I believe I have a fundamental misunderstanding of the concept of the Vitali Covering Lemma.
Definition – A closed bounded interval $[c, d]$ is said to be nondegenerate provided $c < d$.
Definition – A collection $\mathcal{F}$ of closed, bounded, nondegenerate intervals is said to cover a set $E$ in the sense of Vitali provided for each point $x$ in $E$ and $\epsilon > 0$, there is an interval $I$ in $F$ that contains $x$ and has $l(I) < \epsilon$.
The Vitali Covering Lemma – Let $E$ be a set of finite outer measure and $\mathcal{F}$ a collection of closed, bounded intervals that cover $E$ in the sense of Vitali. Then for each $\epsilon > 0$, there is a finite disjoint subcollection $\{I_k\}_{k = 1}^n$ of $\mathcal{F}$ for which $$m^*\left[E \setminus \bigcup_{k = 1}^n I_k\right] < \epsilon.$$
Ok, so it is a true fact that the Vitali Covering Lemma does not apply when the covering collection contains degenerate closed intervals. I don't see why that is the case.
Consider a set $E$ that is covered in the sense of Vitali by a collection $\mathcal{F}$. Let's add some degenerate closed intervals into our collection $\mathcal{F}$. Suppose we add $E_1 = [a, a] = \{a\}$ and $E_2 = [a, -a] = \emptyset$ for $a > 0$ to $\mathcal{F}$.
If the degenerate sets somehow cause a problem, why can't the Vitali Covering Lemma just not "choose" the degenerate sets in the finite disjoint subcollection $\{I_k\}_{k = 1}^n$ when it asserts the existence of such a subcollection?
Best Answer
If the intervals were allowed to be degenerate, we could simply consider the collection of degenerate intervals $[x,x]$ for each $x\in E$. But then, any countable subcollection $\{I_k\}_{k=1}^n$ has measure zero, and thus, does not remove any substantial part of $E$.
The reason the Vitali Covering Lemma can't just not choose the degenerate sets, is because what's left may not actually be enough.