When do a collection of rational balls with changing radius cover $\mathbb{R}$? Motivation: I was looking for a slick open cover of a set in $\mathbb{R}$ by open balls and ended up constructing the following. Let $\{q_i\}_{i\in I}$ be an enumeration of the rationals and let $r_i = \varepsilon/2^i.$ 
If $\bigcup_{i\in I}B_{r_i}(q_i)=\mathbb{R},$ then the Lebesgue measure says:
$$
\mu(\mathbb{R})\leq \sum_{i\in I} \mu(B_{r_i}(q_i))=\varepsilon.
$$
Well, $\mathbb{R}$ doesn't have arbitrarily small measure, so the collection of balls must not be a cover for $\mathbb{R}.$
Question: Is there a way to see that this does not cover $\mathbb{R}?$ Moreover, is there a criteria for $r_i$ so that one can see immediately whether $\{B_{r_i}(q_i)\}$ will cover?
Current Thoughts: I asked my friend this question this morning and he suggested that a necessary and sufficient condition might be that $\liminf r_i>0,$ but I only see why this should be sufficient. And in particular, I am still confused by my questions posed above.
Thank you for any and all help.
 A: Choose any enumeration of the rationals and let $B_i$ denote the interval with 'radius' $r_i$ around the $i$th rational $q_i$.
Then the $B_i$ cover $\Bbb R$ if and only if for every $r\in\mathbb R$, there exists an index $i$ such that the $i$th rational is at most $r_i$ away from $r$.
The problem with your first enumeration is that, even though there are plenty of rationals arbitrarily close to any real, most of these rationals (and for almost all reals all of the rationals) have an associated radius that is too small. Since the cover depends on the choice of the enumeration, there is no easy way to explicitly give a point that is not in your cover. So I think your proof using Lebesgue measure might be the easiest.
About the conjecture of your friend: It is wrong (i.e. $\liminf r_i=0$ is sufficient but not necessary). One could do what @TheoreticalEconomist pointed out in the comments: Set an infinite amount - say every second - of $r_i=0$ and use the rest to cover $\mathbb R$ (you can choose arbitrary radii for the other intervals).
Another, similar, example: Consider any enumeration of the reals such that $r_{i(j)}$ is the $j$-th harmonic number and set $r_{i(j)}=\frac1j$. Then the $\liminf r_i=0$ but The $B_{i(j)}$ alone cover the whole $\mathbb R_+$.
