Approximation of Lebesgue measurable set

Question

Let $E \subseteq \mathbb{R}$ be a set with finite Lebesgue measure. I want to show that for each $\varepsilon > 0$, there exists a disjoint collection of closed intervals $\{I_k\}$ such that $m(E \setminus \bigcup I_k) = 0 $ and $\sum_{k}m(I_k) \leqslant m(E) + \varepsilon$. I know that $E$ can be approximated by a $F_\sigma$ set, but it seems like we are proving a slightly stronger statement.

I considered \begin{equation*} m^{**}(E) = \inf \{ \sum_{k}m(I_k) \mid E \subseteq \bigcup_{k}I_k\} \end{equation*} where the $I_k$'s are disjoint closed intervals. I thought $m^{**}(E) = m^{*}(E)$ (one direction is easy; for the other direction, I use the fact that each closed interval can be realized as the closure of an open interval, and that $m^{*}(E) = \inf \{m(O): E \subset O, O \text{ open}\}$). Is this correct? If so, how can we proceed from there and argue that the infimum can be achieved?

Answer 1

Lemma: Suppose $(a,b)$ is a bounded open interval. Then there exist disjoint closed intervals $I_1,I_2,\dots \subset (a,b)$ such that $\sum_{n=1}^{\infty}m(I_n)=b-a.$ Hence $m[(a,b)\setminus \bigcup_{n=1}^{\infty}I_n]=0.$

Proof: Let $I_1=[a+(b-a)/4, b-(b-a)/4].$ This the closed "middle half" of $(a,b).$ Put $I_1$ aside. We are left with two disjoint open intervals of length $(b-a)/4.$ In each of these intervals, do the same: remove the closed middle halves of length $(b-a)/8.$ We now have disjoint closed subintervals $I_1,I_2,I_3$ of $(a,b)$ whose lengths sum to $(b-a)/2+(b-a)/4.$ Continue to obtain obtain disjoint closed intervals $I_1,I_2,\dots \subset (a,b)$ such that

$$\sum_{n=1}^{\infty}m(I_n)=\sum_{m=1}^{\infty}\frac{b-a}{2^m}=b-a.$$

To the problem at hand, let $\epsilon >0.$ Then there exists an open set $U$ containing $E$ such that $m(U)<m(E)+\epsilon.$ Now $U$ is the countable disjoint union of open intervals $(a_n,b_n).$ For each $n,$ apply the lemma to obtain disjoint closed intervals $I_{nk},k=1,2,\dots$ The collection $\{I_{nk}\}$ then does the job.