What is wrong in this proof: That $\mathbb{R}$ has measure zero Consider $\mathbb{Q}$ which is countable, we may enumerate $\mathbb{Q}=\{q_1, q_2, \dots\}$. For each rational number $q_k$, cover it by an open interval $I_k$ centered at $q_k$ with radius $\epsilon/2^k$.
The total length of the intervals is a geometric progression that sums up to $\epsilon$.
Each real number is arbitrarily close to a rational number since $\mathbb{Q}$ is dense in $\mathbb{R}$. Thus, each real number is in one of the open intervals.
Thus the entire real line is covered by the union of the $I_k$, thus $\mathbb{R}$ is a null set with measure zero.
Clearly there is something wrong in the above proof, however I am not sure where is it?
Thanks for any help.

Heine-Borel theorem prevents such covering.
Actually the answer is ridiculously easy. Suppose you can cover irrational and rational numbers contained in an  interval $[0, 1]$ with a countable union of small intervals and sum of these intervals is less than $1$ then according to Heine-Borel theorem you can take a finite number of these intervals and should cover $[0, 1]$ so you get a contradiction as the sum of your intervals is less than $1$.
