Set of rational numbers $\mathbb{Q}$ is measure $0$.
I approach this question by two sides.
(First)
Like here Showing that rationals have Lebesgue measure zero., $$
\mu(\mathbb Q) = \mu\left(\bigcup_{n=1}^\infty \{q_n\}\right) = \sum_{n=1}^\infty \mu(\{q_n\}) = \sum_{n=1}^\infty 0 = 0.
$$
(Second)
Using the definition of Lebuesgue measure. let's order $\mathbb{Q}=\bigcup_{i=1}^{\infty}\left\{ r_{i}\right\} $. For given
$\epsilon>0$, cover $\mathbb{Q}$ by $\bigcup_{i=1}^{\infty}\left\{ r_{i}-\frac{\epsilon}{2^{i+1}},\ r_{i}+\frac{\epsilon}{2^{i+1}}\right\} $(Let this covers $E_{\epsilon}$).
Then $\mu\left(\mathbb{Q}\right)\leq\sum_{i=1}^{\infty}\epsilon/2^{i}\leq\epsilon$.
So $\mu\left(\mathbb{Q}\right)=0$.
Here, I have a question on the second proof.
If we cover $\mathbb{Q}$ like the second proof, I think the covers $E_{\epsilon}$
cover all real numbers $\mathbb{R}$.
The reason is here : If we let $a\in\mathbb{R}\setminus\mathbb{Q}$,
for any $\delta>0$, there exists $b\in\mathbb{Q}$ such that $|a-b|<\delta$ (by
the density of $\mathbb{Q}$). So $a$ must be in some intervers in
the covers $E_{\epsilon}$. So $\mu\left(E_\epsilon\right)=\infty$ $\forall\epsilon>0$.
What is my fault in my reason?
Best Answer
The fault in your reasoning is where you claim, without justification, "So $a$ must be in some . . ." Unless you come up with some reason for that "so", there is nothing more to be explained here.
Your conclusion is wrong, those intervals do not cover $\mathbb R.$ That is clear from measure theory, as you know. In order to give an explicit example of a real number which is not covered, we have to know how the rationals are enumerated. Since you did not assume anything special about the enumeration, I can use any enumeration I like. Let me define a special enumeration which makes it easy to exhibit an uncovered number. (Also I will take $\epsilon=1.$)
First, let $q_1,q_2,q_3,\dots$ be your favorite enumeration of the rationals. Now I define a new enumeration $\{r_i\}$ recursively. If $r_1,\dots,r_{i-1}$ have already been defined, then I define $r_i$ to be the first $q_j$ such that $q_j\notin\{r_1,\dots,r_{i-1}\}$ and $|q_j-\sqrt2|\gt\frac1{2^{i+1}}.$
Does the sequence $r_1,r_2,r_3,\dots$ contain all the rationals?
Since your sequence $q_1,q_2,q_3,\dots$ contains all the rationals, it will be enough to show that each $q_j$ occurs in the sequence $r_1,r_2,r_3,\dots.\ $ I will prove this by induction on $j.$ Suppose that each of the numbers $q_1,q_2,\dots,q_{j-1}$ occurs in the sequence $\{r_i\}.$ Since $q_j$ is rational while $\sqrt2$ is irrational, we know that $|q_j-\sqrt2|\gt0.$ Thus (with $j$ fixed) if we choose $i$ large enough, we will have both $\{q_1,\dots,q_{j-1}\}\subseteq\{r_1,\dots,r_{i-1}\}$ and $\frac1{2^{i+1}}\lt|q_j-\sqrt2|.$ For such an $i$ we will certainly have $q_j\in\{r_1,\dots,r_{i-1},r_i\}.$
Do the intervals $(r_i-\frac1{2^{i+1}},\ r_i+\frac1{2^{i+1}})$ cover $\mathbb R$?
No, none of those intervals covers $\sqrt2,$ because $|r_i-\sqrt2|\gt\frac1{2^{i+1}}.$