I know that a measurable function is defined on two measurable spaces $(\Omega_1, \mathcal F_1)$ and $(\Omega_2, \mathcal F_2)$.
My problem is that I don't understand why we need the second sigma-algebra $\mathcal F_2$.
What's the harm in defining a measurable function as a function which fulfills the condition: for all $A \subseteq\Omega_2: f^{-1}(A) \in \mathcal F_1$?
I know that we would include non-measurable sets this way but why can't the function map to these kind of sets? My understanding is that it has to do with the integrability of the function f.
Is that correct?
If there is freedom in choosing $\mathcal F_2$ then it is possible to give functions $f:\Omega_1\to\Omega_2$ with $f^{-1}(A)\in\mathcal F_1$ for all subsets of $\Omega_2$ the predicate "measurable".
This simply by choosing $\mathcal F_2=\wp(\Omega_2)$.
This happens e.g. if we are observing discrete random variables that only take values in a countable set $\Omega_2$.
Observe however that - if $\mathcal F_2$ is a proper subcollection of $\wp(\Omega_2)$ - this is a stronger demand than $f^{-1}(A)\in\mathcal F_1$ for all $A\in\mathcal F_2$.
So lots of functions that might be useful might become unmeasurable if we replace a suitable $\mathcal F_2$ by $\wp(\Omega_2)$.
It is at first hand the more basic "measurability" that suffers under this. The "integrability" that you suggests also suffers, but this merely because integral functions are measurable.