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

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 $1$ Commented Aug 7, 2020 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
Commented Aug 7, 2020 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 Commented Aug 7, 2020 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 $[0,1]$. Commented Aug 7, 2020 at 11:39
• 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. Commented Aug 7, 2020 at 12:30

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
Commented Aug 7, 2020 at 13:07