Open cover of $\mathbb Q$ which is pairwise disjoint Using the fact that the Lebesgue measure is countably additive, it is trivial to prove that $\mathbb Q$ has measure zero. However, in order to gain an intuition for this fact, I thought it would be instructive to consider the following equivalent definition of null set:

A set $E\subset \mathbb R$ is a null set if for every $\varepsilon>0$, there is a countable collection $A$ of open intervals such that $E\subset\bigcup A$, and the sum of the lengths of those intervals is less than or equal to $\varepsilon$.

If we enumerate $\mathbb Q$ in some manner, then we can let
$$
A=\left\{\left(q_0-\frac{\varepsilon}{4},q_0+\frac{\varepsilon}{4}\right),\left(q_1-\frac{\varepsilon}{8},q_1+\frac{\varepsilon}{8}\right),\left(q_2-\frac{\varepsilon}{16},q_2+\frac{\varepsilon}{16}\right),\dots\right\} \, ,
$$
where $q_i$ denotes the $i$-th rational. Then, we clearly have $E\subset\bigcup A$, and the the sum of the lengths of the intervals is $\varepsilon/2+\varepsilon/4+\varepsilon/8+\dots=\varepsilon$. While this does make things more intuitive for me, I don't find this explanation totally satisfying because of the "overlap" between the intervals in $A$.
Question: Fix an $\varepsilon>0$. Is there an explicit example of set $A$ with the following properties:

*

*A is a countable collection of open intervals that covers $\mathbb Q$,

*The sum of the lengths of the intervals in $A$ is less than or equal to $\varepsilon$,

*$A$ is pairwise disjoint?

 A: This is really the same argument, just with a little care to make sure that the sets are disjoint. In particular, if the cover is of the form $U_{j}=(a_j,b_j),$ we need to have $a_j,b_j$ irrational, because no open set can contain $a_j$ or $b_j$ and be disjoint from $U_j.$
Enumerate the rationals $(r_i)_{i=1}^{\infty}.$
We will construct an increasing sequence $(i_j)_{j=1}^{\infty}$ of positive integers and a sequence of positive irrational numbers $(\delta_j)_{j=1}^{\infty}$ so that the intervals $U_j=(r_{i_j}-\delta_j,r_{i_j}+\delta_j)$ are disjoint and cover the rationals and $\sum 2\delta_j<\epsilon.$
Let $\alpha<\epsilon/2$ be an irrational number.
Define $i_1=1, \delta_1=\alpha/2.$ Let $U_1=(r_1-\delta_1,r_1+\delta_1).$
Given values for $j=1,\dots,n-1,$ find the smallest $i_{n}$ such that $r_{i_n}\notin \bigcup_{j=1}^{n-1} U_j.$
Define $$\delta_{n}=\min\left(\alpha/2^{n},\left|r_{i_{n}}-r_{i_j}\pm \delta_j\right|\mid j=1,\dots,n-1\right).$$
$\delta_n$ is irrational (and hence non-zero) and non-negative because it is the minimum of a finite set of non-negative values, all irrational.
$\delta_n$ is chosen to be no bigger than $\alpha/2^n,$ but also so that $U_{n}=(r_{i_n}-\delta_n,r_{i_n}+\delta_n)$ is disjoint from the previous $U_j.$
Then we get that the $U_j$ cover all rationals (why?) and are disjoint, and $\sum\mu(U_j)\leq \sum 2\delta_j<\epsilon.$

This works for rationals because we can pick $\delta_i$ irrational to make sure that no rational is on a boundary of one of the $U_j.$ It will be slightly harder in the general countable case. But not much harder.
Show the lemma:

If $X$ is a countable subset of $\mathbb R,$ $x_0\in\mathbb R$ and $\delta>0$ there exists a $0<\delta'\leq \delta$ such that $x_0-\delta',x_0+\delta'\not\in X.$

A: This can be done. A rough sketch: start with a set $A = \{A_i \mid i \in \mathbb{N}\}$ of intervals which have irrational endpoints, cover $\mathbb{Q}$, and have total length $< \epsilon$.
Enumerate the rationals with $q : \mathbb{N} \to \mathbb{Q}$. For each $n$, pick an interval $I_n$ with irrational endpoints as follows: Pick the least $j < n$ such that $q_n \in I_j$ and let $I_n = I_j$, if such a $j$ exists; otherwise, let $j$ be the least $j$ such that $q_n \in A_j$, and let $I_n$ be the connected component of the interior of $A_j \setminus \bigcup\limits_{j < n} I_j$ which contains $q_n$.
Then we see that $\{I_n \mid n \in \mathbb{N}\}$ is a countable set of pairwise disjoint intervals whose union is $\mathbb{Q}$ and whose total length does not exceed that of $A$, which is $< \epsilon$.
