What am I misunderstanding about this constructive proof that $\mu(\mathbb{Q}) = 0$? In class we were given a constructive proof that $\mu(\mathbb{Q}) = 0$, with $\mu$ the Lebesgue measure. Of course it is clear that they have measure zero since they are countable, but this constructive proof doesn't sit well with me.
Let $\{ q_n\}_{n=1}^{\infty}$ be an enumeration of the rationals. Then fix $\varepsilon > 0$ and for each $q_n$ take the interval $A_n = (q_n - \frac{\varepsilon}{2^n}, q_n + \frac{\varepsilon}{2^n})$. Then 
$$\mu^*\left(\{q_n\}_{n=1}^{\infty}\right) \leq \sum_{n=1}^{\infty} \mu (A_n) = \sum_{n=1}^{\infty} \frac{\varepsilon}{2^{n-1}} = 2 \varepsilon$$
So the measure is zero since $\varepsilon$ was arbitrary.
However, the part that doesn't sit well with me is that it seems that eventually it must be that this covering of the rationals by open sets must eventually cover the entire real line. Indeed, if there were some "gap" in the cover, no matter how small, since the rationals are dense then there must be some rational (infinitely many rationals, actually) that are not covered. So then the combined measure of these intervals could not be zero since their union is the real line. What is wrong with my thinking?
 A: What this example tells you is that your intuition about these things is not really reliable. Don't worry too much about that; everybody goes through it.
In particular, the union of the intervals does not cover the entire real line. There are gaps -- they are small (none of them contain an interval), but there are a lot of them, and somehow they manage to add up to something with positive measure.
A: I get your confusion, but a set having "gaps" does not imply the set misses an open interval. For more trivial examples, consider $\mathbb R\setminus\{\sqrt{2}\}$ or $\mathbb R\setminus\mathbb Q$. The first set has a single gap but still contains every rational number. The second set has gaps that are dense in the whole real line, but have zero measure. The cover that has been constructed in your question covers every rational number (and hence many irrational numbers), but misses "most" of $\mathbb R$, as evidenced by the fact it has very small measure.
A: Replace your $A_n$ by $ A_n = \Bigg{(}q_n - \frac{ε}{2^{n+1}}, q_n + \frac{ε}{2^{n+1}} \Bigg{)} $ and this will works.
