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?


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 $< \epsilon$.
