Equalities between the preimage of a r.v. $A$ on the Borel $\sigma$-algebra, the sigma algebra $\sigma(A)$ and $\sigma(A^{-1}(\mathcal{B}))$

Question

Let $A$ be a real-valued random variable on the probability space $(\Omega, F, P)$. I'm trying to show that $\sigma(A) = \sigma(A^{-1}(\mathcal{B})) = A^{-1}(\mathcal{B} = \{A^{-1}(S): S \in \mathcal{B}\}$, where $\mathcal{B}$ is the Borel sigma-algebra on $\mathbb{R}$. Ironic enough, there seems to be a similar post Preimage of generated $\sigma$-algebra and the $\sigma$-algebra generated by the preimage, but as I don't understand how its conclusion is applicable to my specific question, I made this post.

So far I've shown that $\sigma(A) \subset \sigma(A^{-1}(\mathcal{B}))$, and now I don't know how to approach the reverse inclusion. $\sigma(A)$ is defined to be the smallest sigma algebra with which $A$ is a measurable mapping, hence we only know that the pre-image $A^{-1}(S)$ of every $S \in \mathcal{B}$ is in $\sigma(A)$. While this yields quite trivially that $\sigma(A) = A^{-1}(B)$, I'd still like to verify the set equality between $\sigma(A)$ and $\sigma(A^{-1}(B))$, but I don't know how to show that any element of $\sigma(A^{-1}(B))$ is also contained in $\sigma(A)$. I certainly know this to be true for $A^{-1}(B)$, $A^{-1}(B) \subset \sigma(A)$, but I don't see how to make the last step of the claim.

Answer 1

I just realized the missing piece of the proof: In general, if $A$ is a sigma-algebra and $B \subset A$ is any subset of $A$, then $\sigma(B)\subset \sigma(A)$. Therefore, $A^{-1}(B)\subset \sigma(A) \implies \sigma(A^{-1}(B))\subset \sigma(A)$.