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In Vitali's Lemma it uses outer measure rather than measure. What are some results that depend on it this theorem applying to sets with only outer measure rather than measurable sets?

Vitali's Lemma: Let $E$ be a set of finite outer measure and $G$ a collection of intervals that cover $E$ in the sense of Vitali. Then given $\varepsilon> 0$ there is a finite disjoint collection of intervals in $G$ such that $m^*(E - \bigcup_{n=1}^N I_n) < \varepsilon$.

I'm trying to learn this theorem and I keep replacing "outer measure" with "measure" and I want a reason to stop doing that.

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up vote 9 down vote accepted

The short answer (to the question in the title) is so that you don't need to worry about whether E is measurable. If you happen to know that E is measurable then you can drop "outer" everywhere.

There may be a longer answer (better addressing the question as stated in the page) involving specific applications where E is in fact nonmeasurable, but I personally don't know of such applications offhand.

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If you keep working with the Borel sigma algebra, you might encounter problematic examples. For instance, a Hamel basis of R over Q. You can so arrange it that it is contained within a measure zero perfect set, for instance the Cantor set. This has measure zero in Lebesgue measure, but is nonmeasurable in the Borel setup. – Anweshi Jan 12 '10 at 18:14

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