Why do we define Measurable functions by their pre-image? Let $(X,\mathcal{M})$ and $(Y,\mathcal{N})$ be measurable spaces. We call the function $f:X\to Y$ to be $(\mathcal{M},\mathcal{N})$-measurable if $f^{-1}(E)\in \mathcal{M}$ for all $E\in \mathcal{N}$.
Similarly, when we are trying to define "nearness" in functions we call a function to be continuous if the preimage of the open set is open. There might be other types of similar definitions (if so, I'd appreciate more examples!), but my question is why we define the particular function always based on the preimage?
Thanks in advance!
 A: Unless we assume injectivity, it is extremely difficult to "control" the image with the preimage.  In fact, in the case of a constant map, every nonempty preimage has the same image, so the preimage tells us nothing about the image at all.
Inverse functions are forced to be injective, so we do not have this problem when we try to control the preimage in terms of the image.
A: It's not that we "always" define things in terms of preimages, it's just that the preimage-based definitions are the ones that "do the right thing" in the cases you mentioned.
Measurable functions allow Lebesgue integration because we can partition the range into small disjoint sets, measure the preimages of those sets, and add up the resulting areas. The sum is justified because the preimages of disjoint sets are disjoint; it wouldn't work right if you instead tried to define integration using the "forward" property $E\in\mathcal M\implies f(E)\in\mathcal N$ (but you should try it!).
The situation is more subtle for continuity of functions (see this question), but the familiar $\epsilon/\delta$ definition of continuity in metric spaces is equivalent to the preimage-based definition in terms of open sets.
