This is exercise number 28 from chapter 1 of Stein & Shakarchi's Princeton Lectures in Analysis III: Real Analysis.

Let $E$ be a subset of $\mathbb{R}$ with $m_*(E) > 0$, where $m_*(E)$ denotes the outer (or sometimes reffered to as the exterior) measure of $E$.

Prove that for each $\alpha \in (0, 1) \subset \mathbb{R}$, there exists an open interval $I$ such that

$$m_*(E \cap I) \geq \alpha \cdot m_*(I) $$

Can someone verify my proof of the above statement?

I have already read Let $E$ be a subset of $\mathbb{R}$ with $m_*(E) > 0.$ There exists an open interval $I$ so that $m_*(E \cap I) \geq \alpha m_*(I).$ and the poster of that question does not prove existence of an open set $O$ that contains $E$ such that $m_*(E) \geq \alpha m_*(O)$.


First we state a lemma from page 13 of the stated textbook.

Lemma: If $E \subset \mathbb{R}^d$, then $m_*(E) = \text{inf} \space m_*(O)$, where the infimum is taken over all open sets $O$ containing $E$.

Let $\alpha \in (0, 1)$ be arbitrary. Choose $\epsilon = \dfrac{m_*(E)(1 - \alpha)}{\alpha}$. Certainly $\epsilon$ is a positive quantity since $m_*(E) > 0$ by assumption.

By the lemma, we can choose an open set $O$ with $E \subset O$ and $$ \begin{align*} m_*(O) &\leq m_*(E) + \epsilon \\ &= m_*(E) + \frac{m_*(E)(1 - \alpha)}{\alpha} \\ &=\frac{\alpha m_*(E) + m_*(E) - \alpha m_*(E)}{\alpha} \\ &= \frac{m_*(E)}{\alpha} \end{align*} $$

Thus, we have that $m_*(O) \leq \dfrac{m_*(E)}{\alpha}$ which is equivalent to $\alpha m_*(O) \leq m_*(E)$.

Now, since $O$ is an open subset of $\mathbb{R}$, Theorem 1.3 says that $O$ can be written uniquely as a countable union of disjoint open intervals.

Let $\bigcup_{j = 1}^{\infty} I_j = O$ denote the disjoint union where each $I_j$ is an open interval.

For the sake of contradiction, suppose for each $j$ that $m_*(E \cap I_j) < \alpha m_*(I_j)$.

Consider: $$ \begin{align*} &m_*(E) \\ \\ = \space &m_*(E \cap O) \\ = \space &m_*(E \cap (\bigcup_{j = 1}^{\infty}I_j)) \\ = \space &m_*(\bigcup_{j = 1}^{\infty}(E \cap I_j)) \space \text{since intersections distribute over unions} \\ \leq \space &\sum_{j = 1}^{\infty} m_*(E \cap I_j) \\ < \space &\alpha \sum_{j = 1}^{\infty} m_*(I_j) \space \text{by the contradiction assumption} \\ = \space &\alpha m_*(\bigcup_{j = 1}^{\infty}I_j) \space \text{since each} \space I_j \space \text{is disjoint and measurable} \\ = \space &\alpha m_*(O) \end{align*} $$

Thus, we have that $m_*(E) < \alpha m_*(O)$ which is a contradiction. Hence, one of the intervals $I_j$ has the desired property.

Are all of these steps legitimate? I have just started teaching measure theory to myself and it would be great if someone could verify this proof.

  • $\begingroup$ Looks good to me. And I just worked this problem the other day in almost this same fashion. $\endgroup$ – Walton Jun 18 '18 at 20:34

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