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?

• Is their union actually the real line? Commented Nov 27, 2015 at 19:09
• The proof shows that there are many reals not covered. It may help you to construct an example in which say $\sqrt{2}$ is not covered. Commented Nov 27, 2015 at 19:10
• I think the key observation is that the centers of your open covers "move with" $n$, meaning that you are explicitly covering each $q \in \mathbb{Q}$. But I do not believe this covering actually obtains $\mathbb{R}$. Commented Nov 27, 2015 at 19:11
• if you remove a single (irrational) number from the real line, you have a 'gap' which does not contain a rational number. Of course in your cover there is no gap which contains an interval, but there are (uncountably many) gaps, each of which does not contain any rational number. Commented Nov 27, 2015 at 19:12
• The bestes answer is here: math.stackexchange.com/questions/1525184/… Commented Nov 27, 2015 at 20:07

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.

• So when you say the "gaps" are small, are they singular points?
– MT_
Commented Nov 27, 2015 at 19:18
• @Soke: Yes -- or at least there is no connected gap with more than one point in it. Commented Nov 27, 2015 at 19:22
• Then there must be uncountably many gaps or else the remaining set would have measure zero as well - how do we get this result from construction? I would think that the number of gaps would be no more than the number of "gaps" in $\mathbb{R} \setminus \mathbb{Q}$, which would be countable since $\mathbb{Q}$ is. Does it not make sense to think of it like that since there is no notion of a "next largest" rational number?
– MT_
Commented Nov 27, 2015 at 19:25
• @Soke: No no no -- $\mathbb R\setminus \mathbb Q$ has the same kind of gaps: there's no connected gap with more than one point, and there are uncountably many of them, namely one for each irrational number. Note that with infinity, you cannot think of a gap as "the distance between two neighboring rationals": there is no such thing as neighboring rationals! Commented Nov 27, 2015 at 19:26
• Yeah, that's what I figured. Thanks for the help -
– MT_
Commented Nov 27, 2015 at 19:28

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.

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.