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A collection of sets $O_{\lambda}$ open covers a set $A$ in $\bf R$ that is bounded and measurable. Assume that $A$ can not be open covered by a finite subcollection of $O_{\lambda}$, let $\epsilon>0$. Is there a finite subcollection of $O_{\lambda}$, denote $O_n$ (don’t have to be disjoint), such that $m^*(A\setminus\bigcup O_n)<\epsilon$?

Also, is it equivalent to say "the subset of $A$ that can not be finitely open covered by some subcollection of $O_{\lambda}$ has measure 0" ? edit: (this paragraph is not mathematically constructed right, attribute to Wong pointing it out, but would happy to take other good alternate rephrases)

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  • $\begingroup$ I think it might be helpful to mention vitali's lemma as a good comparison. Vitali's lemma, as far as i understand, from royden's real analysis, states the conditions roughtly the same, except that, the sets Oλ covers A in a sense of vitali, and on top of that, the finite subcollection of Oλ are disjoint. Which are two very very interesting differences $\endgroup$
    – pasta
    Commented Jul 9, 2022 at 0:09
  • $\begingroup$ I've added some formatting, please check to see that I haven't messed anything up. $\endgroup$ Commented Jul 9, 2022 at 0:37
  • $\begingroup$ yep it's perfectly correct, thx! $\endgroup$
    – pasta
    Commented Jul 9, 2022 at 0:43
  • $\begingroup$ Isn't it obvious? Instead of general $A$ it's enough to consider simply $\ A:=\bigcup_\lambda\,O_\lambda.$ $\endgroup$
    – Wlod AA
    Commented Jul 9, 2022 at 1:20
  • $\begingroup$ i am not sure if i understand what you mean by considering a specific example of A. Thanks for an editing of better formatting btw. $\endgroup$
    – pasta
    Commented Jul 9, 2022 at 1:23

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