$A$ is Lebesgue measurable iff $\forall \epsilon > 0$, there exists open set $G$ such that $|G \setminus A| + |A \setminus G| < \epsilon$

Question

In Sheldon Axler's Measure Theory book, I came across this problem -

Suppose $A \subset \mathbb{R}$ and $|A| < \infty$. Prove that $A$ is Lebesgue measurable if and only if for every $\epsilon > 0$ there exists a set $G$ that is the union of finitely many disjoint bounded open intervals such that $|A \setminus G| + |G \setminus A| < \epsilon$.

Here, $|A|$ stands for the outer measure of $A$.

The definition of Lebesgue measurable sets are as follow -

$A$ is Lebesgue measurable iff -

1. For each $\epsilon > 0$, there exists a closed set $F \subset A$ with $|A \setminus F| < \epsilon$.

2. There exists a Borel set $B \subset A$ such that $|A \setminus B| = 0$.

3. For each $\epsilon > 0$, there exists an open set $G \supset A$ such that $|G \setminus A| < \epsilon$

4. There exists a Borel set $B \supset A$ such that $|B \setminus A| = 0$

I have proved the $\implies$ part, but I am stuck in the converse part. Any help is appreciated.

Answer 1

I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
This exercise is Exercise 6 on p.60 in Exercises 2D in this book.

Exercise 6
Suppose $A\subset\mathbb{R}$ and $|A|<\infty$. Prove that $A$ is Lebesgue measurable if and only if for every $\epsilon>0$ there exists a set $G$ that is the union of finitely many disjoint bounded open intervals such that $|A\setminus G|+|G\setminus A|<\epsilon$.

Exercise 6-1
Suppose $A\subset\mathbb{R}$ and $|A|<\infty$. Prove that if for every $\epsilon>0$ there exists a set $G$ that is the union of finitely many disjoint bounded open intervals such that $|A\setminus G|+|G\setminus A|<\epsilon$, then $A$ is Lebesgue measurable.

Proof of Exercise 6-1
Let $\epsilon$ be an arbitrary positive real number.
Then, there exists a set $G$ that is the union of finitely many disjoint bounded open intervals such that $|A\setminus G|+|G\setminus A|<\frac{\epsilon}{2}$.
$|A\setminus G|\leq |A\setminus G|+|G\setminus A|<\frac{\epsilon}{2}$.
$|G\setminus A|\leq |A\setminus G|+|G\setminus A|<\frac{\epsilon}{2}$.
By the definition of the outer measure (please see p.14 in the book), there exist open intervals $I_1,I_2,\dots$ such that $A\setminus G\subset I_1\cup I_2\cup\dots$ and $l(I_1)+l(I_2)+\dots<\frac{\epsilon}{2}.$
Of course, $I_1,I_2,\dots$ are bounded open intervals.
Let $G':=I_1\cup I_2\cup\dots$.
By 2.8 on p.17 in the book, $|G'|\leq\sum_{k=1}^\infty |I_k|=\sum_{k=1}^\infty l(I_k)<\frac{\epsilon}{2}.$
$|(G\cup G')\setminus A|=|(G\setminus A)\cup (G'\setminus A)|\leq |G\setminus A|+|G'\setminus A|\leq |G\setminus A|+|G'|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$
by 2.8 on p.17 and 2.5 on p.16.
$G\cup G'$ is an open set.
By 2.71(e) on p.52 in the book, $A$ is Lebesgue measurable.

Exercise 6-2
Suppose $A\subset\mathbb{R}$ and $|A|<\infty$. Prove that if $A$ is Lebesgue measurable, then for every $\epsilon>0$ there exists a set $G$ that is the union of finitely many disjoint bounded open intervals such that $|A\setminus G|+|G\setminus A|<\epsilon$.

Proof of Exercise 6-2
Let $\epsilon$ be an arbitrary positive real number.
By 2.71(e) on p.52 in the book, there exists an open set $H\supset A$ such that $|H\setminus A|<\frac{\epsilon}{2}.$
Since $|A|<\infty$, there exists a positive real number $M$ such that $|A\setminus [-M, M]|<\frac{\epsilon}{4}.$ (Please see copper.hat's proof or Dasherman's proof.)
$A=(A\cap [-M,M])\cup (A\setminus [-M,M])$ is a union of disjoint Lebesgue measurable sets.
$|A|=|A\cap [-M,M]|+|A\setminus [-M,M]|<|A\cap [-M,M]|+\frac{\epsilon}{4}.$
So, $|A|-|A\cap [-M,M]|<\frac{\epsilon}{4}.$
By 2.71(b) on p.52 in the book, there exists a closed set $F\subset A\cap [-M,M]$ with $|(A\cap [-M,M])\setminus F|<\frac{\epsilon}{4}.$
Therefore, $|A|-|F|=(|A|-|A\cap [-M,M]|)+(|A\cap [-M,M]|-|F|)<\frac{\epsilon}{4}+\frac{\epsilon}{4}=\frac{\epsilon}{2}.$
By 0.59 on p.30 in "Supplement for Measure, Integration & Real Analysis" by Sheldon Axler, we can write $H=(a_1,b_1)\cup (a_2,b_2)\cup\dots.$
Since $|H|<\infty$, $(a_1,b_1),(a_2,b_2),\dots$ are bounded open intervals.
$\{(a_1,b_1),(a_2,b_2),\dots\}$ is an open cover of $F$.
$F$ is a closed bounded subset of $\mathbb{R}.$
By 2.12 on p.19 in the book, there exists a finite subcover $\{(a_{i_1},b_{i_1}),\dots,(a_{i_n},b_{i_n})\}$ of $F$.
Let $G:=(a_{i_1},b_{i_1})\cup\dots\cup (a_{i_n},b_{i_n}).$
Then, $F\subset G\cap A\subset A\subset G\cup A\subset H.$
So, $|G\cup A|-|G\cap A|\leq |H|-|F|=(|H|-|A|)+(|A|-|F|)<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$
$G\cup A=(G\cap A)\cup (A\setminus G)\cup (G\setminus A)$ is a union of disjoint Lebesgue measurable sets.
$|G\cup A|=|G\cap A|+|A\setminus G|+|G\setminus A|.$
So, $|A\setminus G|+|G\setminus A|=|G\cup A|-|G\cap A|<\epsilon.$

Answer 2

For the sake of future readers, I am going to post another solution to the second direction of the problem. @tchappy ha's solution to the other direction looks very straightforward and well-written.

Problem: Suppose $A\subset\mathbb{R}$ and $|A|<\infty$. Prove that if $A$ is Lebesgue measurable, then for every $\epsilon>0$ there exists a set $G$ that is the union of finitely many disjoint bounded open intervals such that $|A\setminus G|+|G\setminus A|<\epsilon$.

Proof: Since $A$ is Lebesgue measurable, there is an open set $G$ containing $A$ such that $|G\setminus A|=m(G\setminus A)< \frac{\epsilon}{2}.$ Note that since $A$ is Lebesgue measurable (as well as $G$ and all intervals), instead of outer measure, we can work with the Lebesgue measure $m$. By this theorem, we may assume $G= \bigcup\limits ^{\infty}_{i=1}I_i$, where $I_i$'s are disjoint open intervals. Now assume for every $n$, we have: $$m(A \setminus(I_1 \cup ... \cup I_n ))> \frac {\epsilon}{2}.$$ Considering $E_n=A \setminus(I_1 \cup ... \cup I_n )$, by this famous theorem, we have:

$$0=m(\varnothing)=m (\bigcap_{n=1}^{\infty} E_n) = \lim_{n \to \infty} \mu(E_n) \geq \frac{\epsilon}{2},$$ which is a contradiction. Hence, there is $n_0$ such that:

$$m(A \setminus(I_1 \cup ... \cup I_{n_0} )) \leq \frac {\epsilon}{2}.$$

Now, observe that: $$m(A \setminus(I_1 \cup ... \cup I_{n_0} ))+m((I_1 \cup ... \cup I_{n_0} ) \setminus A)\leq m(A \setminus(I_1 \cup ... \cup I_{n_0} ))+m(G \setminus A) \\ < \frac {\epsilon}{2}+\frac {\epsilon}{2}= \epsilon. $$

We are done.