable. By the minimal property of $\sigma(X)$ we can conclude that $\sigma(X) \subset \sigma(X^{-1}{\cal A})$. QED