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Consider the Lebesgue outer measure $$ \bar{m}(X) = \inf_{A \supset X}\bigg\{\sup_{P\subset A}\quad m(P)\bigg\} $$ where $X = [0,1]\cap \mathbb{Q}$ and $P = \bigcup [a_i,b_i]$ is a suitable union of intervals. My question is: suppose that $\bar{m}(X)=0$: can you exhibit one of those $A$'s? Thanks

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up vote 3 down vote accepted

Since $[0,1]$ is Borel and $\mathbb Q$ is Borel, so is their intersection.

Therefore these sets are in fact Lebesgue measurable, so the outer Lebesgue measure is equal to the Lebesgue measure.

We have, if so $\overline m(X)=0$, therefore the intersection is of shrinking intervals.

Consider $\mathbb Q=\{q_n\mid n\in\mathbb N\}$ an enumeration of the rationals in $X$ and $\epsilon>0$, let $[a_i,b_i]$ be an interval around $q_i$ such that $b_i-a_i<\frac{\epsilon}{2^i}$, and let $A=\bigcup [a_i,b_i]$.

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I am a bit confused about your answer. Your $A$ contains $[0,1]$, right? So its outer measure should be $\geq 1$ ? – user13823 Jul 28 '11 at 2:13
@Sxy: No, why would it contain $[0,1]$? Note that the sum of the interval's length is only $\epsilon$. So for a very very small $\epsilon$ we have that $A$ contains almost nothing out of $[0,1]$. However if you take $\epsilon$ to be $42$ then it might have some larger measure. – Asaf Karagila Jul 28 '11 at 2:15
Sorry, you are right. My misunderstanding was related to a bad density argument I had in mind: the set of elements $\{q_i\}$ is dense in $X$, then each interval surrounding a single $q_i$ must contain some little part of $[0,1]$. So I jumped to the conclusion that those intervals had to cover $[0,1]$ completely. – user13823 Jul 28 '11 at 7:01
$X=\mathbb{Q}\cap[0,1]$ is measurable and countable, therefore, $m(X)=0$. But, take intervals $I_k$ that cover $X$. Then $1=m([0,1])=m^*(\overline{X})\leq m^*\left(\overline{\bigcup I_k}\right)\leq m^*\left(\bigcup \overline{I_k}\right)\leq \sum m^*(\overline{I_k})=\sum m^*(I_k)$ – leo Sep 30 '11 at 5:59
Where I've put the overline for closure. Then for any covering by intervals $I_k$, we have $\sum m^*(I_k)\geq 1$ and this implies that $m(X)\geq 1$. What is the problem? What am I missing? Thank you. – leo Sep 30 '11 at 6:04

Any cover of the rationals would be a collection of open sets containing all rationals in [0,1]. A specific example would be: take any enumeration {$q_1,q_2,..,q_n,...$} of all rationals in $\mathbb Q \cap [0,1]$, and the use the open sets $O_n:=(q_n-\frac{1}{2^n},q_n+\frac {1}{2^n})$

To determine the outer measure, you may want to scale each interval by a fixed $\epsilon>0$ and then add the widths of all the intervals (see what happens when you let $\epsilon>0$ become small).

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