The outer Jordan content $J_{*}(E)$ of a set $E$ in $\mathbb{R}$ is defined by $$ J_{*}(E)=\inf \sum^{n}_{j=1}|I_j| $$ where the $\inf$ is taken over every finite covering $E\subset \underset{j\in \mathbb{N}_n}{\bigcup}I_j, $ by closed intervals $I_j$.
($b$) Exhibit a countable subset $E\subset [0,1]$ so that $J_{*}(E)=1$ where $m_{*}(E)=0.$
{Proposition}($14.b.1$): Let $Q'=(0,1)\cap \mathbb{Q}$. Let $\mathcal{X}=\{I_j\}^{N}_{1} $ of closed intervals be a cover of $Q'$. Then $(0,1)$ is covered by $\mathcal{X}.$
Proof.Let the assumptions be as above. Let $x\in Q'\subset (0,1)$. Then $\exists \delta >0$ such that $B_\delta(x)\subset (0,1)$. We will prove $\exists \varepsilon >0$ such that $B_\varepsilon(x)$ is covered by $\mathcal{X}$. For every $k\in \mathbb{N}$, the interval $(x-\frac{\delta}{k})\subset B_{\frac{\delta}{k}}(x)$. We will denote these intervals by $A_k$. Since $\mathbb{Q}$ is dense in the reals, $A_k\cap \mathbb{Q}$ is non empty for every $k$. Pick a point in this intersection and denote it as $y_k\in Q'.$ Therefore, we have a sequence $\{y_k\}_k\subset Q'$ which converges to $x$ since for each $k$ we have $$ d_{A_k}(x,y_k)=|x-y_k|< \frac{\delta}{k} $$ Therefore, each point in the sequence must be contained by elements of $\mathcal{X}$. But there are only a finite number of elements in $\mathcal{X}$. Hence, there exists at least one element, say $I\in \mathcal{X}$, which contains a subsequence of $\{y_k\}_k$. This must be since if each interval contained only a finite number of points in the sequence, then not all of the sequence would be covered.
Hence, $\big\{y_{k_m}\big\}_{k_m}\subset I$. Since $I$ is closed, it must contain the limit point $x$. We can repeat the same procedure for the intervals $(x+\frac{\delta}{k})$ and sequence $\{z_k\}_k \subset Q'$ associated with it. Therefore, there $\exists I'\in \mathcal{X}$ which also contains $x$. Thus, $\exists k'\in \mathbb{N}$ such that $B_{\frac{\delta}{k'}}(x)\subset I\cup I'$. Which must be true since for some $k'$, we'd have $[y_k',x]\subset I$ and $[x,z_k']\subset I' \Rightarrow [y_k',z_k']\subset I\cup I'.$ (It is by definition of an interval that all those points in between are also contained). Set $\varepsilon = \frac{\delta}{k'}.$ Then $B_{\varepsilon}(x)$ is covered by $\mathcal{X}$.
{Claim}($i$): Denote $\underset{x\in Q'}{\bigcup}B_{\varepsilon_x}(x)$ by $\mathcal{Q}$. We claim $\mathcal{Q}=(0,1).$ (Where $\varepsilon_x$ is the required radius so that $\mathcal{X}$ covers the associated ball)
Proof. It is clear by construction that $\mathcal{Q}\subset (0,1)$. So let $x\in (0,1).$ Then $\exists \delta>0$ such that $B_\delta(x)\subset (0,1)$. Therefore, $B_{\frac{\delta}{k}}(x)$ is definitely contained in $(0,1)$, for any $k>2$. By the density of $\mathbb{Q}, \exists$ $y\in B_{\frac{\delta}{k}}(x)$ such that $y\in Q'$. let $p\in B_{\frac{\delta}{k}}(y)$. Then we have $$ d_\delta(x,p)\leq d(x,y)+d(y,p) $$ $$<\delta/k + \delta/k<\delta $$ Hence, $B_{\frac{\delta}{k}}(y)\subset B_\delta(x)$. Thus, we have $x\in B_{\frac{\delta}{k}}(y)$. We can then choose $k$ such that $\frac{\delta}{k}\leq \varepsilon_ y$. Which would yield $B_{\frac{\delta}{k}}(y)\subset \mathcal{Q}$, exactly as desired.
Now, Since $\mathcal{Q}=(0,1)$ by {Claim}($i$), it follows that $\mathcal{X}$ covers $(0,1)$ and our proof is complete.
{Proposition}($14.b.2$): Let $Q'=(0,1)\cap \mathbb{Q}$. Then $J_{*}(Q')=1.$
Proof. Let $\mathcal{X}=\{I_j\}^{N}_{1}$ be a cover of $Q'$. By {Proposition}($14.b.1$), $ \mathcal{X}$ must cover $(0,1)$. Therefore, $$ |(0,1)|=1\leq \sum^{N}_{j=1}|I_j| $$ $$\Rightarrow 1\leq J_{*}(Q') $$ Now we will prove the reverse inequality. $I=[0,1]$ covers $Q'$. Therefore, $$ J_{*}(Q')\leq |I|=1 $$ Thus, we have that $J_{*}(Q')=1$. This completes our proof.
{Proposition}($14.b.3$): $m_{*}(Q')=0.$
Proof. Let $x\in Q'$ and $I_{x,n}=[x-1/n, x+1/n]$. Fix $\frac{\varepsilon}{2N}>0$. Let $S_{x,n}$ be an open interval such that $$ |S_{x,n}|\leq (1+\frac{\varepsilon}{2N})|I{x,n}| $$ Let $\mathcal{X}=\{I_{x,n}\}_{x\in Q'}$ and $\mathcal{Y}=\{S_{x,n}\}_{x\in Q'}$. These are countable collections since $Q'$ is countable. Both of these collection cover $Q'$. Since $Q'\subset [0,1]$, compactness implies we only need a finite subcover of $\mathcal{Y}.$ Let $$ \mathcal{Y}'=\{S_{x_j,n}\}_{j\in \mathbb{N}_{N}} $$ Then we have that $$ m_{*}(Q')\leq \underset{j\in \mathbb{N}_N}{\sum}|S_{x_j,n}| $$ $$ \leq \underset{j\in \mathbb{N}_N}{\sum}(1+\frac{\varepsilon}{2N})|I_{x_j,n}| $$ $$ \leq \underset{j\in \mathbb{N}_N}{\sum}|I_{x_j,n}|+\frac{\varepsilon}{2N}\underset{j\in \mathbb{N}_N}{\sum}|I_{x_j,n} $$ $$=\frac{2N}{n}+\frac{\varepsilon}{n} < \frac{2N}{n} + \varepsilon $$ As $\varepsilon$ is arbitrary, we have that $$ m_{*}(Q')\leq \frac{2N}{n} $$ Letting $n\rightarrow \infty$, we have that $m_{*}(Q')=0$ as desired.
I just wanted to check if my proof is correct. Thank you for any feedback.