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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?

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    $\begingroup$ You don't really need Vitali's covering lemma for this: just note that an arbitrary union of (nontrivial) closed intervals has countably many connected components (because they have nonempty disjoint interiors), and every connected subset of reals is an interval, and thus measurable. $\endgroup$ – tomasz Mar 20 '19 at 18:37
  • $\begingroup$ @tomasz If my collection of closed intervals is uncountable, why are there only countably many connected components in the union? Why must the interiors be disjoint? $\endgroup$ – Junglemath Mar 20 '19 at 19:12
  • $\begingroup$ Because connected components are disjoint, by definition, so their interiors certainly are, too! $\endgroup$ – tomasz Mar 20 '19 at 19:34
  • $\begingroup$ I still don't see why it's a countable collection. Plus, what if two intervals overlap? How are their interiors disjoint? $\endgroup$ – Junglemath Mar 20 '19 at 20:59
  • $\begingroup$ If they overlap, they are in the same connected components. A disjoint family of open subsets of reals cannot be uncountable. $\endgroup$ – tomasz Mar 21 '19 at 11:17
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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.

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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.

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