# $A\subset\mathbb{R}$ such that $|(-n,n)\cap A|+|(-n,n)\setminus A|=|(-n,n)|$ for all $n\in\mathbb{Z}^+$ implies that $A$ is Lebesgue measurable

I have proved the following statement and I would like to know if it is correct, thanks - I have doubts because it seems I have proved it without never having used the hypothesis:

"$$A\subset\mathbb{R}$$ such that $$|(-n,n)\cap A|+|(-n,n)\setminus A|=|(-n,n)|$$ for all $$n\in\mathbb{Z}^+$$ implies that $$A$$ is Lebesgue measurable"

By the measurable cover lemma we can find $$B\supset A$$ Borel such that $$|B|=|A|$$ and $$C\supset\mathbb{R}\setminus A$$ Borel such that $$|C|=|\mathbb{R}\setminus A|$$ so $$\mathbb{R}\setminus C\subset A$$. So, $$|B\setminus A|\leq |B\setminus (\mathbb{R}\setminus C)|=|B|-|\mathbb{R}\setminus C|=|B|-|\mathbb{R}\setminus (\mathbb{R}\setminus A)|=|B|-|A|=0$$ so we have found a Borel set $$B$$ such that $$B\supset A$$ and $$|B\setminus A|=0$$ thus $$A$$ is Lebesgue measurable, as desired. $$\square$$

Lemma (measurable cover): if $$A\subset\mathbb{R}$$ then there exist a Borel set $$B$$ such that $$A\subset B$$ and $$|A|=|B|$$

Proof: If $$|A|=\infty$$ by taking $$O=(-\infty,\infty)$$ we have that $$A\subset O$$, $$|O|=\infty=|A|$$ and $$O\in\mathcal{B}$$ so we are done.

If $$|A|<\infty$$ then by definition of outer measure if we take $$\varepsilon>0$$ there exist open intervals $$I_1,I_2,\dots\subset\mathbb{R}$$ such that $$A\subset\bigcup_{k=1}^{\infty}I_k$$ and $$\sum_{k=1}^{\infty}\ell(I_k)\leq |A|+\varepsilon$$ because $$|A|+\varepsilon$$ cannot be a lower bound for $$\{\sum_{k=1}^{\infty}\ell(I_k):I_1,I_2,\dots\text{ are open intervals such that }A\subset\bigcup_{k=1}^{\infty}I_k\}$$. This implies that for every $$n\geq 1$$ there exists $$O_n=\bigcup_{k=1}^{\infty}I_{k_n}$$ such that $$|O_n|\leq\sum_{k=1}^{\infty}\ell(I_{k_n})<|A|+\frac{1}{n}$$ so if we set $$O:=\bigcap_{n=1}^{\infty}O_n$$ we have that $$A\subset O\subset O_n$$ which implies that $$|A|\leq |O|\leq |O_n|\leq |A|+\frac{1}{n}$$ for every $$n\geq 1$$, thus $$|A|=|O|$$. The claim now follows by noticing that by (2.25 c) each $$O_n$$ and thus also $$O$$, is a Borel set, as desired.

NOTE: $$|\cdot|$$ refers to outer measure

• Or does $|X|$ mean the Lebesgue measure? I assumed it meant cardinality. Commented Sep 12, 2021 at 20:09
• @ThomasAndrews thank you for your interest in my question; $|\cdot |$ refers to outer measure Commented Sep 12, 2021 at 20:09
• Edit your question to make that clear. It is non-standard. Commented Sep 12, 2021 at 20:10
• @ThomasAndrews done Commented Sep 12, 2021 at 20:11
• @Lorenzo: $\mathbb{R}\setminus A\subset C$ and $|\mathbb{R}\setminus A |=|C|$ does not imply $|C|=|\mathbb{R}\setminus A|$ if $A$ is not measurable in the sense of Caratheodory. All you can say with certainty is $|\mathbb{R}\setminus C|\leq |A|$. Commented Sep 19, 2021 at 14:34

(1) if $$A\subset\mathbb{R}$$ is such that $$|(-n, n)\cap A|+|(-n, n)\setminus A|= 2n$$ for every $$n\in\mathbb{Z}^+$$ then for every $$n\in\mathbb{Z}^+$$ there exist Borel sets $$F_n$$ and $$G_n$$ such that $$F_n\subset (-n,n)\cap A\subset G_n$$ and $$|G_n\setminus F_n|=0$$.
Proof (1). Fix $$n\geq 1$$. By measurable cover lemma we can find a Borel set $$G_n\supset (-n,n)\cap A$$ such that $$|G|=|(-n,n)\cap A|$$ and also a Borel set $$H_n\supset (-n,n)\setminus A$$ such that $$|H_n|=|(-n,n)\setminus A|$$. As $$H_n\supset (-n,n)\setminus A$$ it is $$(-n,n)\setminus H_n\subset A$$ so if we set $$F_n:=(-n,n)\setminus H_n$$ we have that $$F_n\subset A$$ and $$F_n$$ is Borel so $$|(-n,n)\cap F_n|+|(-n,n)\setminus F_n|=|(-n,n)|$$ which implies $$\fbox{|(-n,n)\cap F_n|}=|(-n,n)|-|(-n,n)\setminus F_n|= |(-n,n)|-|H_n|=$$ $$|(-n,n)|-|(-n,n)\setminus A|\overset{hypothesis}{=}\fbox{|(-n,n)\cap A|}$$ hence $$|G_n\setminus F_n|=|G_n|-|F_n|=|(-n,n)\cap A|-|(-n,n)\cap A|=0$$, as desired. $$\square$$
Proof of the original statement: By (1) we know that for every $$n\in\mathbb{Z}^+$$ there exist Borel sets $$F_n$$ and $$G_n$$ such that $$F_n\subset (-n,n)\cap A\subset G_n$$ and $$|G_n\setminus F_n|=0$$ so if we set $$F:=\bigcup_{n=1}^{\infty}F_n$$ and $$G:=\bigcup_{n=1}^{\infty}G_n$$ we have that they are Borel sets and $$|A\setminus F|=|(\bigcup_{n=1}^{\infty}((-n,n)\cap A))\setminus F|\leq |(\bigcup_{n=1}^{\infty}G_n)\setminus F|=|G\setminus F|=|(\bigcup_{n=1}^{\infty}G_n)\setminus (\bigcup_{n=1}^{\infty}F_n)|\leq |\bigcup_{n=1}^{\infty}(G_n\setminus F_n)|=|\bigcup_{n=1}^{\infty} \emptyset|=|\emptyset|=0$$ so we have found a Borel set $$F\subset A$$ such that $$|A\setminus F|=0$$ so $$A$$ is Lebesgue measurable by definition of Lebesgue measurable set, as desired. $$\square$$