all maps are measurable? A map $T:X  \rightarrow Y$ is $A, B$- measurable if the pre-image of every set in $B$ is a set in $A$, where $A,B$ are $\sigma$-algebras in $X,Y$. 
But I was thinking that isn't any map $T: X \rightarrow Y$ measurable then? Because since pre-images behave so nicely, the pre-image of all sets from $B$ make up a $\sigma$-algebra in $X$, and thus the definition is satisfied?
 A: It is true that preimages of $B$ form a sigma-algebra in $X$. However, there is no a priori reason why this sigma-algebra should be contained in $A$.
A: That's not true, of course. 
Here's an example which doesn't require us to be in any special setting: Let $X$ be any set and let $\mathcal A$ be a $\sigma$-algebra on $X$ that is a strict subset of $\mathcal P(X)$. Let $A \in \mathcal P(X) -\mathcal A$. Then: $1_A - 1_{A^C}$ is not measurable. 
A part of what you are saying is true, though. The set $f^{-1}(\mathcal B)$ does form a $\sigma$-algebra on $X$, but it simply isn't necessarily $\subseteq \mathcal A$, the presumed $\sigma$-algebra on $X$.
IMO one must refer to a measurable space as $(X, \mathcal A)$ rather than just $X$ when one is still a beginner. This allows one to avoid possible misconceptions. What I mean is: had you written $T: (X, \mathcal A) \to (Y, \mathcal B)$, you wouldn't have had this confusion, I think.
A: Hint: The more sets you have in the $\sigma$-algebra of $X$ and the less sets you have in the $\sigma$-algebra of $Y$, the more "likely" your function is to be measurable.
I'm sure you can think of some trivial nonmeasurable examples now.
