5

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

2 Answers2

3

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).

Quinn Culver
  • 4,471
gary
  • 4,027
2

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 $A=\bigcup [a_i,b_i]$.

Asaf Karagila
  • 393,674
  • I am a bit confused about your answer. Your $A$ contains $[0,1]$, right? So its outer measure should be $\geq 1$ ? –  Jul 28 '11 at 02:13
  • @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 02:15
  • 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. –  Jul 28 '11 at 07:01
  • $X=\mathbb{Q}\cap[0,1]$ 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 05:59
  • 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 06:04
  • @leo: First of all a measurable set has the same outer measure; $\mathbb Q\cap [0,1]$ is measurable so it has the same outer measure. Secondly to be of measure zero is to be covered by arbitrarily small measure. You can choose your intervals so their sum as small as you want. As for the closure, there is nowhere a request for the closure of a set to have the same measure as the original set. – Asaf Karagila Sep 30 '11 at 06:32
  • yes I know that if a set is measurable his measure is his outer measure. And I know that $m(\mathbb{Q}\cap [0,1])=0$. What I'm trying to show is that for any covering by intervals of $\mathbb{Q}\cap[0,1]$, we have $\sum m^*(I_k)\geq 1$. My question is: why this does not contradicts the fact of $m(\mathbb{Q}\cap [0,1])$ – leo Sep 30 '11 at 19:41
  • :please forgive my ignorance, this is true only for a finite covering. – leo Oct 01 '11 at 00:17
  • @leo: You are trying to cover $[0,1]$, not $\mathbb Q\cap[0,1]$. Given $\epsilon>0$ take an enumeration of these rationals and cover the $n$-th rational of the enumeration by an interval of length $\frac{\epsilon}{2^{n+1}}$, then the whole set of rationals is covered by the union of these intervals, whose measure is at most $\sum\frac{\epsilon}{2^{n+1}}\le\epsilon$. – Asaf Karagila Oct 01 '11 at 06:16
  • yes that's the thing. Thanks a lot. – leo Oct 01 '11 at 19:58
  • Let $A_n$ be the covering of rationals ${ q_n }_{n \in \mathbb{N}}$ in $[0,1]$ by intervals $A_n = (q_n - \varepsilon/2^n , q_n + \varepsilon/2^n)$. Since every number in $[0,1]$ is in an open neighborhood of a rational, is $\bigcup A_n = [0,1]$? To contradict that we need to build an irrational number $r$ such that $\forall n : |r - q_n| > \varepsilon/2^n$. Can you construct such a number (or prove it exists)? – LinAlgMan Aug 24 '13 at 19:33
  • @LinAlgMan: The measure argument works just fine. The interval has measure $1$ but the union of the intervals has a smaller measure. – Asaf Karagila Aug 24 '13 at 21:06