As mentioned in the title, the question is how to show (using Vitali Covering Lemma) that the union of any collection of closed, bounded non-generate intervals is measurable.
The Vitali Covering Lemma as stated in my notes is: Let $E$ be a subset of $\mathbb{R}$ with $m^*(E)<\infty$ and let $\mathcal{V}$ be a Vitali covering of $E$. Then, for each $\epsilon>0$, there is a finite disjoint sequence $(J_k)_{k=1}^n$ of intervals in $\mathcal{V}$ such that $m^*(E\setminus \cup_{k=1}^n J_k)<\epsilon$.
I am stuck with what to choose as "$E$". I was about to choose $E$ as the union of the collection of intervals, but realised that $m^*(E)$ may be infinite, and hence the conditions for using the lemma are not met. I also thought about choosing $E$ as a single interval, but Vitali doesn't seem to help here since an uncountable union of finite intervals is still uncountable. (Initially, I was worried that Vitali covering may not exist, but I managed to figure that a Vitali covering always exists, just take the set of all closed, bounded, nondegenerate intervals).
Thanks for any help.
Hints that may or not help:
[Using Vitali] If $\cal I$ is the collection of intervals (open, closed, half-open -- just not single points) then construct your own Vitali cover of $E= \bigcup\{I:I\in {\cal I}\}$. For each point $x\in E$ there must exist an arbitrarily small closed interval $J$ (not necessarily from $\cal I$) that contains $x$ and that is contained in the set $E$. Etc.
[Not using Vitali] Show that any union of such intervals is equal to a countable union of intervals and so a Borel set.
The reason for the exercise asking for a less general conclusion using a sledge hammer to obtain it, is that in higher dimensions everything changes.
[P.S. Find a different restatement of the Vitali theorem (without $\epsilon$) that is more suited to the purpose here.]
[Added since the poster asked for further help:
Let $\cal V$ be the collection of all closed intervals that are contained in $E$. Verify that $\cal V$ is a Vitali cover of $E$. Claim by the Vitali covering theorem that there is a countable collection ${\cal V}_0\subset {\cal V}$ whose union is all of $E$ except for a set of measure zero. Conclude that $E$ is measurable.]