Intuitive, possibly graphical explanation of why rationals have zero Lebesgue measure

Solution 1:

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:

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 $\mathbb R^+$).

Solution 2:

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.

Solution 3:

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.