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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.

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(I'm cutting the proof here, but Yeh pretty much bounds each of the 3 pieces on the RHS separately after this).

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  • $\begingroup$ Do you know that a Lebesgue measurable subset of finite measure is approximable, in measure, by compact subsets? $\endgroup$ – zhw. Mar 14 '17 at 22:51
  • $\begingroup$ Sorry, I have not heard of it $\endgroup$ – Jason Mar 14 '17 at 23:34
  • $\begingroup$ It seems to be the modern teaching style to avoid mentioning inner measure. If $m$ is a measure for which closed sets are measurable, the inner measure of $S$ is $m^i(S)=\sup \{m(C): C=\overline C\subset S\}.$ $\endgroup$ – DanielWainfleet Apr 13 '18 at 18:26
  • $\begingroup$ When $m$ is Lebesgue measure and $m^o$ is outer Lebesgue measure and $m^i$ is inner Lebesgue measure we have $S\in dom (m)$ iff $m^i(A\cap S)=m^o(A\cap S)$ for every bounded interval $A.$ It follows that $S\in dom (m)$ iff there exists an $F_{\sigma}$-set $C$ and a $G_{\delta}$-set $D$ such that $C\subset S\subset D$ and such that $m(C)=m^i(S)=m^o(S)=m(D).$ $\endgroup$ – DanielWainfleet Apr 13 '18 at 19:41
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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 $m(E\Delta (\cup G))<\epsilon.$

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  • $\begingroup$ $E\Delta (\cup G)\subset P\cup Q\cup R$ where $Q=E\setminus (-a,a)$ and $Q=(\cup (F)\setminus (\cup G)$ and $R=E\setminus (\cup F)$. $\endgroup$ – DanielWainfleet Apr 13 '18 at 19:22
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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.

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