Let $E$ be a subset of $\mathbb{R}$ with $m_*(E) > 0.$ There exists an open interval $I$ so that $m_*(E \cap I) \geq \alpha m_*(I).$ Let $E$ be a subset of $\mathbb{R}$ with $m_*(E) > 0.$ Prove that for each $0 < \alpha < 1,$ there exists an open interval $I$ so that $m_*(E \cap I) \geq \alpha m_*(I).$
Loosely speaking, this estimate shows that $E$ contains almost a whole interval.
[Hint: Choose an open set $O$ that contains $E$, and such that $m_*(E) \geq \alpha m_*(O)$.
 Write $O$ as the countable union of disjoint open intervals, and show that one of these intervals must satisfy the desired property.]
Attempt: Suppose $O = \cup_{i=1}^{\infty } I_i $ for all $i$. And let  Suppose by contradiction that the  conclusion is not true. 
then $m_*(E \cap I) < \alpha m_*(I).$ Then $m_*(E) = m_*(E \cap O) = m_*\left( \cup_{i=1}^{\infty } E \cap I_i\right) \leq \Sigma_{i = n}^{\infty} m_*(E \cap I_i) < \alpha \sum_{i = n}^{\infty} m_*(I_i)$ . So we have a contradiction.
Can someone please help me? I don't know if this makes sense.
Any feedback or better approach would really help.
Thank you!
 A: Yes, your solution is fine, but I would add a couple details especially if you're not fully convinced. 
Proof: Let $O \subseteq \mathbb{R}$ be an open set such that $E\subseteq O$ and $m_*(E) \geq \alpha m_*(O)$. Write $O$ as the countable union of disjoint open intervals $I_k$, i.e. $$O = \bigcup_{k=1}^\infty I_k$$ 
Thus $$ E = E\cap O = E \cap \left( \bigcup_{k=1}^\infty I_k \right) = \bigcup_{k=1}^\infty E \cap I_k $$
So by the countable subadditivity of the exterior measure we have $$ m_*(E) \leq \sum_{k=1}^\infty m_*(E\cap I_k)$$
Suppose toward a contradiction that $m_*(E\cap I_k) < \alpha m_*(I_k)$ for all $k$. Then 
$$ m_*(E) < \sum_{k=1}^\infty \alpha m_*(I_k) = \alpha\sum_{k=1}^\infty  m_*(I_k) = \alpha m_*(O) \leq m_*(E)$$ 
Note: The second equality holds since $I_k$ are disjoint and open. It is not true in general for the exterior measure that $m_*(A) + m_*(B) = m_*(A \cup B)$ even if $A$ and $B$ are disjoint. In this case however, since they're disjoint open intevals, they are disjoint measurable sets and hence additivity holds. 
Thus $m_*(E) < m_*(E)$ which is a contradiction. Hence there is at least one $I_k$ such that $m_*(E\cap I_k) \geq \alpha m_*(I_k)$.  
