# what is the measure of the set $[0,\infty)$?

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 $$0 < \alpha <1$$ ?

• You will need to look up the definition of "Lebesgue measurable set". You will find that $[0,\infty)$ satisfies that definition, so it is a measurable set. Commented Jul 8, 2019 at 15:04

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.