Show that the Vitali Covering Lemma does extend to the case in which the covering collection
consists of nondegenerate general intervals.
I do not understand what does "nondegenerate general intervals" mean exactly.
Thank you.
Question about Vitali Covering
lebesgue-measuremeasure-theory
Best Answer
Here's the problem if your intervals are not closed: you have a set $E$ covered in the sense of Vitali, and you are drawing elements of the cover one at a time, making sure a certain requirement (that is not necessary to specify for this discussion) is satisfied. Now, at the $k\ th$ step, you have a sequence $I_1,\cdots, I_k$ of $closed$ pairwise disjoint sets. If these cover $E$ you are done. If not, to proceed to the $k+1 \ th$ step, you need, in order to satisfy the abovementioned requirement, to be able to find an $x\in E\setminus\bigcup^k_{j=1} I_j$ such that $there\ is\ a\ neighborhood\ of\ x$ disjoint from $I=\bigcup^k_{j=1} I_j$. This is always possible since $I$ is closed. In other words, if $I$ were not closed, you'd be stuck right here, trying to do the next step in the construction. It is not hard to see that if you include degenerate intervals, the Vitali theorem fails, but that otherwise, if you use general (non-degenerate) intervals for $\mathscr F$, you can always extract a Vitali cover: if $I\in \mathscr F$ is a general interval, there is an open interval $J$ such that $\overline J\subseteq I$. Then, $\mathscr F'=\{\overline J:J\subseteq I; I\in \mathscr F\}$ is a bona fide Vitali cover of $E$. To see that degenerate intervals will not work, consider the cover $\mathscr F=\{[x,x]\}_x$ of $E$ and see what happens if you remove $any$ countable collection from $\mathscr F$.