I am reading Real Analysis by Yeh and have a question about the following result (Thm 3.25 in the book)
Theorem If $E \in \mathfrak{M}_L$ and $\mu_L(E)< \infty$, then $\forall \epsilon>0$ $\exists$ finitely many open intervals $I_1, \ldots, I_N$ s.t. $\mu_L(E \triangle \cup_{n=1}^NI_n)< \epsilon$.
Notations: $\mathfrak{M}_L$ is the $\sigma$-algebra of Lebesgue measurable subsets of $\mathbb{R}$, $\mu_L$ denotes Lebesgue measure. If we're not sure $E \in \mathfrak{M}_L$, then $\mu_L^*(E)$ denotes the Lebesgue outer measure of E. For intervals $I$, Yeh sometimes uses $\ell(I)$ to denote their length. The symmetric difference of 2 sets is $A \triangle B:=(A \setminus B) \cup (B \setminus A)$.
I didn't know how to prove this, so I read Yeh's proof. After a while I tried to prove it again but came up with something simpler that seems to work, so wanted to ask if I might have missed something. My attempt is below, followed by Yeh's proof (it's long, so I of course want something simpler in my notes if possible). Thanks in advance for any help.
My attempt: Fix $\epsilon>0$. Use the definition of outer measure as an infimum to pick a sequence $(I_n)$ of open intervals s.t.
$E \subseteq \cup_{n=1}^ \infty I_n$ and $\sum_{n=1}^\infty \mu_L(I_n) \leq \mu_L(E) + \epsilon$.
Since $\mu_L(E)< \infty$, $\sum_{n=1}^\infty \mu_L(I_n)$ converges, so pick $N \in \mathbb{N}$ s.t. $\sum_{n=N+1}^\infty \mu_L(I_n)< \epsilon$.
Then
\begin{split}
\mu_L(E \triangle \cup_{n=1}^NI_n) &= \mu_L \big( (E \setminus \cup_{n=1}^N I_n) \cup ( \cup_{n=1}^N I_n \setminus E) \big) \\
& \leq \mu_L(E \setminus \cup_{n=1}^N I_n)+\mu_L( \cup_{n=1}^N I_n \setminus E) \\
& \leq \mu_L(\cup_{n=1}^\infty I_n \setminus \cup_{n=1}^N I_n)+\mu_L( \cup_{n=1}^\infty I_n \setminus E) \\
& \leq \mu_L(\cup_{n=N+1}^\infty I_n)+\mu_L( \cup_{n=1}^\infty I_n)- \mu_L(E) \\
& \leq \sum_{n=N+1}^\infty \mu_L(I_n)+ \sum_{n=1}^\infty \mu_L(I_n)- \mu_L(E) <2 \epsilon
\end{split}
where we have repeatedly used that a measure is monotone and subadditive. In the 4th line, we have also used an earlier fact that $A,B \in \mathfrak{M}_L$, $A \subseteq B$, $\mu_L(A)< \infty$ implies $\mu_L(B \setminus A)= \mu_L(B)- \mu_L(A)$. Since $\epsilon$ is arbitrary, this is the desired result.
(I'm cutting the proof here, but Yeh pretty much bounds each of the 3 pieces on the RHS separately after this).
Best Answer
Let $m$ be Lebesgue measure on $\Bbb R.$ Let $E\in dom (m)$ with $m(E)<\infty.$ We have $m(E)=\sum_{n\in \Bbb Z}m(E\cap (n,n+1]).$ Let $\epsilon>0.$
There exists $a\in \Bbb Z^+$ such $m(E\cap (-a,a))=\sum_{n=-a}^{a-1}m(E\cap (n,n+1])>m(E)-\epsilon /3.$
If U is an open subset of $\Bbb R$ then $U=\cup F$ for some countable family $F$ of pair-wise disjoint convex open sets. If $U$ is a bounded open set then each $f\in F$ is a bounded open interval.
Let $U$ be open with $U\supset E$ and $m(U\setminus E)<\epsilon /3.$ Let $V=U\cap (-a,a+\epsilon /3).$ Let $F$ be a countable family of pair-wise disjoint bounded open intervals such that $\cup F=V.$
Now $\sum_{f\in F}m(f)=m(\cup F)=m(V)<\infty,$ so there exists a finite $G\subset F$ such that $m(\cup G)=\sum_{g\in G}m(g)>-\epsilon /3+\sum_{f\in F}m(f)=m(V)-\epsilon /3.$
You may now confirm that $m(E\Delta (\cup G))<\epsilon.$