Problem: True or False: Let $E \subset [0, 1] \subset \mathbb{R}$ be a countable subset. Then, for any $\epsilon> 0$, there is a finite cover of E by open intervals $\{I_k\}_{k=1}^{n}$ such that $$ \sum_{k=1}^{n} m(I_k) < \epsilon $$
This sounds like a quite easy problem, but I don't how to solve it. This one is quite similar to the one on Folland: Let $E \subset \mathbb{R}$ be Lebesgue measurable set and assume that there exists $0 < \alpha < 1$ such that $m(E \cap I) \leq \alpha m(I)$ for all open intervals $I$. Then, $m(E) = 0$.
Proof: If $m(E) > 0$, then let $O$ be an open set that contain $U$ and $O = \bigcup_{i=1}^{\infty} I_{i}$, where $I_i$ is an open interval. Then we have $$ m(O) = \sum_{i=1}^{\infty}m(I_i) \geq \frac{1}{\alpha} \sum_{i=1}^{\infty}m(E \cap I_i) \geq \frac{1}{\alpha} m(E) $$ By regularity, we can always make $O$ such that $ m(E)\leq m(O) < \frac{1}{\alpha}m(E)$.
Can any one help me with that?
False. Take $E$ to be dense in $[0, 1]$. If a finite cover by open intervals $\{I_k\}^n_{k = 1}$ with total length $L < 1$ exists, then the complement $[0, 1] \setminus \bigcup_{k = 1}^n I_k$ contains an open interval, which by denseness contains a point of $E$, which is absurd.