Consider the following definiton:
Definition: Let $(X, \mathcal{A})$ be a measurable space. A function $f:X \rightarrow \mathbb{R}$ is measurable if {$x \in X:f(x)>a $}$ \in \mathcal{A}$ for all $a \in \mathbb{R}$.
This definition seems kinda odd to me. I heard that the motivation of measurable functions is to have functions which "preserve measurable sets". Now I cannot follow on why the above definiton does that. Wouldn't it be more natural to define measurable function as "If $f(E)$ is a measurable set, then so is $E$".
I do not really see the connection with the intention and the definition.
Could someone pleas explain to me why we use the definition we use and not something like what I wrote. Could it be that both versions are equivalent and the official definition is just easier to work with?
If $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ are measurable spaces, a function $f:X\to Y$ is measurable if for all $B\in\mathcal{B}$, the inverse image $$f^{-1}(B)=\{x\in X:f(x)\in B\}$$ is in $\mathcal{A}$. If $Y=\mathbb{R}$ and $\mathcal{B}$ is the smallest $\sigma$-algebra containing all open sets (the Borel $\sigma$-algebra)m then this definition is equivalent to the one you gave. However, proving the equivalence is not straightforward. Usually, one proves this by some "monotone class argument."