Showing that rationals have Lebesgue measure zero. I have been looking at examples showing that the set of all rationals have Lebesgue measure zero. In examples, they always cover the rationals using an infinite number of open intervals, then compute the infinite sum of all their lengths as a sum of a geometric series. For example, see this proof.
However, I was wondering if I could simply define an interval $(q_n - \epsilon, q_n + \epsilon )$ around each rational number $q_n$. Since there is a countable number of such intervals, the Lebesgue measure must be bound above by a countable sum of their Lebesgue measure (subadditivity). i.e. $\mu(\mathbb{Q}) \leq \mu(\bigcup^{\infty}_{i=1} (q_n - \epsilon, q_n + \epsilon)) = \sum^{\infty}_{i=1} \mu((q_n - \epsilon, q_n + \epsilon))$.
Then, argue that each individual term is $\mu((q_n - \epsilon, q_n + \epsilon)) < 2\epsilon$ and is thus zero since the epsilon is arbitrary?

Define an interval $(q_n−\frac{ϵ}{2^n},q_n+\frac{ϵ}{2^n})$
around each rational number $q$
For $\epsilon>0$, $\mu((q_n−\frac{ϵ}{2^n},q_n+\frac{ϵ}{2^n}))=2\left(\frac{\epsilon}{2^n}\right)$
and $\sum^{\infty} _{n=1} 2\frac{ϵ}{2^n}=2ϵ$
Since $ϵ$ is arbitrary, so  $μ(Q) = 0$.
I think this simple trick could work
