Approximating a Lebesgue measurable set by a finite union of intervals 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).
 A: 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  $m(E\Delta (\cup G))<\epsilon.$ 
A: Your proof is correct.
Yeh's proof is longer (more complicated?) because he wants to use (ii) of Theorem 3.22 ($E$ measurable iff $\forall\ \epsilon > 0\ \exists$ an open set $O \supseteq E$ such that $\mu_L(O \setminus E) < \epsilon$). Directly using the definition of measurability on $E$ as you have done also works.
