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.

  • 1
    $\begingroup$ Just a guess I think it means excluding such intervals like $[a,a]$, which is a point. $\endgroup$ – twnly Nov 11 '18 at 22:49
  • $\begingroup$ Ok, could you please give a hint about showing that. $\endgroup$ – Ahmed Nov 12 '18 at 0:13

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 $\mathscr F$.

  • $\begingroup$ Oh I see thank you, could you provide me an example $\endgroup$ – Ahmed Nov 12 '18 at 5:02
  • $\begingroup$ I think I was not clear enough in my answer, so I elaborated on it! $\endgroup$ – Matematleta Nov 13 '18 at 2:13

HERE IS WHAT I SEE FROM the explanation above:

Let $\epsilon>0$ be given. Let $E \subseteq \mathbb{R}$ such that $m^*(E)<\infty$. Let $\mathcal{F}$ contains nondegenerate general intervals. Then, $\mathcal{F}$ contains all kind of intervals such that the length of those intervals is not zero since they are non degenerate intervals. Let $x \in E$, then we can find $I$ open interval in $\mathcal{F}$ such that $x \in I$, so since $I$ is open interval then we can find a neighborhood $J$ of $x$ such that $x \in J$ and $l(J)<\epsilon.$ Therefore, $\mathcal{F}$ is indeed a cover of $E$. Now, we need to show that there exists a finite collection of disjoint bounded intervals $\{I_i\}_{i=1}^n$ of $\mathcal{F}$ such that $$m(E \sim \bigcup_{i=1}^nI_i)<\epsilon.$$ Define $\mathcal{F}_0=\{J \in \mathcal{F}:J \text{ is closed and bounded}\}.$ To see that $\mathcal{F}_0$ is a cover of $E$, let $x \in E$. Then, since $\mathcal{F}$ is a Vitali cover there exists $I$ such that $x \in I$ and $l(I)<\epsilon$. Since $l(I)<\epsilon$, and $\mathcal{F}_0$ contains all closed and bounded intervals, we can find $I' \in \mathcal{F}_0$ such that $I \subset I'$, so $I \in \mathcal{F}_0.$ Thus, $\mathcal{F}_0$ is indeed a Vitali cover of $E$. Now, by applying Vitali's lemma, we see that $$m(E \sim \bigcup_{i=1}^nI'_i)<\epsilon.$$ Clearly, $\{I'\}_{i=1}^n \subset \mathcal{F}_0 \subset \mathcal{F}$, so the Vitali Covering Lemma does extend to the case in which the covering collection consists of nondegenerate general intervals.

  • $\begingroup$ why are you assuming that $F$ contains all nondegenarate intervals? may be there is some outside it. $\endgroup$ – Emptymind Nov 3 '19 at 9:11

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