I'm trying to prove that the function $f$ of here is Lebesgue measurable. So we must prove that the set: $$ E_{\alpha} = \left\{x : f(x) > \alpha \right\} $$
is measurable for every $\alpha \in \mathbb{R}$
if $\alpha \geq1 $ so $E_{\alpha } = \emptyset$ because there is no value of function so $E_{\alpha}$ is measurable.
if $\alpha \leq 0$ so $E_{\alpha} = [0,\infty)$ because all the values of the functions are greather than 0. But is this set measurable even thou that the length of the interval is $\infty$?
and... some help with the case $0 < \alpha <1 $ ?
Hint:
We have that $[-n, 0)$ is measurable, then
$(-\infty , 0) = \bigcup_{n \in \mathbb{N}}[-n, 0)$ is measurable. Hence, $(\infty, 0)^{c} = [0, \infty)$ is measurable
To prove that $E_{\alpha}$, $0 < \alpha < 1$, observe that $[0, 1] = \bigcup_{n \in \mathbb{N}}[\frac{1}{2^{n+1}}, \frac{1}{2^{n}}]$.
Hence, exist $m \in \mathbb{N}$ such that $\alpha \in [\frac{1}{2^{m+1}}, \frac{1}{2^{m}}]$
See that $E_{\alpha} \subset \mathbb{R} \setminus \mathbb{Q}$, and $\mathbb{R} \setminus \mathbb{Q}$ is measurable.
To $x \in E_{\alpha}$, $f(x) > \alpha$, then $[x] \leqslant m$.
Thus, $E_{\alpha} = (\mathbb{R} \setminus \mathbb{Q}) \cap [0, m+ 1]$ that is measurable.