Prove sum of the lengths of intervals in a finite covering of $\mathbb{Q}\cap [0,1]$ is $\geq 1$ 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.
 A: Let $B$ be the set of rational numbers in the interval $[0,1]$, and let $\{I_k\}_{k=1}^n$ be a finite collection of open intervals that covers $B$. Prove that $\sum_{k=1}^n m^*(I_k) \geq 1$.

Proof: 
First, we have $B = \mathbb{Q}\cap [0,1] \subset [0,1]$ and $B \subset \bigcup_{k=1}^n I_k$, where each $I_k$ is an open interval. It follows immediately that since $B$ is countable has outer measure zero and
$$B \subset [0,1] \Rightarrow m^*(B) \leq m^*([0,1]) = 1,$$
$$B \subset \bigcup_{k=1}^n I_k \Rightarrow m^*(B) \leq m^*(\bigcup_{k=1}^n I_k) \leq \sum_{k=1}^{\infty} m^*(I_k).$$
Now, since $0 \in B$, there exist one of the $I_k$'s that contains $0$, denote such interval by $(a_1,b_1)$. By a similar reasoning, there exist one of the $I_k$'s that contains $1$ (Note that $1 \in B$), denote such interval by $(a_N,b_N)$. We thus have:
$$a_1 < 0 < b_1$$
$$a_N < 1 < b_N$$
Now, lets assume the contrary, that is $\sum_{k=1}^n m^*(I_k) < 1$. In particular, we are assuming the following:
$$\sum_{k=1}^n m^*(I_k) < m^*([0,1])$$
We claim that if $b_1 \geq 1$, we obtain a contradiction, since:
$$a_1 < 0 < 1 \leq b_1$$
and
$$ l((a_1,b_1)) = b_1 - a_1 \leq \sum_{k=1}^n m^*(I_k) < 1,$$
but $b_1 - a_1 > 1$. Otherwise, $b_1 \in [0,1)$ and since $b_1 \notin (a_1,b_1)$, there exist an interval in the collection $\{I_k\}_{k=1}^{n}$ which we label as $(a_2,b_2) \ni b_1$ and $$a_2 < b_1 < b_2.$$
We claim that if $b_2 \geq 1$, we obtain a contradiction, since:
$$b_2 - a_1 < (b_2 -a_2)+(b_1 -a_1) \leq \sum_{k=1}^n m^*(I_k) < 1$$
but $b_1 -a_0 > 1$ because $$a_0 < 0 < 1 \leq b_1.$$
We can continue this selection process until it terminates, as it must since there are only $n$ intervals in the collection $\{I_k\}_{k=1}^{n}$. We thus obtain a sub-collection $\{(a_k,b_k)\}_{k=1}^N$ of $\{I_k\}_{k=1}^{n}$ for which $a_1 < 0,$ while $a_{k+1} < b_k$ for $1 \leq k \leq N-1$ and  $b_N>1$. We thus have:
$$\begin{align*}
b_N-a_1 &< b_N - (a_N-b_{N-1})- \cdots-(a_2-b_1) -a_1 \\ 
 &\leq (b_N-a_N)+(b_{N-1}-a_{N-1})+\cdots +(b_1-a_1) \\ 
 &= \sum_{k=1}^N l(I_k) \leq \sum_{k=1}^n m^*(I_k)<1 
\end{align*}$$
a contradiction, since $$l((0,1)) = 1 < b_N-a_1$$ and $$a_1 < 0 < 1 < b_N.$$
Thus, we conclude that: $$\sum_{k=1}^n m^*(I_k) \geq 1.$$
A: 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 $\ge 1$.
A: Your proof does not look right because the key point is that there are only finitely many intervals - this is not true for countably many intervals.
In your case, arrange the intervals $I_k = (a_k,b_k)$ in the obvious way such that $0\in (a_1,b_1), 1\in (a_n,b_n)$. Then, since the rationals are dense, $a_{k+1} \leq b_k$ for all $k$. Replacing $I_k$ by smaller intervals if need be, we may assume that $a_{k+1} = b_k$ for all $k$. Then
$$
\sum_{i=1}^n l(I_k)
$$
becomes a telescoping sum equal to $b_n - a_1 \geq 1$
A: All you need is this (which, by the way, is not true for an infinite sequence of intervals):

If $\sum_{k=1}^{n}l(I_{k})< 1$, then there is an interval $J \subset [0,1]$, of strictly positive length, which has no elements in common with any $I_k$ 

This $J$, with strictly positive length, must contain rational numbers, which means that the $I_k$ don't cover $B$.
A: $A=\mathbb{Q}\cap[0,1] $ 
This is a sequence of inequalities.
As $A$ is countable so $m^*(A)=0$. Let $\{I_n\}$ be a finite sequence of intervals covering $A$ 
$1=m^*([0,1])=m^*(\bar{A})\leq m^*(\bar{\cup I_n})\leq m^*(\cup{\bar{I_n}})\leq \sum m^*(\bar{I_n})=\sum l(\bar{I_n})=\sum l(I_n)$ $m^*$ is the outer measure of a set.
Hence the result follows.
A: It's obvious we need only consider bounded open intervals . For any non-empty finite set $C$ of non-empty bounded open intervals let $$d(C)=\min \{a |\exists b [(a,b\in C]\}.$$ $$\text {And then let } e(C)=\max \{b|(d(C),b)\in C\}.$$ $$\text {And for any } D\subseteq C \text { let }  D^*=\{s\in C | s \not \subseteq \cup D\}.$$ $$\text {Now let } D_0=\phi  \text { and let } D_{n+1}=D_n\cup \{(d(D_n^*),e(D_n^*)\}\text { if } \cup D_n\ne \cup C,$$ $$\text { or } D_{n+1}=D_n \text { if } \cup D_n=\cup C.$$ I leave it to you to show that  $$(1)... \exists n (\cup D_n=\cup C)$$ and that if  $m$ is the least $n$ satisfying (1) and if  $Q\cap [0,1]\subset \cup C$  , then $$(2)...D_{m}=\{(u_j,v_j) : j=1,...,m\}$$ $$\text {where } u_{j+1}\leq v_j<v_{j+1} \text { whenever }1\leq j\leq m-1$$ $$\text {and (of course) } u_1<0\wedge v_{m}>1.$$ $$\text { Since } D_{m}\subset C \text { we have } \sum_{s \in C} l(s)\geq \sum_{s\in D_m} l(s)=\sum_{j=1}^{j=m} (v_j-u_j)\geq$$  $$ (v_1-u_1)+\sum_{j=1}^{j=m-1} (v_{j+1}-v_j)=v_m-u_1>1.$$.....Footnote:In (2) we have $(u_1,v_1)=(d(C),e(C))$ ,  and when $j<m$ we have $(u_{j+1},v_{j+1})=(d(D_j^*,e(D_j^*))$.And in the last  "summation",if $m=1$ we take the value of the summation to be $0$.  
