Studying the Lebesgue measure on the line I've found the following argument which concludes that $m(\mathbb{R}) < +\infty$ (where $m$ denotes the Lebesgue measure on $\mathbb{R}$). Obviously it must be flawed, but I haven't been able to find the flaw so far.

Recall that for a Lebesgue measurable set $A$ we have by definition that $$ m(A)=\inf\left\{\sum_{n=1}^\infty(b_n-a_n):\cup_{n=1}^\infty(a_n,b_n]\supset A\right\}. $$ Pick your favorite summable sequence of positive terms, $\{a_n=1/n^2\}_{n=1}^\infty$ for instance. We know the rational numbers are countable, so we can index them in a sequence $\{q_n\}_{n=1}^\infty$. Now consider the intervals $$ I_n=\left(q_n-\frac{a_n}{2},q_n+\frac{a_n}{2}\right]. $$ As the rationals are dense in $\mathbb{R}$ we must have $\mathbb{R}\subset\cup_{n=1}^\infty I_n$ but then, having into account the definition of Lebesgue measure we have $$ m(\mathbb{R})\leq\sum_{n=1}^\infty\left(\big(q_n+\frac{a_n}{2}\big)-\big(q_n-\frac{a_n}{2}\big)\right)=\sum_{n=1}^\infty a_n<+\infty. $$ For instance with $a_n=1/n^2$ we get $m(\mathbb{R})\leq\pi^2/6$.

As I said, I know this is flawed but I've spent almost two hours trying to find the flaw so any help would be appreciated!

  • 7
    $\begingroup$ This strange set you constructed is discussed in this video. It is known for its paradoxical property of being of finite measure, but having all rationals as inner points. Still, it is not all of $\Bbb R$. $\endgroup$
    – M. Winter
    Sep 11, 2017 at 17:14
  • 9
    $\begingroup$ A closed subset of $\mathbb{R}$ containing all rational numbers must equal $\mathbb{R}$ itself. Without closedness, however, nothing can be said. You have just constructed one such non-trivial example. $\endgroup$ Sep 11, 2017 at 17:22
  • 1
    $\begingroup$ @SangchulLee This should be the answer. $\endgroup$
    – Zach Boyd
    Sep 11, 2017 at 17:30
  • $\begingroup$ @ZachBoyd, I tried to improve it to an answer :) $\endgroup$ Sep 11, 2017 at 18:20
  • $\begingroup$ @M.Winter Thank you your comment/answer and thank you for the video too, I've found it really instructive and interesting!! :) $\endgroup$ Sep 11, 2017 at 18:43

4 Answers 4


"As the rationals are dense in $\mathbb{R}$ we must have $\mathbb{R}\subset \bigcup_{n=1}^\infty I_n$."

This is false. Pick your favorite irrational number $x$. For every $n$, there exist infinitely many rationals $q_m$ such that $q_m$ is within $a_n/2$ of $x$. But the least such $m$ may be much larger than $n$. In particular, it does not follow that there exists some $n$ such that $x$ is within $a_n/2$ of $q_n$, which is what is necessary to have $x\in \bigcup_{n=1}^\infty I_n$.


One can explicitly demonstrate the failure of the argument for $\mathbb R_+$. Generalize to $\mathbb R$ as needed.

Consider the following maps $f:\mathbb N_+^2 \rightarrow \mathbb N_+$ and $g: \mathbb N_+^2 \rightarrow \mathbb Q_+$ \begin{eqnarray} f(\{k,m\}) &=& (k + m - 2 )(k + m - 1) + m \\ g(\{k,m\}) &=& k/m. \end{eqnarray} You can verify that $f$ is a bijection and $g$ is onto, and thus $h(n) = g(f^{-1}(n))$ maps from $\mathbb N_+$ onto a dense subset of $\mathbb R_+$. Now consider the set $$ U = \bigcup_{n=1}^\infty \left(h(n) - \frac{a_n}{2},h(n) + \frac{a_n}{2}\right). $$ Using the sequence $\{a_n = n^{-2}\}_{n=1}^\infty$, I can prove $\phi = (1+\sqrt{5})/2\notin U$.

First, note that $h$ orders the ratios $k/m$ first by increasing $k+m$, then by increasing $m$. Because of this, if $h(n)$ is a continued fraction convergent of $\phi$, then $|h(n') - \phi| > |h(n) - \phi|$ for all $n' < n$. The continued fraction convergents of $\phi$ are of the form $F_{i+1}/F_i$, where $F_i$ are Fibonnaci numbers, and occur when $n_i = (F_{i+1} + F_{i} - 2)(F_{i+1} + F_{i} - 1)/2 + F_{i}\ge F_{i}F_{i+1}$. One can prove from the Fibonacci formula that the error on these convergents is always larger than $0.4F_{i}^2$.

Now consider $n_i \le n < n_{i+1}$. Then we have $a_n = n^{-2} \le n_i^{-2}\le F_{i}^{-2}F_{i+1}^{-2}$ and $|h(n) - \phi| > |h(n_{i+1})-\phi| > 0.4 F_{i+1}^{-2}$. If $\phi\in U$, there must be an $n$ with $|h(n) - \phi| < a_n/2$ for some $n$. This would then imply $0.4 < F_{i}^{-2}/2$, which can only be satisfied for $F_i = 1$. Thus such an $n$ would have to satisfy $n < n_3 = 8$. Since it can be checked that none of these works, $\phi \notin U$.

  • 2
    $\begingroup$ Wow! I really appreciate the effort in writing this concrete example, and that's why I am upvoting this, but you didn't need to post another answer since the existing ones already addressed my question completely. In particular my fake proof above shows that $\cup_nI_n$ doesn't contain the real line by contradiction. I hope this answer helps future readers who want specific examples though. $\endgroup$ Sep 12, 2017 at 2:48
  • 6
    $\begingroup$ I have a fondness for disproof by constructing counterexamples. It feels more concrete than a proof by contradiction. Also, upon further reflection, I have a suspicion that when using $h$ to index the rationals and assuming $a_n$ is monotone, $\phi \in U$ implies $\sum_{n=1}^\infty a_n = \infty$. That sounds hard to prove, though. $\endgroup$ Sep 12, 2017 at 6:04

(Extended from a comment) Recall that a subset $\mathcal{D}$ of $\mathbb{R}$ is called dense if its closure $\bar{\mathcal{D}}$ is all of $\mathbb{R}$. From this definition, you can readily check that the followings are equivalent for dense subsets $\mathcal{D} \subset \mathbb{R}$:

$$ \text{$\mathcal{D}$ is closed in $\mathbb{R}$} \qquad \Leftrightarrow \qquad \text{$\mathcal{D} = \mathbb{R}$}. $$

In particular, you can't tell much how porous your set $\mathcal{D}$ will be only by knowing the density of $\mathcal{D}$. There are lots of dense subsets of $\mathbb{R}$ which is not closed (and hence not all of $\mathbb{R})$.

Example. If we pick a sequence $(r_n)_{n=1}^{\infty}$ of positive numbers satisfying $\sum_{n=1}^{\infty} r_n < \infty$, then

$$ U = \bigcup_{n=1}^{\infty} (q_n - r_n, q_n + r_n)$$

is an open set such that $\mathbb{Q} \subset U$ but $\operatorname{Leb}(U) \leq \sum_{n=1}^{\infty} 2r_n < \infty$. So there are open dense subsets of $\mathbb{R}$ with finite length. Notice also that $\operatorname{Leb}(U)$ can be made arbitrarily small by choosing suitable $(r_n)$. Another interesting feature of this example is that the boundary $\partial U$ has infinite length! In this way, you may think of $U$ as tiny pores of a highly porous material, such as a sponge.

  • 1
    $\begingroup$ (+1) Thank you for your comment and answer, I've found them really helpful! I accepted Alex Kruckman's answer though because he gets exactly to the point I was missing, and because we're only allowed to accept one answer. :( $\endgroup$ Sep 11, 2017 at 18:29

The rationals are all contained in $\bigcup I_n$ but not all irrationals are.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .