Show that a simple function is measurable if its parts are all measurable

Question

I understand that a simple function $s:\mathbb{R}^2 \to \mathbb{R}$ is any function which assumes only a finite number of distinct values. It can also be written as a linear combination of indicator functions $s= \sum_{k=1}^N \alpha_k f_{A_k}$ with disjoint sets $A_k$ and distinct $\alpha_k \in \mathbb{R}$. I'm trying to prove that a simple function, defined above, is measurable if and only if all of the sets $A_k$ are measurable. Since I am still somewhat new to these types of proofs, I am having difficulty being rigorous, though I have a general idea of how to explain this.

Answer 1

Let $B$ a Borel subset of $\Bbb R$. Let $I:=\{k\in [N],\alpha_k\in B\}$ (this set may be empty). Then $s^{-1}(B)=\bigcup_{k\in I}A_k$. So, if the $A_k$ are all measurable, so will be $s$.

Now, assume $s$ measurable, and let $k\in [N]$. Then $s^{-1}(\{\alpha_k\})$ is measurable, as so is $\{\alpha_k\}$. But $s^{-1}(\{\alpha_k\})=A_k$, which concludes the proof.