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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?

MT_
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    Is their union actually the real line? – Ben Longo Nov 27 '15 at 19:09
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    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. – André Nicolas Nov 27 '15 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 $\mathbb{R}$. – Kevin Sheng Nov 27 '15 at 19:11
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    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. – Thomas Nov 27 '15 at 19:12
  • @AndréNicolas Ok, that helps. I think the problem I was having is that I was thinking of this as a finitely determined process, in which having "gaps" in the intervals on the real line would indeed be indicative of us missing rationals, but we can construct such a gap between two rationals if the constructive process is countably infinite (so we can say get closer, and closer, and closer...) – MT_ Nov 27 '15 at 19:14
  • @Thomas How do the gaps come to have an uncountable cardinality? – MT_ Nov 27 '15 at 19:19
  • I can kind of foresee that you won't like the answer, but if the number of gaps were countable, you could cover them in a similar vein as described in your question and would end up with a cover of the reals of volume less than any given $\varepsilon$. Which is impossible ;-) – Thomas Nov 27 '15 at 19:28
  • @Thomas Yeah it definitely makes sense that these gaps, each being singletons, would be uncountable (else we will have decomposed $\mathbb{R}$ into two sets with measure zero), but a priori from construction it doesn't seem to follow. I'll let it simmer a bit. – MT_ Nov 27 '15 at 19:30
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    The bestes answer is here: http://math.stackexchange.com/questions/1525184/apparent-paradox-covering-the-real-line – Michael Hoppe Nov 27 '15 at 20:07
  • These questions all ask which irrationals are omitted from the covering: 1 2 3 4. Nearly everyone seems to find this mind-boggling. – MJD Nov 28 '15 at 00:00

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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.

hmakholm left over Monica
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  • So when you say the "gaps" are small, are they singular points? – MT_ Nov 27 '15 at 19:18
  • @Soke: Yes -- or at least there is no connected gap with more than one point in it. – hmakholm left over Monica Nov 27 '15 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_ Nov 27 '15 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! – hmakholm left over Monica Nov 27 '15 at 19:26
  • Yeah, that's what I figured. Thanks for the help - – MT_ Nov 27 '15 at 19:28
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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.

Jason
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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.

igorkf
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