As I wrote in the title, I was wondering if it was true that given $f$ a Lebesgue-measurable function,and $E$ a measurable set, we have that $f^{-1} (E)$ is also measurable, and (if not) to show a proof or a counterexmple. I have thought that since the counnterimage of every borelian set is a borelian, I had to use some "patological" sets, such as Cantor set or Vitali set.
P.S. I would like to prove it in $\mathbb{R}$, without considering the possibility that $f(a)= \pm \infty$ for some $a$.
Since preimages of closed sets are measurable, you should focus on sets of measure zero. In particular, if you can find an injective measurable function $f$ that sends a set of positive measure $P$ to a set of zero measure, you can find a counterexample by restricting $f$ to a nonmeasurable set $N \subset P$ and setting $E = f(N)$.
A standard example uses the Cantor function. Let $c : [0,1] \to [0,1]$ be the Cantor function and let $g(x) = c(x) + x$. Then $g : [0,1] \to [0,2]$ is continuous and strictly increasing. Look at the function $f = g^{-1}$.