I know that rationals, being a countable set, have zero Lebesgue measure. I think one way to prove it is to find an open set containing rationals that has measure less than $\epsilon$ for every $\epsilon >0$ fixed. You can do it by taking the rational points in sequence and taking intervals of length $\epsilon/2^n$. Then the union of these intervals has measure less or equal than $\epsilon$.

However I was wondering: how can I explain this intuitively? If one thinks of a dense subset, such as $\mathbb{Q}$ in $\mathbb{R}$, one thinks of something that is "so close" to the original set that it is undistinguishable, in a certain way. I think the most intuitive explanation would be that when you take those intervals, you are "scaling down" their lengths faster than how a given sequence of rational points approach a non rational one.

But this may sound a bit confusing, tricky, so I was wondering: is there a simple, intuitive, possibly graphical way of explaining to someone with very little background in math why rationals have measure zero?

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    $\begingroup$ It’s geometrically clear that a single point has measure zero. On the other hand, measures are something which are countably additive; this is also intuitive geometrically. So if you take a countable collection of distinct points, that must be measure zero as well. The lesson here is to not confuse topological notions with measure-theoretic ones. The two are very much distinct. $\endgroup$
    – shalop
    Jun 6, 2020 at 3:53
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    $\begingroup$ With respect to "with very little background in math", this video might help to achieve a related intuition: youtube.com/watch?v=p-xa-3V5KO8 ? $\endgroup$
    – badroit
    Jun 6, 2020 at 23:59

4 Answers 4


You could utilize one of the well known ways to count the rational numbers, namely consider the integer lattice $\mathbb Z^2$ and the subset $\{(a,b)\mid a\geq 1\ \wedge\ b\geq 0\}$ as illustrated here:

enter image description here

This corresponds to the positive rationals, namely $(a,b)\mapsto\frac ba$. It is a surjective covering and it is now simple to see how we might cover all those points using circles of a finite total area $\varepsilon$ for any given $\varepsilon >0$. In the image above, I have done this using circles of exponentially decreasing sizes, which corresponds to using the well known sum $$ 2=\sum \frac n{2^n} $$ as a finite bound which can then be scaled down ad infinitum.

Thus we can project this representation onto $\mathbb R$ and make a similarly effective covering there.

BTW one way to project this onto the number line $\mathbb R^+$ would be to draw a vertical line at $x=1$. Then given any rational number $q$ one could draw the line from the origin $(0,0)$ through $(1,q)$ and project the circle from the first lattice point this line passes through onto the vertical line $x=1$. This projected circle around $(1,q)$ would effectively translate into an interval around $q$ on the positive $y$-axis (corresponding to $\mathbb R^+$).


This is a really hard question; I think in general intuition for this sort of thing tends to come with experience, as you get used to the concepts. Having said that, I'll try to articulate the way that I think about it.

I guess the way of viewing $\mathbb{Q}$ as a subset of $\mathbb{R}$ is a load of dots on a continuous line. Obviously these dots are very close together (in fact the whole thing is nonsense because they're dense in $\mathbb{R}$), but intuitively the mental image does help to capture some of the relevant properties, particularly with an eye to the Lebesgue measure.

I would suggest constructing this set in steps, according to increasing denominator. Start with $\mathbb{Z}$. It seems pretty clear to me that this should have measure zero, since the dots are spaced out, and hence they occupy an "infinitely small" proportion of $\mathbb{R}$. Rigorously, we can prove that $\mathbb{Z}$ has measure zero by putting an interval of width $\epsilon 2^{-\lvert n \rvert}$ around each $n$.

For each $n\geq 1$, define $S_n = \{\frac{a}{b}\mid a,b\in\mathbb{Z}, b \leq n\}$ to be the set of rational numbers with denominator at most $n$. Thus, $\mathbb{Z} = S_1$. For each $n$, the elements of $S_n$ have some minimum gap between them (the lowest common multiple of the denominators less than or equal to $n$), hence the same argument we used for $\mathbb{Z}$ shows that $S_n$ has measure zero for each $n$.

At each step, we have a set of measure zero. If we continue this process infinitely, we will eventually reach every rational number (i.e. for every rational number $x$, there is a finite $n$ with $x \in S_n$), so in some sense $\mathbb{Q}$ is the "limit" of these null sets, and hence it is itself null. We can certainly make this "some sense" rigorous, since $\mathbb{Q}$ is the countable union of the $S_n$, but I'm not sure that's useful for the intuition.

Obviously what I've done here is not very sophisticated, but I think it is a bit easier to visualise than just invoking countability of $\mathbb{Q}$, since we are actually "zooming in" on $\mathbb{Q}$ in an explicit way.

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    $\begingroup$ This is a bit dangerous. The "dots on a line" idea would suggest that the irrationals have measure 0. $\endgroup$
    – J. Mini
    Jun 6, 2020 at 12:13
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    $\begingroup$ @J.Mini I agree. That's why I've been careful to highlight that everything I say is nonsense and I've sketched justifications for all my actual claims. The case with irrationals is different though because, as I say in the fourth paragraph, every rational number is actually in $S_n$ for some finite $n$, so this method of "populating" $\mathbb{Q}$ does reach every rational number. In contrast, it never reaches any irrational number. "Dots on a line" is just a way of visualising properties that we know to be true because we have proven them; it is not a substitute for proof. $\endgroup$ Jun 6, 2020 at 12:21

This isn't a geometric answer, but you can get a lot of intuition for Lebesgue measure by thinking about it probabilistically. Specifically:

The measure of a subset $S\subseteq [0,1)$ is the same as the probability that a randomly chosen point in $[0,1)$ will be an element of $S$.

For example, the set $S = [0,1/4] \cup [5/8,3/4]$ has measure $3/8$ because there is a $3/8$ chance that a randomly chosen number between $0$ and $1$ lies in $S$. Thus you can understand why the set $\mathbb{Q}\cap [0,1)$ has measure zero by thinking about why a randomly chosen real number between $0$ and $1$ has zero probabilty of being rational.

To understand the latter, observe that one method for producing a random real number between 0 and 1 is to repeatedly roll a 10-sided die (with faces labeled 0 through 9) to decide on the decimal digits of the number. For example, if you happen to roll the sequence $$ 3,\quad 1,\quad 4,\quad 1,\quad 5,\quad 9,\quad 2,\quad 6,\quad\ldots $$ then you have randomly selected the number $0.31415926\ldots$, or $\pi/10$. Since a real number has an infinite sequence of decimal digits, you have to roll the die an infinite number of times, but at the end you have produced a random real number.

Such a randomly produced number is rational if and only if the sequence of digits that you roll is eventually repeating, and if you think about it this is extremely unlikely. For example, it's basically impossible (probability zero) that you will eventually start rolling the same digit forever. It's just as unlikely that you will eventually start rolling the same pair of digits over and over, or the same sequence of three digits over and over, and so on. To me, this is a very intuitive argument that $\mathbb{Q}\cap[0,1)$ has measure zero.

Open Sets Containing $\mathbb{Q}\cap(0,1)$

I can't resist mentioning that you can use this same point of view to understand why there are open sets of small measure in $(0,1)$ that contain $\mathbb{Q}\cap (0,1)$. Given an $n\geq 2$, we say that a real number $x\in(0,1)$ with decimal digits $d_1,d_2,d_3,\ldots$ is $\boldsymbol{n}$-repetitive if there exists a $k\in\mathbb{N}$ so that $$ (d_{k+1},d_{k+2},\ldots,d_{2k}) = (d_{2k+1},d_{2k+2},\ldots,d_{3k}) = \cdots = (d_{nk+1},d_{nk+2},\ldots,d_{nk+k}) $$ That is, $x$ is $n$-repetitive if at any point in the decimal expansion the digits so far consist of a block of digits of some length $k$ followed by $n$ identical blocks of digits of length $k$. For example, the number $$ 0.157\,432\,432\,432\,432\,761\,398\,\ldots $$ is $4$-repetitive because of the $157$ followed by four repetitions of $432$. (Note that this concept is well-defined even though some numbers such as $1/2=0.4999\ldots=0.5000\ldots$ have more than one decimal expansion, since in this case both expansions are always $n$-repetitive.) I claim that:

(1) Every rational number in $(0,1)$ is $n$-repetitive for every $n\in\mathbb{N}$.

(2) For each $n\in\mathbb{N}$, the set of $n$-repetitive numbers in $(0,1)$ is an open set.

(3) The probability that a number in $(0,1)$ is $n$-repetitive goes to $0$ as $n\to\infty$.

For (1), if $x$ is rational then its digits must consist of an initial block of length $i$ followed by a repeated block of length $j$, and it follows that $x$ satisfies the $n$-repetitive condition on its first $k+nk$ digits for any $k\geq i$ that is a multiple of $j$.

For (2), observe that every $n$-repetitive number is contained in an open interval of $n$-repetitive numbers. In particular, if $x$ satisfies the $n$-repetitive condition using its first $k+nk$ digits, then so does any other number with the same first $k+nk$ digits, and this determines an open interval around $x$ of $n$-repetitive numbers. (If $x$ has two different decimal expansions then a slightly different argument is required.)

For (3), observe that the probability that a number $x\in(0,1)$ is $n$-repetitive using its first $k+nk$ digits is exactly $10^{-k(n-1)}$. It follows that $$ P(x\text{ is }n\text{-repetitive}) \leq \sum_{k=1}^\infty 10^{-k(n-1)} = \frac{1}{10^{n-1}-1}. $$ For example, the probability that a number is 2-repetitive is less than $1/9$, and the probability that a number is 3-repetitive is less than $1/99$.

Of course, this construction of open neighborhoods of $\mathbb{Q}\cap (0,1)$ with small measure is much more complicated than the usual one, but it always feels somehow more concrete to me, in the sense that we have described very explicitly which numbers are in the set using the digits of the decimal expansion, and it's somehow much more obvious that this set takes up only a small portion of the unit interval.

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    $\begingroup$ I think the trouble here is the lack of finite procedures for picking random real numbers. You need a good intuition for how to continue things to infinity, which isn't in any greater supply than intuition for the OP's question. $\endgroup$
    – hobbs
    Jun 6, 2020 at 0:52
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    $\begingroup$ @hobbs You're completely right, of course. Ultimately there's something irreducibly complex about Lebesgue measure, and the best we can do is to look at that complexity from all sorts of different points of view and try to get a sense of its shape. I think we gain a better understanding of the statement "the rationals have measure zero" by knowing that it's connected to the statement "a random sequence of digits will almost surely not repeat", but the latter statement is neither completely obvious nor the only helpful way of looking at the former. $\endgroup$
    – Jim Belk
    Jun 6, 2020 at 9:02
  • $\begingroup$ I think your comment about the irreducible complexity of the Lebesgue measure and the fact that we still can gain something from knowing the probabilistic interpretation of measure should be a part of your nice answer. =) $\endgroup$
    – user21820
    Jun 8, 2020 at 7:23

You know that the rationals are a countable set. So pick an epsilon. Lay out an interval of width 1/2 epsilon around the first ration. An interval of 1/4 epsilon around the second one. An interval of 1/8 epsilon around the third rational. Each rational gets an interval that is half the size of the previous one. To get an upper bound for the measure of the rationals, sum all of those intervals. The sum is.... epsilon, no matter how small you originally chose it. In other words, any positive number, no matter how small, is an upper bound for the measure of the rationals. So the measure can not be a positive number. It must be zero or negative. I haven't seen very many sets with negative measure, so it must be zero.

  • 1
    $\begingroup$ Yeah, that’s indeed how you prove it. But it doesn’t seem intuitive, doesn’t it? It is very simple and it just requires you to know a little bit of calculus and measure theory, but if you are explaining density of $\mathbb{Q}$ to someone and then state this result, it feels odd. How come that I can find an open set containing all rationals which has measure zero? It doesn’t seem intuitively right, even though we both know it is $\endgroup$
    – tommy1996q
    Jun 10, 2020 at 9:17

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