# What's the difference between “$\exists F_\epsilon :m^*(E\setminus F_\epsilon)<\epsilon$ for each $\epsilon$,” and “$\exists F:m^*(E\setminus F)=0$”?

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 $m^*(E\setminus F)=0$?

• What is $E/F$? Do you mean $E\setminus F$? – Amit Kumar Gupta Mar 21 '13 at 7:13
• I think $m^*$ is a more usual notation for outer measure. – copper.hat Mar 21 '13 at 7:14

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 $m^*(E\setminus F) > \frac{\inf F}{2}> 0$.

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.

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.