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

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.

• Are you allowed to use the fact that the Lebesgue measure is a complete measure? Commented Apr 3, 2021 at 20:56
• @dem0nakos No, we have not been taught that yet Commented Apr 4, 2021 at 5:40

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, $$|A\setminus G|+|G\setminus A|=|G\cup A|-|G\cap A|<\epsilon.$$