We define a simple function, $s=\sum_k \alpha_k\chi_{A_k}$ where the sets $A_k$ are all disjoint and the numbers $\alpha_k \in [-\infty,\infty]$
I am looking to prove that if we know that all the $A_k$ are measurable then s is measurable too .
My approach : Since by definition the simple function can attain almost countable values , arrange the $\alpha_k$ in ascending order. Correpondingly arrange the $A_k$ in such a way that $s^{-1}(x) = \alpha_k$ implies that $ x \in A_k$ Now for any t , we have that
1.) $t \in [-\infty,\alpha_1] $ or
2.)$t \in [\alpha_k,\alpha_{k+1}]$ or
3.) $t \in [\alpha_n,\infty] $
In case 1.) we get ${(x|s^{-1}(x) < t )}$ = $\phi$ which we know is measurable
In case 2.) we get ${(x|s^{-1}(x) < t )}$ = $\cup_{n=1}^{k}A_k$ since each $A_k$ is measurable then so is their finite union
IN case 3.) we get ${(x|s^{-1}(x) < t )}$ = $\cup_{n=1}^{n}A_k$ since $A_k$ is measurable so is their countable union
Hence these three cases together exhaust all the possibilities and prove that s is a measurable function.
Please check my proof and indicate if it is incorrect somewhere or if it could be improved in any way.
This is correct but you don't need to do separate case.
$f^{-1}(- \infty, t) = \bigcup_{i}A_{i}$ where $I$ is the set $k$ such that $\alpha_k < t$, and this set is mesurable in all case.