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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

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  • $\begingroup$ Or does $|X|$ mean the Lebesgue measure? I assumed it meant cardinality. $\endgroup$ Commented Sep 12, 2021 at 20:09
  • $\begingroup$ @ThomasAndrews thank you for your interest in my question; $|\cdot |$ refers to outer measure $\endgroup$
    – lorenzo
    Commented Sep 12, 2021 at 20:09
  • $\begingroup$ Edit your question to make that clear. It is non-standard. $\endgroup$ Commented Sep 12, 2021 at 20:10
  • $\begingroup$ @ThomasAndrews done $\endgroup$
    – lorenzo
    Commented Sep 12, 2021 at 20:11
  • $\begingroup$ @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 $|\mathbb{R}\setminus C|\leq |A|$. $\endgroup$
    – Mittens
    Commented Sep 19, 2021 at 14:34

1 Answer 1

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We first prove that

(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. $\square$

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