I have been looking at examples showing that the set of all rationals have Lebesgue measure zero. In examples, they always cover the rationals using an infinite number of open intervals, then compute the infinite sum of all their lengths as a sum of a geometric series. For example, see this proof.

However, I was wondering if I could simply define an interval $(q_n - \epsilon, q_n + \epsilon )$ around each rational number $q_n$. Since there is a countable number of such intervals, the Lebesgue measure must be bound above by a countable sum of their Lebesgue measure (subadditivity). i.e. $\mu(\mathbb{Q}) \leq \mu(\bigcup^{\infty}_{i=1} (q_n - \epsilon, q_n + \epsilon)) = \sum^{\infty}_{i=1} \mu((q_n - \epsilon, q_n + \epsilon))$.

Then, argue that each individual term is $\mu((q_n - \epsilon, q_n + \epsilon)) < 2\epsilon$ and is thus zero since the epsilon is arbitrary?

  • $\begingroup$ First, your less-than sign should be replaces with an equals sign. $\:$ Second, you only get "each individual term" approaches zero. $\;\;\;$ $\endgroup$
    – user57159
    Sep 28 '13 at 21:42
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    $\begingroup$ Your measure is countably additive. Points have measure zero, and $\Bbb Q$ is countable. $\endgroup$
    – Pedro Tamaroff
    Sep 28 '13 at 21:48
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    $\begingroup$ Using the countable additivity of Lebesgue measure doesn't seem in the spirit of the problem, or the example proofs given. The originally approach does in fact work, with slight modification. Notice we run into the problem that $\sum_{n=1}^\infty 2\epsilon =\infty$. We need a convergent sum. The fix clearly presents itself: simply take the intervals to be $(q_n-\epsilon_n, q_n+\epsilon_n)$ where $\epsilon_n=\epsilon\cdot 2^{-n}$. Then $\sum_{n=1}^\infty \epsilon 2^{-n}=4\epsilon$, which now goes to $0$ with $\epsilon$. $\endgroup$ Jul 22 '14 at 7:35

For $\epsilon > 0$, $\mu((q_n - \epsilon, q_n + \epsilon)) = 2\epsilon$ and $\sum_{n=1}^\infty 2\epsilon = \infty$, so this reasoning doesn't work.

A simple method to show that $\mu(\mathbb Q) = 0$ is to notice that the measure of a single point is $0$, thus: $$ \mu(\mathbb Q) = \mu\left(\bigcup_{n=1}^\infty \{q_n\}\right) = \sum_{n=1}^\infty \mu(\{q_n\}) = \sum_{n=1}^\infty 0 = 0 $$

  • $\begingroup$ @AymanHourieh then could you not do that for ANY finite subset as well? $\endgroup$ Nov 8 '18 at 13:57
  • $\begingroup$ @Euler_Salter you absolutely can, which is a proof that any countable (in particular finite) subset has measure zero $\endgroup$
    – Tyler6
    Feb 5 '19 at 6:30
  • $\begingroup$ @AymanHourieh I know you posted this 7 years ago, but I was about to ask a similar question. Just to be 100% certain, the argument that the measure of a single point is zero is essentially similar to the OP's original method, right? Except that in this case, you aren't summing from $1$ to $\infty$. $\endgroup$ Aug 31 '20 at 15:56
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    $\begingroup$ @roundsquare The crucial difference is that I take "intervals" consisting of a single point each, hence of measure zero. The OP takes intervals of positive measure. $\endgroup$ Aug 31 '20 at 20:18

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