Explanation of the first step in the proof of Vitali theorem. The theorem and the first part of its proof is given below:



But I do not understand the first statement in the proof, why countable subadditivity of outer measure lead us to suppose that $E$ is bounded? could anyone explain this for me, please?
 A: Theorem 17: Any set of real numbers with positive outer measure contains a subset that fails to be measurable.
Theorem 17': Any bounded set of real numbers with positive outer measure contains a subset that fails to be measurable.
The part of the proof that follows the first sentence proves theorem 17'. We therefore just have to deduce theorem 17 from theorem 17'. 
Deduction of theorem 17 from theorem 17': Let $E$ be a set of real numbers with positive outer measure. Let $\mu$ denote outer measure. If $\mu(E \cap [n,n+1)) = 0$ for each $n \in \mathbb{Z}$, then $\mu(E) = \mu(\cup_{n \in \mathbb{Z}} E\cap [n,n+1) ) \le \sum_{n \in \mathbb{Z}} \mu(E\cap [n,n+1)) = 0$, a contradiction. In other words, there is some $n \in \mathbb{Z}$ with $\mu(E\cap [n,n+1)) > 0$. Applying theorem 17' to $E \cap [n,n+1)$ (which we may, since $\mu(E\cap[n,n+1)) > 0$ and since $E \cap [n,n+1) \subseteq [n,n+1)$ is bounded), we get some non-measurable set $F \subseteq E\cap [n,n+1)$. Since $F$ is then a non-measurable subset of $E$ as well, theorem 17 is deduced.
