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.

"Each real number is arbitrarily close to a rational number since ℚ
 is dense in ℝ
. Thus, each real number is in one of the open intervals."
Wrong. 
I'll construct an enumeration specifically to show that your cover does not contain every real number necessarily.
$\mathbb{Q}$ is the union of $A_n=\{x\in \mathbb{Q}: \epsilon2^{-n}<|x-\sqrt2|<\epsilon2^{-n+1}\}$
Now, enumerate each $A_n$ as $a_{n,1},a_{n,2},\ldots$. Then, enumerate each $\mathbb{Q}$ as $e_k=a_{m,n-m}$ regularly using each integer $k$ can be uniquely decomposed as $n(n+1)/2 +m$, where $0\le m \le n$. You can see that $\sqrt2$ is not in any of the interval you've constructed.
