(Use the Vitali covering lemma) to show that the union of closed intervals of real numbers is measurable

Question

I have seen similar questions before but the replies given were less than satisfactory.

Let $\mathcal{F} = \{F_\alpha\}_{\alpha \in J}$ be an arbitrary family of closed intervals of real numbers. Let $E = \displaystyle \bigcup_{\alpha \in J} F_\alpha$, and let $E_n = E \cap [-n, n]$. Let $V$ be a Vitali covering of $E$; then $V$ is a Vitali covering of $E_n$. By the Vitali Lemma, we can find a finite, pairwise disjoint collection $\{V_i\}_{i = 1}^m$ of $V$ such that $m^\ast(E_n \smallsetminus \displaystyle \bigcup_{i = 1}^m V_i) < \varepsilon$. Since each $V_i$ is an interval of real numbers, and intervals are Lebesgue measurable, it follows that $\displaystyle \bigcup_{i = 1}^m V_i$ is measurable. Thus $$\bigcup_{n = 1}^ \infty E_n$$ is a countable collection of measurable sets and is therefore measurable. Moreover, the Lebesgue measure of this union is equal to the Lebasgue measure of $E$, which is what we sought to prove.

Does this make any sense?

Answer 1

I cannot write a comment because I have no reputation... Yes, that is correct. $E_n$ is measurable iff $\exists \mathcal{O}$ so that $m*(E_n/\mathcal{O})<\epsilon$ for any $\epsilon>0$. So your proof is complete.

Answer 2

No, your proof is not correct. In your proof, you need to show that the union of Vi is in En, and your conclusion can hold. Thus, you cannot find an arbitrary Vitali cover of E. Did you realize that you did not even use the fact that E is the union of a family of closed interval? Now, you can select all closed intervals included in E. Then the collection of them is a Vitali cover of E (you can show it). And the union the the select sequence is in E.

Answer 3

As mentioned by @salixliu,

Now, you can select all closed intervals included in E. Then the collection of them is a Vitali cover of E (you can show it). And the union the the select sequence is in E.

Then by the Vitali Covering Lemma, for each $\epsilon>0$, we can find a finite disjoint subcollection $\{I_k\}^n_{k=1}$ of the above Vitali cover for which \begin{equation} m^*[E\sim\bigcup^n_{k=1}I_k]<\epsilon. \end{equation}

$\bigcup^n_{k=1}I_k$ is closed since it is finite union of closed sets. Also $\bigcup^n_{k=1}I_k\subset E$ by construction.

Finally we apply Theorem 11 in Sec. 2.4 of Real Analysis by Roden & Fitzpatrick: $E$ is measurable iff for each $\epsilon>0$, there is a closed set $F$ contained in $E$ for which $m^*(E\sim F)<\epsilon$.

As you can see, you don't need to decompose $E$ into countable union of $E_n$.

Using Theorem 11 is inspired by @Al-FahadAl-Qadhi.