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

Question

For $E\subseteq\mathbb{R}^n$ I was given this definition of Lebesgue measure:

$m(E)=\sup \{m(K) \mid K\subseteq E \text{ compact}\} = \inf \{m(A) \mid A\supseteq E \text{ open}\}$ if they are equal, where:

$m(K)=\inf\{m(P) \mid P\supseteq K \text{ multi-interval}\}$

$m(A)=\sup\{m(P) \mid P\subseteq A \text{ multi-interval}\}$

(a multi-interval is a finite union of intervals. An interval $I$ is $I=\prod_{j=1}^n [a_j,b_j)$. So we might write $P=\bigsqcup_{j=1}^k I_k$, with the natural measure).

Now for $n=1$ and $E:=\mathbb{Q}\cap [0,1]$ I expect $m(E)$ to be zero. But when I imagine an open set $A\supseteq E$, it never gets a less-than-one measure. Can somebody give me a hint please?

PS: I don't want to use other ways to find the measure of $E$: I'm looking for a better understanding of the definitions.

Enmerate the rationals as $q_n$ and place around the $n$-th a ball of radius $r_n/2$ chosen small enough so that $\sum r_n$ converges to a number smaller than $1$ — Alessandro Codenotti, Aug 07 '20 at 11:28
But the balls will necessarily intersect and therefore (1) the measure won't be $\sum r_n$, and (2) I will still cover all the $[0,1]$ interval. Isn't it so? — rod, Aug 07 '20 at 11:34
They will overlap, but that means that the actual measure is even less than $\sum r_n$ so there is no issue. And no, not the whole interval will be covered — Alessandro Codenotti, Aug 07 '20 at 11:36
The balls don't necessarily cover $[0,1]$. Take for example $c = 1/\pi \in [0,1]$. As $c$ is irrational, for any rational number $q \in [0,1]$ you can take the ball $B(q,(q-c)/4)$. Then $c$ lies in none of those balls, so those balls do not cover $[0,1]$. — Hetebrij, Aug 07 '20 at 11:39
Yes, that is true, thank you. May I ask you guys if you get to visually see what's happening with the measure provided by the open sets given by the union of those balls? Cause I can't. — rod, Aug 07 '20 at 11:41
@tomasz can you draw a mental picture that clarifies by itself the inf of measures attaining to zero? — rod, Aug 07 '20 at 11:46
Well, you draw these countably many balls of measures adding up to less than infinity, and then you make them progressively smaller. At the limit, you have just ${\mathbf Q}$, which is of course of measure zero (although that is no longer open). — tomasz, Aug 07 '20 at 11:47
@tomasz I am sorry, I don't see it :( and you can't say $\mathbf{Q}$ has obviously zero measure if that's what I'm trying to see. I should see the limit of measure attaining to zero. — rod, Aug 07 '20 at 11:54
@roddik: Well, the measure is countably additive, and points have measure zero, the rationals are countable... — tomasz, Aug 07 '20 at 12:01
@tomasz Yes, that is why I added the "PS" at the end of the question: I already know the measure is zero. I need to understand the definition... — rod, Aug 07 '20 at 12:05
@roddik: The countable additivity is the essence of the definition! — tomasz, Aug 07 '20 at 12:17
@tomasz I agree: but this definition does not refer to countable additivity, so it must work without asking its help. This apparently means that the complexity of $\mathbb{Q}$ does indeed not allow the visual representation I was looking for. — rod, Aug 07 '20 at 12:23
The visual representation is the same for every countable set. So take $\mathbb{N}$ for simplicity. You can wrap your mind about that, then you can see more clearly what is going on with your set. — Mushu Nrek, Aug 07 '20 at 12:30
@MushuNrek You must be right: I guess we can't do better than that. — rod, Aug 07 '20 at 12:43
Since the measure in question is zero, you need only focus on showing that there are a countable collection of $n$-intervals that contain the rationals and whose volumes sum to less than $\epsilon$. — copper.hat, Aug 07 '20 at 13:05
@AlessandroCodenotti your answer was right. I can choose it if you want. — rod, Aug 08 '20 at 17:18

rod · Accepted Answer · 2021-10-25T19:27:40.643

Ok, I honestly think I figured out the answer I'd want to receive.

The confusing point is: every open set $A\supseteq E$ is "basically E". The insight that I find satisfying is the following. This is basically the answer provided by @Alessandro Codenotti in the comments section.

The measure of an open set is defined as a $sup$ of measures of $\textit{finite}$ unions of intervals. Since $\textit{any}$ union of open sets is open, I might cover $E$ with a "big" union of $\textit{small}$ open intervals in such a way that any time I try to fill this covering with a countable (actually finite but then I'm taking the limit as a $sup$) number of intervals I am forced to take them so tiny that I can't cover most of $[0,1]$).

So the heart of this definition is the double approximation. In other words "one gets zero not when he approximates with open sets from outside but when he approximates those open sets from inside with multi-intervals".

Of course, once you realize it is actually a countably additive measure, everything becomes easier (you basically only need to know $m(x)=0$ $\forall x\in\mathbb{R}$).

An explicit example is the following.

Take $0<\epsilon<1$ and take an enumeration $E=\{q_n\}_n$ and then pose $\forall n$ $I_n:=(a_n,b_n)\ni q_n$ such that $b_n-a_n<\frac{\epsilon}{2^n}$. Now $U:=\cup_n I_n\supseteq E$ open, so $m(U)=\sup\{m(P) \mid P\subseteq U \text{ multi-interval}\}$.

The point is that each time I try to fill $U$ with a multi interval, I am forced to take the single intervals $<\frac{\epsilon}{2^n}$ for some $n$, meaning the series will be $<\epsilon$.

A final notable remark is: it might appear that $\cup I_n \supseteq [0,1]$, but it's NOT the case! This is intuitively because each $q_n$ has indeed a sphere around itself, but the radius of the sphere decreases dramaticallyfast, and there are not enough rationals to keep up with this decay.

Please leave an upvote (on the answer/question) if you found this insight useful!

I thank all the people who helped me, they all were really useful but this time I only feel satisfied by this reasoning. — rod, Aug 07 '20 at 13:07

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

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