I learned that a measurable function can be defined as:
Let ( X, Σ ) and ( Y, T ) be two measurable spaces. A function f: X → Y is called measurable if for every E ∈ T the pre-image of E under f is in Σ.
I read that f doesn't have to be surjective, but couldn't find why:
Given the definition, $f^{-1}(Y) \in Σ$.
Since every $E \in\Sigma$ is a subset of $X$, $f^{-1}(Y) = \{x\in X|f(x)\in Y\} \subset X$. Doesn't that mean that for any $y \in Y$ there $\exists x: f(x)=y$ and therefore f is surjective?
Also, doesn't this mean that $f(X)=Y$?
Where did I make a mistake here?
You need to revise the definition of surjective and function, $f^{-1}(Y) \subset X $ is true for any function. The notation $f^{-1}(Y)$ means: the elements of $X$ that land in $Y$ when $f$ is applied. But $Y$ is the image of $f$, all elements of $X$ land in $Y$! What you really want for $f$ to be surjective is that $f(X)=Y$. The two things are not the same. Just try to picture in your head what happens with a constant function.