Let $(I_j)_{j = 1}^n$ be a finite collection of intervals that covers the rationals in $[0, 1]$. Prove that $\sum_{j = 1}^n \ell (I_j) \geq 1$. (Here, $\ell (I)$ denotes the length of the interval $I$.)
I thought the following: consider $\mathfrak{m}^\ast$ the exterior Lebesgue measure on $\Bbb R$. We have that $\Bbb Q \cap [0, 1] \subseteq \bigcup_{j = 1}^n I_j$. Since $\Bbb Q$ is dense in $\Bbb R$, we have $[0, 1] \subseteq \bigcup_{j = 1}^n I_j$, hence: $$\mathfrak{m}^\ast [0, 1] \leq \sum_{j = 1}^n \mathfrak{m}^\ast I_j = \sum_{j = 1}^n \ell (I_j) \implies 1 \leq \sum_{j = 1}^n \ell (I_j)$$
as we wanted.
However, I'm not sure that $\Bbb Q$ being dense really implies that $[0, 1] \subseteq \bigcup_{j = 1}^n I_j$. Also, I would rather not use that $\Bbb Q$ is dense. Can someone give me some light here, please? Alternative solutions are welcome too. Thanks.
Hint: Suppose (without loss of generality; why?) that the intervals $\{I_i\}_{i=1}^n$ are closed. If $\sum_{i=1}^n \ell(I_i) = 1-\epsilon < 1$, then $[0,1]\setminus \left(\cup_{i=1}^n I_i\right)$ is an open set of measure $\epsilon$, so it contains an open ball.
Edit: To be more precise, $A := [0,1]\setminus \left(\cup_{i=1}^n I_i\right)$ is an open set of measure at least $\epsilon$: $$m(A) = m([0,1]) - m\left([0,1]\cap \cup_{i=1}^n I_i\right) \geq 1-\sum_{i=1}^n \ell(I_i)$$ by finite subadditivity.
Alternatively, it is no loss of generality to assume the $I_i$ are disjoint and contained in $[0,1]$. We may simply take $\{\tilde{I}_j\}_{j=1}^{\tilde{n}}$ in $[0,1]$ by chopping off anything else, and disjoint by redistributing overlap whenever $I_i \cap I_j \neq \emptyset$ to new sets $\tilde{I}_k$, $k > n$. Then the contradiction will still follow because we prove $\sum_{i=1}^{\tilde{n}} \ell(\tilde{I}_j)$ and $$\sum_{i=1}^n \ell(I_i) \geq \sum_{j=1}^{\tilde{n}} \ell(\tilde{I}_j).$$