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

Consider the Lebesgue outer measure $$ \bar{m}(X) = \inf_{A \supset X}\bigg\{\sup_{P\subset A}\quad m(P)\bigg\} $$ where $X = [0,1]\cap \mathbb{Q}$ and $P = \bigcup [a_i,b_i]$ is a suitable union of intervals. My question is: suppose that $\bar{m}(X)=0$: can you exhibit one of those $A$'s? Thanks

Any cover of the rationals would be a collection of open sets containing all rationals in [0,1]. A specific example would be: take any enumeration {$q_1,q_2,..,q_n,...$} of all rationals in $\mathbb Q \cap [0,1]$, and the use the open sets $O_n:=(q_n-\frac{1}{2^n},q_n+\frac {1}{2^n})$

To determine the outer measure, you may want to scale each interval by a fixed $\epsilon>0$ and then add the widths of all the intervals (see what happens when you let $\epsilon>0$ become small).

Since $[0,1]$ is Borel and $\mathbb Q$ is Borel, so is their intersection.

Therefore these sets are in fact Lebesgue measurable, so the outer Lebesgue measure is equal to the Lebesgue measure.

We have, if so $\overline m(X)=0$, therefore the intersection is of shrinking intervals.

Consider $\mathbb Q=\{q_n\mid n\in\mathbb N\}$ an enumeration of the rationals in $X$ and $\epsilon>0$, let $[a_i,b_i]$ be an interval around $q_i$ such that $b_i-a_i<\frac{\epsilon}{2^i}$, and let $A=\bigcup [a_i,b_i]$.