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.
Best Answer
Here is a slight variation on your construction.
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 $q_k$ which does not contain $\pi.$
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. In particular, $\pi$ is in one of the open intervals.
Now, how can $\pi$ be in one of the open intervals, when each interval was specifically chosen to exclude $\pi$?