outer Jordan content

hbghlyj

Real Analysis (Princeton Lectures in Analysis, Volume 3) Elias M. Stein, Rami Shakarchi, Chapter 1 Exercise 14, page 41

The purpose of this exercise is to show that covering by a finite number of intervals will not suffice in the definition of the outer measure $m_*$.
The outer Jordan content $J_*(E)$ of a set $E$ in $\mathbb{R}$ is defined by
$$
J_*(E)=\inf \sum_{j=1}^N\left|I_j\right|,
$$
where the inf is taken over every finite covering $E \subset \bigcup_{j=1}^N I_j$, by intervals $I_j$.
(a) Prove that $J_*(E)=J_*(\bar{E})$ for every set $E$ (here $\bar{E}$ denotes the closure of $E)$
(b) Exhibit a countable subset $E \subset[0,1]$ such that $J_*(E)=1$ while $m_*(E)=0$.

hbghlyj

First direction $(\leq)$ :
Let $E \subset \mathbb{R}$. Since $E \subset \bar{E}$, if $\varphi=\left\{I_j\right\}_{j=1}^N$ covers $\bar{E}$, then $\varphi$ covers $E$ as well. Thus, $J_*(E) \leq J_*(\bar{E})$.
Second direction $(\geq)$ :
To see $J_*(E) \geq J_*(\bar{E})$, it will suffice to show that if $\varphi=\bigcup_{j=1}^N I_j$ covers $E$, then $\bar{\varphi}=\bigcup_{j=1}^N \bar{I}_j$ covers $\bar{E}$. This is sufficient because, by the definition of $\left|I_j\right|$ for $I_j \subset\mathbb R,\left|I_j\right|=\left|\bar{I}_j\right|$.

Let $\varphi$ cover $E$. Suppose $\bar{\varphi}$ does not cover $\bar{E}$, then $\exists x \in \bar{E}$, such that $x \notin \bar{\varphi}$. By definition, $\bar{E}=E \cup E^{\prime}$ where $E^{\prime}$ is the set of all limit points of $E$. Thus $\forall \varepsilon>0,(x-\varepsilon, x+\varepsilon) \cap E \neq \emptyset$, but $E \subset \bar{\varphi}$ which implies $(x-\varepsilon, x+\varepsilon) \cap \bar{\varphi} \neq \emptyset$. Therefore $x$ is a limit point of $\bar{\varphi}$.

However, since $\bar{\varphi}$ is the finite union of closed intervals, $\bar{\varphi}$ is closed and must contain all of its limit points, so $x \in \bar{\varphi}$, which is a contradiction. Therefore, if $E \subset \varphi$, then $\bar{E} \subset \bar{\varphi}$, which implies $J_*(E) \geq J_*(\bar{E})$.
Thus $J_*(E)=J_*(\bar{E})$.

hbghlyj

Let $E=\mathbb{Q} \cap[0,1]$. Since $E$ is dense in $[0,1]$, $J_*(E)=J_*(\bar{E})=J_*([0,1])=1$. However for each rational $q_j$, $m_*\left(\left\{q_j\right\}\right)=0$ means $\sum_{j=1}^{\infty} m_*\left(\left\{q_j\right\}\right)=0$.

Moreover, there exists an infinite sequence of sets $E_n=(\mathbb{Q} \cap[0,1]) \backslash S_n$ where $S_n$ is countably infinite, such that $J_*\left(E_n\right)=1$ while $m_*\left(E_n\right)=0$. Let $S_n=\left\{\frac{n}{p}: p=n \cdots \infty\right\}$. Between any two rationals $\frac{n}{p}, \frac{n}{p+1} \in E_n$, there is a subset of rationals, $Q_p=E_n \cap\left(\frac{n}{p}, \frac{n}{p+1}\right)$, that is dense in the interval $\left[\frac{n}{p}, \frac{n}{p+1}\right]$ which means $E_n$ is dense in $[0,1]$. Hence $J_*\left(E_n\right)=1$ and $m_*\left(E_n\right)=0$.

Michelle Hummel, Matthew Dahlgren, Mahshid Rahnamay-Naeini, Martha Byrne, Ming Gong. September 21, 2010