How to prove this set as a null set? Suppose $E \subset \mathbb{R}^1$, and $\exists q: 0<q<1 $, s.t. for every interval $(a, b)$, there is a sequence of open intervals $\{I_n, n \ge 1 \}$:
$E \cap (a, b) = \cup_{n=1}^{\infty} I_n, \sum_{n=1}^{\infty} |I_n|< (b - a) q.$
Prove $m^*(E) = 0.$ (Here $m^*$ means the outer Lebesgue measure.)
In this problem, I tried to prove it by proving $E$ as a uncountable set, but Cantor Set is a counter-example. Then I proved that E has no interval, which means there is no interior points in $E$. However, this cannot help me to prove the initial problem, since the irrational set between $(0,1)$ is another counter-example.
I actually don't know so many ways to prove a set as a null set, so I' m confused.
 A: I'll outline for you how to prove that $E\cap (a,b)$ has outer measure zero for any $(a,b)$. By countable sub-additivity this automatically implies that $E$ has outer measure zero.
The trick is by induction. I claim that for each $k> 0$ there exists a countable collection of open intervals $\{I^{k}_n\}_{n\in \mathbb{N}}$ such that $E \cap (a,b) \subset \cup_n I^k_n$ and that $\sum_{n} |I^k_n| < (b-a) q^k$.
If the claim is true, then taking $k \to +\infty$ gives you that $m^*(E\cap (a,b))$ has measure zero.
To prove the claim: Suppose $I_n^{k-1}$ has been constructed. For each $n$, apply the hypothesis to find a sequence of intervals $J_m^n$ such that
$$ E \cap I_n^{k-1} = \cup_m J_m^n, \qquad \sum |J_m^n| \leq |I_n^{k-1}| q $$
Since union of countable sets is countable, we have that the set $\{ J^n_m : n,m\in \mathbb{N}\}$ is a countable collection of intervals.
Since
$$ E\cap (a,b) \subseteq E \cap \cup I_n^{k-1} = \cup (E\cap I_n^{k-1}) \subseteq \cup_{n} \cup_m J_m^n $$
we have that $\{ J_m^n\}$ covers $E\cap (a, b)$. Finally, we have that
$$ \sum_{n,m} |J_m^n| \leq \sum_{n} |I_n^{k-1}| q \leq q^{k-1} \cdot q = q^k $$
So you can just relabel/enumerate the elements of the countable set $\{J_m^n\}$ as $\{I^k_n\}$.
