Prove that any interval is measurable. The goal is to show that an interval is Lebesgue measurable. As it's been pointed out in my notes that it is enough to show that an interval of the form $(a,\infty)$ is measurable. So here's my attempt.
Note that $(a,\infty)=\bigcup_{n=1}^\infty (a,a+n)$. Let $\epsilon>0$ and $n\in\mathbb{N}$. Then $(a,a+n)$ is measurable iff there exists an open set $U_\epsilon$ such that $(a,a+n)\subseteq U_\epsilon$ and $m^*(U\setminus(a,a+n))<\epsilon$. Taking $U_\epsilon=(a,a+n)$ we have $(a,a+n)$ is measurable. Therefore $(a,\infty)$ is measurable. Hence any interval is measurable. (Here $m^*(A)$ denotes Lebesgue outer measure of the set $A$).
Is this argument alright?
 A: let $A \subset \mathbb R$ be an arbitrary set and $A \subset \cup_{i = 1}^{\infty}(a_{i},b_{i})$. It suffices to construct $(c_i,d_i)$ and $(e_i,f_i)$ such that $A \cap (a,\infty) \subset \cup_{i=1}^{\infty}(c_i,d_i)$, $A \cap (-\infty,a] \subset \cup_{i=1}^{\infty}(e_i,f_i)$, and:

$\sum_{i=1}^{\infty}(d_i - c_i) + \sum_{i=1}^{\infty}(f_i - e_i) \leq \sum_{i=1}^{\infty}(b_i - a_i)$

Since then we will have $m^*(A) \geq m^*(A\cap(a,\infty)) + m^*(A\cap(a,\infty)^c)$ meaning $(a,\infty)$ is Lebesgue measurable.
To construct $c_i,d_i,e_i,f_i$, simply split any intervals containing $a$ into $(a_i,a)\cup (a-\frac{\epsilon}{2^i},b_i) =: (e_i,f_i)\cup (c_i,d_i)$ for $\epsilon > 0$, and set the other intervals to either match the $(a_i,b_i)$ or be the empty set, depending on which interval $(a_i,b_i)$ lies in, then we have:
$\sum_{i=1}^{\infty}(d_i - c_i) + \sum_{i=1}^{\infty}(f_i - e_i) = \sum_{i=1}^{\infty}(b_i - a_i) + \sum_{i\in\{{\text{indices} | a \in (a_i,b_i)}\}}\frac{\epsilon}{2^i} \leq \sum_{i=1}^{\infty}(b_i - a_i) + \sum_{i\in\mathbb N}\frac{\epsilon}{2^i} = \sum_{i=1}^{\infty}(b_i - a_i) + \epsilon$
Since $\epsilon$ was arbitrary, we have the required identity.
