If $f : X \rightarrow [ -\infty, +\infty]$ is a measurable function is $\{x : f(x) < \alpha \}$ a measurable set?
Where $\alpha \in \bar{R}$.
It seems to me that it entirely depends on the range of $f$, for instance if the range of $f$ includes $- \infty$ then the set $\{f(x) : f(x) < \alpha \}$ will be of the form $[- \infty, \beta ]$ or $[- \infty, \beta )$, ($\beta \in \bar{R}$) these sets are not open so they do not map back to a measurable set.
But I think I am wrong, why is that?
Yes, by definition. $\{x\mid f(x)<\alpha\}$ is the preimage $f^{-1}([-\infty,\alpha))$, and the definition of $f$ being measurable is that $f^{-1}(A)$ is measureable whenever $A$ is measurable (not just when $A$ is open). Since $[-\infty,\alpha)$ is clearly measurable, its preimage must be too.
It doesn't matter that not all of $A$ may be hit by $f$: the preimage of $A$ is still the preimage of $A$, even if it may also be the preimage of some other, smaller, set.
Bonus comment: $[-\infty, \alpha)$ is actually open in $\overline{\mathbb R}$!
Extra bonus comment: In the usual case where the $\sigma$-algebra assumed for the codomain of $f$ is the Borel algebra, then it is sufficient for $f$ be measurable that the preimage of every open set is measurable. Then, automatically, the preimages of all other Borel subsets of $\overline{\mathbb R}$ will also be measurable sets. (This can be proved directly from the definition of Borel set and $\sigma$-algebra).