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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!

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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 $m_*(E\cap I_k) \geq \alpha m_*(I_k)$.

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why exists $O$ such that $m(E)\geq \alpha m(O)$?

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  • $\begingroup$ He left out some details, but he's using the definition of the lebesgue measure to choose $O$ such that $E$ is contained in $O$ and $m(O) \leq m(E) + \epsilon$ $\endgroup$ – Dionel Jaime Apr 8 at 5:22

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