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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. $$\mu(\mathbb{Q}) \leq \mu(\bigcup^{\infty}_{n=1} (q_n - \epsilon, q_n + \epsilon)) = \sum^{\infty}_{n=1} \mu((q_n - \epsilon, q_n + \epsilon))$$

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?

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  • $\begingroup$ 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, 2013 at 21:42
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    $\begingroup$ Your measure is countably additive. Points have measure zero, and $\Bbb Q$ is countable. $\endgroup$
    – Pedro
    Sep 28, 2013 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, 2014 at 7:35

3 Answers 3

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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: $$ \mu(\mathbb Q) = \mu\left(\bigcup_{n=1}^\infty \{q_n\}\right) = \sum_{n=1}^\infty \mu(\{q_n\}) = \sum_{n=1}^\infty 0 = 0 $$

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  • $\begingroup$ @AymanHourieh then could you not do that for ANY finite subset as well? $\endgroup$ Nov 8, 2018 at 13:57
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    $\begingroup$ @Euler_Salter you absolutely can, which is a proof that any countable (in particular finite) subset has measure zero $\endgroup$
    – Tyler6
    Feb 5, 2019 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, 2020 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, 2020 at 20:18
  • $\begingroup$ @AymanHourieh Thanks for providing the answer. I'm thinking about the difference between point and interval construction regarding your reply to roundsquare. I think it's much more about how you construct the interval, if I construct each interval by the relative terms in this series $\sum_{i=1}^{\infty}\frac{\epsilon}{2^{i}}$ I think it should be OK? $\endgroup$
    – LJNG
    Feb 12, 2023 at 1:59
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Define an interval $(q_n−\frac{ϵ}{2^n},q_n+\frac{ϵ}{2^n})$ around each rational number $q$

For $\epsilon>0$, $\mu((q_n−\frac{ϵ}{2^n},q_n+\frac{ϵ}{2^n}))=2\left(\frac{\epsilon}{2^n}\right)$ and $\sum^{\infty} _{n=1} 2\frac{ϵ}{2^n}=2ϵ$

Since $ϵ$ is arbitrary, so $μ(Q) = 0$.

I think this simple trick could work

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    $\begingroup$ I think that this is the better answer because to show that $\mu(\{x\})=0$, one usually employs the argument that $\mu(\{x\})\le \mu((x-\epsilon/2,x+\epsilon/2))=\epsilon$ and then let's $\epsilon$ go to $0$. But strictly this only shows that $\mu(\{x\})\le \epsilon$ for arbitrarily small but positive $\epsilon>0$ because if $\epsilon = 0$, then $x\notin (x-\epsilon/2,x+\epsilon/2)$ anymore. Hence when only bounding $\mu(\{x\})$ by arbitrarily small $\epsilon>0$, then one can not just take the infinite sum and conclude that the result is still arbitrarily small. This answer here solves that. $\endgroup$
    – exchange
    Sep 5, 2023 at 11:23
  • $\begingroup$ It seems like this argument can be extended to show the measure of all definable reals is also zero, since it only hinges on countability, yes? I agree this is a better answer because it doesn't rely on knowing that measures are restricted to countable summation. The other approach is an appeal to the definition of measure, whereas this answer nearly justifies it. One might fairly ask, "Well, why can't you do the same thing with the reals to show they don't have measure either?" and the answer, "You're not allowed to add that many singletons because the definition forbids it," seems arbitrary. $\endgroup$ Dec 29, 2023 at 19:50
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Let $\{r_n\}_n$ be an enumeration of the rationals and fix $\epsilon > 0$ and consider the open covering of the rationals

$$\Bbb{Q} \subset \bigcup_n (r_n - \frac{\epsilon}{2^{n+1}}, r_n + \frac{\epsilon}{2^{n+1}}).$$

Use this together with monotonicity to see that the measure of the rationals is less than or equal to $\epsilon$.I.e., $$ \begin{align} m(\Bbb{Q})&\leq m(\bigcup_n (r_n - \frac{\epsilon}{2^{n+1}}, r_n + \frac{\epsilon}{2^{n+1}}) && \text{monotonicity}\\ &\leq \sum_n m((r_n - \frac{\epsilon}{2^{n+1}}, r_n + \frac{\epsilon}{2^{n+1}})) && \text{sub-additivity}\\ &=\epsilon. \end{align} $$

As $\epsilon$ can be made arbitrarily small we are done. EDIT: I like using $2^{-(n+1)}$ instead of $2^{-n}$ to get back epsilon.

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