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Let $E\subseteq \mathbb{R}$ be given, and let $m^*$ denote the outer measure. For each $\epsilon$, there exists a closed set $F_\epsilon\subseteq E$ such that $m^*(E\setminus F_\epsilon)<\epsilon$. Does this imply that there exists a closed set $F\subseteq E$ such that $m^*(E\setminus F)=0$?

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  • $\begingroup$ What is $E/F$? Do you mean $E\setminus F$? $\endgroup$ – Amit Kumar Gupta Mar 21 '13 at 7:13
  • $\begingroup$ I think $m^*$ is a more usual notation for outer measure. $\endgroup$ – copper.hat Mar 21 '13 at 7:14
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There exists a $F_\sigma$ set such that this is true, take $\cup_{n=1}^\infty F_{\frac{1}{n}}$.

However, the statement is not true in general. Take $E = (0,1)$. Then taking $F_\epsilon = [\frac{\epsilon}{3},1-\frac{\epsilon}{3}]$ results in $m^* (E \setminus F_\epsilon) < \epsilon$.

Suppose $F \subset E$ is closed. Let $I=[\inf F, \sup F]$. Clearly, $F \subset I$, and $0< \inf F, \sup F < 1$, hence $m^*(E\setminus F) > \frac{\inf F}{2}> 0$.

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No. If $E$ is closed then the statement clearly holds, so the only possible counterexample must be non-closed. What's the simplest possible example of that? A bounded open interval. No closed subset $F$ of a bounded open interval $E$ can be such that $m^*(E\setminus F)=0$. This is because $E\setminus F$ is open, thus is either empty or contains an open interval. It can't be empty since that would imply $E=F$ which would make $E$ clopen, which is impossible (the only clopen subsets of $\mathbb{R}$ are the empty set and the whole space). So it contains an open interval, and hence has positive outer measure.

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It might be useful to see how these statements can be rewritten to better show the difference in their logical form.

The 1st statement becomes: $\;\; \left(\forall \, \epsilon > 0 \right) \left( \exists F \subseteq {\mathbb R} \right ):$ $\;\;\;F$ is closed and $m^{*}\left(E-F\right) < \epsilon $

The 2nd statement becomes: $\;\; \left( \exists F \subseteq {\mathbb R} \right )\left(\forall \, \epsilon > 0 \right): $ $\;\;\;F$ is closed and $m^{*}\left(E-F\right) < \epsilon $

Of course, the fact that the 2nd statement is a logically stronger $\exists \; \forall$ uniform statement doesn't mean that, in this specific context, we get a mathematically stronger statement. However, we do in fact get a mathematically stronger statement, as the other answers show.

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