See Lebesgue outer measure of $[0,1]\cap\mathbb{Q}$

Lebesgue measure: $$ m(A) = \inf \left\{ \sum |I_n| : A \subset \bigcup I_n \right\} $$

We know that $m(\mathbb{Q} \cap [0,1]) = 0$.


Enumerate the rationals $\mathbb{Q} \cap [0,1] = \{ q_n \}_{n \in \mathbb{N}}$.

Let $A_n$ be the covering of rationals $\{ q_n \}_{n \in \mathbb{N}}$ in $[0,1]$ by intervals $A_n = (q_n - \varepsilon/2^n , q_n + \varepsilon/2^n)$. Then, $$ m(\mathbb{Q} \cap [0,1]) \le m( \bigcup A_n ) \le \sum_n |A_n| = \sum_n \frac{\varepsilon}{2^n} = \varepsilon $$ and since it is true for every $\epsilon > 0$ as small as we want, we have $m(\mathbb{Q} \cap [0,1]) = 0$. $\square$


Every number in $[0,1]$ is contained in an open neighborhood of a rational number. We reach a paradox if $$\bigcup_{n \in \mathbb{N}} A_n = [0,1] $$ since then $$m([0,1]) \le m( \bigcup A_n ) \le \sum m(A_n) = \varepsilon $$ and then $m([0,1])=0$ and not $1$ as it should be.

To contradict this we need to show $$ \bigcup A_n \ne [0,1] \ . $$ We need to build an irrational number $r$ such that $\forall n : |r - q_n| > \varepsilon/2^n$. Can you construct such a number (or prove it exists without using Lebesgue measure argument)?

Moreover, I think we need to show the set $[0,1]-\bigcup A_n$ has $\aleph = 2^{\aleph_0}$ elements. Otherwise, $m([0,1]-\bigcup A_n) = 0$ and we reach a contradiction.

  • 1
    $\begingroup$ But we don't know how you enumerated the rationals... $\endgroup$
    – njguliyev
    Commented Aug 24, 2013 at 19:56
  • $\begingroup$ The set of rational is countable. I can give you a concrete enumeration, e.g. $\frac{a}{b} \mapsto (a,b) \mapsto a+b$, we order them by the sum $a+b$ and if $a+b = a'+b'$ then the inner oreder is determined by $a < a'$). I don't think this is the issue here. $\endgroup$
    – LinAlgMan
    Commented Aug 24, 2013 at 20:00
  • $\begingroup$ math.stackexchange.com/questions/346854/… Look at Did response here for an explicit construction. $\endgroup$ Commented Aug 24, 2013 at 20:00
  • $\begingroup$ I saw an answer here: math.stackexchange.com/questions/213831/… $\endgroup$
    – LinAlgMan
    Commented Aug 24, 2013 at 20:00

2 Answers 2


If I may choose the enumeration of the rationals:

Enumerate the rationals in increasing order of the denominator in the reduced form, rationals with the same denominator in increasing order, so

$$q_1 = \frac01,\, q_2 = \frac11,\, q_3 = \frac12,\, q_4 = \frac13,\, q_5 = \frac23,\, q_6 = \frac14,\,q_7 = \frac34,\,\ldots$$

Then for $q_n = \frac{r_n}{s_n}$, we have $s_n \leqslant n$.

For $x = \frac{1}{\sqrt{2}}$, and a rational $0 \leqslant \frac{r}{s} \leqslant 1$, we have

$$\left\lvert\frac{r}{s} -\frac{1}{\sqrt{2}}\right\rvert = \frac{\left\lvert \frac{r^2}{s^2} - \frac12\right\rvert}{\left\lvert\frac{r}{s} +\frac{1}{\sqrt{2}}\right\rvert} \geqslant \frac{\lvert 2r^2-s^2\rvert}{4s^2} \geqslant \frac{1}{4s^2}.$$

For $0 < \varepsilon < \frac29$, we have $\frac{1}{4n^2} > \frac{\varepsilon}{2^n}$ for all $n \geqslant 1$, hence $1/\sqrt{2}$ is not contained in any $(q_n - \varepsilon 2^{-n},\,q_n + \varepsilon 2^{-n})$.

By a similar reasoning, we find $2^{\aleph_0}$ irrationals that aren't covered:

For an irrational $x \in (0,\,1)$ with continued fraction expansion

$$x = [a_0,\, a_1,\, a_2,\, \dotsc],$$

convergents $\frac{r_n}{s_n}$, and the complete quotients $\alpha_n$ - that means

$$x = \frac{\alpha_{n+1}r_n + r_{n-1}}{\alpha_{n+1}s_n + s_{n-1}},$$

and thus the continued fraction expansion $\alpha_n = [a_n,\, a_{n+1},\, \dotsc]$, whence $a_n < \alpha_n < a_n + 1$ - we have

$$\begin{align} \left\lvert x - \frac{r_n}{s_n}\right\rvert &= \left\lvert \frac{\alpha_{n+1}r_n + r_{n-1}}{\alpha_{n+1}s_n + s_{n-1}} - \frac{r_n}{s_n}\right\rvert\\ &= \left\lvert \frac{\alpha_{n+1}(r_n s_n - s_nr_n) + (r_{n-1}s_n - r_ns_{n-1})}{(\alpha_{n+1}s_n+s_{n-1})s_n}\right\rvert\\ &= \left\lvert\frac{(-1)^{n}}{(\alpha_{n+1}s_n+s_{n-1})s_n}\right\rvert\\ &> \frac{1}{\bigl((a_{n+1}+1)s_n + s_{n-1}\bigr)s_n}\\ &\geqslant \frac{1}{(a_{n+1}+2)s_n^2}.\tag{1} \end{align}$$

For a rational number $\frac{r}{s}$ with $s \leqslant s_n$, $n \geqslant 1$, we have

$$\lvert sx - r\rvert \geqslant \lvert s_nx - r_n\rvert$$

with equality if and only if $s = s_n$ and $r = r_n$.

This immediately leads to

$$\left\lvert x - \frac{r}{s}\right\rvert > \frac{1}{(a_{n+1}+2)s\cdot s_n}.\tag{2}$$

Thus, for any fixed $M \in \mathbb{Z}^+$, let $A_M$ be the set of all irrational numbers in $(0,\,1)$ whose continued fraction expansion has all partial quotients $\leqslant M$.

For $M \geqslant 2$, the set $A_M$ has cardinality $2^{\aleph_0}$.

For $x \in A_M$, the denominators of the convergents satisfy

$$s_{n+1} = a_{n+1}s_n + s_{n-1} \leqslant M\cdot s_n + s_{n-1} \leqslant (M+1)\cdot s_n.\tag{3}$$

Choosing the minimal $n$ with $s_n \geqslant s$ in $(2)$, we deduce

$$\left\lvert x - \frac{r}{s}\right\rvert > \frac{1}{(M+2)(M+1)s^2}$$

from $(3)$ for $x \in A_M$.

If $\varepsilon$ is small enough, so that

$$\frac{1}{(M+2)^2n^2} > \frac{\varepsilon}{2^n}$$

for all $n$, the set

$$U_\varepsilon = \bigcup_{n=1}^\infty \left(q_n - \frac{\varepsilon}{2^n},\,q_n + \frac{\varepsilon}{2^n}\right)$$

doesn't intersect $A_M$.

  • $\begingroup$ Thank you. The next step is to show there are $\aleph$ such numbers. Otherwise, if $[0,1]-\bigcup A_n$ is countable then $$m( [0,1]-\bigcup A_n ) = 0 $$ and thus $m(\bigcup A_n) = 1$ which is again a contradiction. $\endgroup$
    – LinAlgMan
    Commented Aug 24, 2013 at 20:58
  • $\begingroup$ @LinAlgMan, why do you need more evidence? The valid (and simple) proof that the measure is $0$ and the evidence that the paradox you thought you saw doesn't hold together so well should really be quite convincing enough, no? $\endgroup$
    – dfeuer
    Commented Aug 24, 2013 at 22:59
  • $\begingroup$ Because I'm arguing with someone and can't convince him. I try to give as analogue the Cantor set. He claims that since the rationals are densed in $[0,1]$ then the numbers that are not contained in the covering $\bigcup A_n$ must be in the endpoints of the intevals $A_n$ and there are only $\aleph_0$ such endpoints. Therefore, $\bigcup A_n$ covers almost all $[0,1]$. $\endgroup$
    – LinAlgMan
    Commented Aug 25, 2013 at 9:14
  • 2
    $\begingroup$ @LinAlgMan: Are you arguing with Wolfgang Mueckenheim? $\endgroup$
    – Asaf Karagila
    Commented Aug 25, 2013 at 10:49
  • 1
    $\begingroup$ @LinAlgMan I've added a proof that (for the chosen enumeration of the rationals; it holds for arbitrary enumerations, but proving that without using measure theory is probably hard) for all small enough $\varepsilon > 0$, there are $2^{\aleph_0}$ irrationals not covered. Is that good enough to convince your someone? [Well, since there are null sets of cardinality $2^{\aleph_0}$, it would still be possible that $A_M$ is a null set.] $\endgroup$ Commented Aug 25, 2013 at 10:49

It will be difficult to prove the statement using topological arguments alone! Consider an analogous covering with wider intervals: $|B_n|=\epsilon/n$. Then $\bigcup B_n$ might or might not cover the whole interval $[0,1]$; it depends on the enumeration of rationals $\{q_n\}$.

The intervals $B_n$ look a lot like $A_n$. But in the situation $|A_n|=\epsilon/2^n$, we know that $\bigcup A_n$ doesn't cover the whole interval, regardless of the enumeration $\{q_n\}$, precisely because of a measure argument. I suspect that any proof of this fact will have to be a thinly disguised version of the same argument. (But by all means, prove me wrong!)


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