TITLE: Prove sum of the lengths of intervals in a finite covering of $\mathbb{Q}\cap [0,1]$ is $\geq 1$
QUESTION [5 upvotes]: I have the following proof, but I don't think it's right. Could somebody please tell me what's wrong with it, and how to fix it?  Thanks :)
Let $B$ be the set of rational numbers in the interval $[0,1]$ Let $I_{k}$, $k=1,2,\cdots, n$ be a finite collection of open intervals that covers $B$.  Prove that $\sum_{k=1}^{n}l(I_{k})\geq 1$.
($l(I)$ denotes the length of the interval $I$).

$\underline{\mathbf{\text{Proof}}}$: Suppose $a,b \in [0,1]$ such that $a < b$. By the density of the rational numbers in $\mathbb{R}$, $\exists q_{i} \in \mathbb{Q}$ such that $\forall a_{i} < b_{i}$, $q_{i}\in(a_{i},b_{i})$.
Since the rationals are dense in $\mathbb{R}$ and hence in $[0,1]$, we can cover $[0,1]$ as follows:
$[0,1] \subseteq \cup_{k=1}^{\infty}I_{k}=\cup_{k=1}^{\infty} (a_{k},b_{k}) $.
(In the case where the $(a_{k},b_{k})$ do not completely cover $[0,1]$ completely, $\exists$ a point between $(a_{i},b_{i})$ and $(a_{i+1}, b_{i+1})$ that remains uncovered.  However, because there are countably many intervals, there will be only countably many such points.  Thus, the set of points, call it $S$, being a countable set will have $m^{*}(S)=0$.  So, these points will be negligible in terms of their effect on the sum of the lengths of the intervals.)
By Heine-Borel, since $[0,1]$ is closed and bounded, every such open cover has a finite subcover. So, $[0,1] \subset \cup_{k=1}^{n}(a_{k},b_{k}) = \cup_{k=1}^{n}I_{k}$.
Now, the outer measure of an interval is its length.  So $m^{*}([0,1])=l([0,1]) = 1$, and by the definition of the outer measure, $1 = m^{*}([0,1]) = \inf_{k=1}^{n}l(I_{k}) \leq \sum_{k=1}^{n}l(I_{k})$

Q.E.D.
I also realize this question has been asked before on here, but I am terribly dissatisfied with the answers that have thus far been given.

REPLY [2 votes]: Consider the corresponding closed intervals. Their union covers $\mathbb{Q}\cap[0,1]$ and is closed  ( a finite union of closed sets). Therefore their union contains $[0,1]$. So the sum of the length of the intervals is $\ge 1$.