Length of finite sequence of open intervals whose union contains $X = \mathbb{Q} \cap [0,1]$ $X = \mathbb{Q} \cap [0,1]$ and $(I_n)$ is a finite sequence of open intervals such that $X \subseteq \bigcup_{n=1}^{N} I_n$. Prove that: $\sum_{n=1}^{N} \ell (I_n) \geq 1$. 
I’m not sure how to get started with approaching this question. Any help is appreciated. 
 A: The measure of $\bigcup_nI_n$ is at most the sum of the lengths of the intervals. If this is less than $1$, the measure is less than $1$, and thus its complement in $[0,1]$ has positive measure. Thus it contains infinitely many points. Since there are only finitely many $I_n$, at least two of these points have no $I_n$ between them. Between these two points lies a rational that's not covered by the $I_n$.
A: Note that $[0,1] = \overline{X} $ and $\overline{\cup_k I_k} = \cup_k \overline{I_k}$ hence
$[0,1] \subset \cup_k \overline{I_k}$. Then $1_{[0,1]} \le \sum_k 1_\overline{I_k}$ and
hence integrating gives
$1 \le \sum_k l(\overline{I_k}) $. Since $l(I_k)= l(\overline{I_k})$ we are finished.
Note: A slightly different approach is to pick $\epsilon>0$ and let
$I_k' = (\inf I_k-\epsilon, \sup I_k+\epsilon)$, note that the $I_k'$ cover $[0,1]$. Then $1_{[0,1]} \le \sum_k 1_{I_k'}$ and integrating gives
$1 \le \sum_l l({I_k'}) = \sum_k (l(I_k) +2 \epsilon)$. Since $\epsilon>0$
was arbitrary, we have the desired result.
A: What I tried to prove is that the union of $I_n$ is basically at least $[0,1] \setminus Z $ where $Z$ is of measure 0. I'm not 100% sure it's correct but as the other answer follows another idea(which I think is much evident) I will post just for the sake of diversity.

 You can just take the closure of $\bigcup_{n} I_n$ and it's a compact space as it's closed and bounded. We have added to the previous set a finite number of points so the measure is still the same. Take the covering of open balls $B_{\epsilon_x}(x)$ where $x \in$ the interior(add what's necessary). As it's compact it has a Lebesgue number $\epsilon$. Use this number and that $\mathbb Q$ is dense in $[0,1]$ to finish the proof. 

