I am not sure if I understand it. Is that correct?
Show that $f(x)=2x$ is measurable.
$1) \quad\alpha < 0$
$\{x: f(x) < \alpha \}=(-\infty,\frac{\alpha}{2})$
$2) \quad \alpha=0$
$\{x:f(x)<0\} = (-\infty, 0)$
$3)\quad \alpha >0 $
$\{x: f(x) < \alpha \}=(-\infty,\frac{\alpha}{2})$
These sets are measurable so function is measurable.
This is correct, using the theorem stating that if $f^{-1}(B)$ is measurable for all $B$ in some generator of the $\sigma$-algebra being used, then $f$ is measurable. Since $\{ (-\infty,a) \mid a\in \mathbb R\}$ is a generator for the Borel-$\sigma$-algebra on $\mathbb R$, $f$ is measurable with respect to this $\sigma$-algebra. Note too that you didn't need to make a distinction between $\alpha < 0$, $\alpha = 0$ and $\alpha > 0$ since in all cases, $f^{-1}((-\infty,a)) = (-\infty, a/2)$.
You can use the fact that continuous functions are always measurable. A short and simple answer can be found here: f a real, continuous function, is it measurable?
