Prove that if $B$ is the set of rationals in $[0,1]$ with a finite subcover, then: $1 \leq \sum_{k=1}^n m^*(I_k)$ I have the following homework problem:
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 $\sum_{k=1}^n m^*(I_k) \geq 1$.

This is my approach, First we have that $B = \mathbb{Q}\cap [0,1]$. Now, $B \subset \bigcup_{k=1}^nI_k$, where $I_k$ are open intervals.In particular $0 \in B \wedge 1 \in B$.Now, since $B \subset \bigcup_{k=1}^nI_k$ then there exist positive numbers, $\epsilon_0, \epsilon_1$ such that:
$$(0-\epsilon_0,0+\epsilon_0)\subset \bigcup_{k=1}^nI_k$$
$$(1-\epsilon_0,1+\epsilon_0)\subset \bigcup_{k=1}^nI_k.$$ 
Clearly, $-e_0 \notin [0,1]$ and $e_1 \notin [0,1]$.I am not sure, if so far I am in the right track.
 A: There is a mistake in your argumentation.
There are open covers of the rationals in $[0, 1]$ that don't cover the whole interval. For example, let $x_i$ be an enumeration of all those rationals and take the sets of the form $(x_i - 2^{-i - 10}, x_i + 2^{-i - 10})$. The subadditivity of the outer measure shows that these sets don't cover the whole interval $[0, 1]$.
Note that $[0, 1]$ isn't even always covered by the $I_k$. Consider the open intervals $\left(-1, \frac{1}{\sqrt{2}}\right), \left(\frac{1}{\sqrt{2}}, 2\right)$.
However, there are at most finitely many numbers that aren't covered by the union of the intervals. More precisely: Show that there are at most $n - 1$ numbers that aren't covered by the union of the $I_k$. Conclude that there are real numbers $x_1, \ldots, x_k$ so that each of the intervals $(x_i, x_{i + 1})$ is a subset of the union.
A: Another  way to tackle this problem is the following :
 Let $ \left \{ I_{k} \right \}_{k=1}^{n} $ such that $ \mathbb{Q} \cap [0,1] \subseteq \bigcup^{n}_{k=1} {I_{k}} $ 
Taking closures implies that $ [0,1] \subseteq \bigcup^{n}_{k=1} \overline{I_{k}} $ Thus $ 1 \leq m(\bigcup^{n}_{k=1} \overline{I_{k}}) \leq \sum_{k=1}^{n}m(\overline{I_{k}})=\sum_{k=1}^{n}m({I_{k}}) $
$ \Longrightarrow \sum_{k=1}^{n}m({I_{k}}) \geq 1 $ which is exactly what we wanted to prove . 
