Exercise 1.1.18 (Topological formulation of Jordan measurability)

We prove the above properties using the definition of Jordan measurability directly. The fundamental observation here is that the assertion is already true for the elementary sets by definition, i.e., for any box (and therefore elementary set) $B$ we have $m(B)=m(\overline{B})$.

(i)

To show that both $E$ and $\overline{E}$ have the same outer Jordan mesure, we have to demonstrate that $$\inf <!--\; FUNCTION\; APPLICATION\; -->\left\{m(B):\begin{array}{c}E\subset B\hfill \\ B\text{ is elementary}\hfill \end{array}\right\}=\inf <!--\; FUNCTION\; APPLICATION\; -->\left\{m(B^{\prime }):\begin{array}{c}\overline{E}\subset B^{\prime }\hfill \\ B^{\prime }\text{ is elementary}\hfill \end{array}\right\}$$

Notice that to prove the above assertion it suffices to show that the set of outer closed covers of both $E$ and $\overline{E}$ coincide (we only look at the closed covers, since they, by definition, have the same elementary measure $m(\overline{B})=m(B)$ as their non-closed versions anyway)

$$\left\{B\subset {ℝ}^d:\begin{array}{c}E\subset B\hfill \\ B\text{ is elementary and closed}\hfill \end{array}\right\}=\left\{B^{\prime }\subset {ℝ}^d:\begin{array}{c}\overline{E}\subset B^{\prime }\hfill \\ B^{\prime }\text{ is elementary and closed}\hfill \end{array}\right\}$$

• $\supseteq$ Let $B^{\prime }$ be a closed elementary outer cover of $\overline{E}$. Since $E\subseteq \overline{E}$ we naturally obtain $E\subseteq B^{\prime }$, $B^{\prime }$ elementary.
• $\subseteq$ Let $B$ be a closed elementary outer cover of $E$. Since $\overline{E}=E\cup \mathit{\partial E}$, it suffices to demonstrate that $B$ covers $\mathit{\partial E}$. Thus, let $x\in \mathit{\partial E}$ be arbitrary. Lets look at the distance $d(x,B)=\inf <!--\; FUNCTION\; APPLICATION\; -->\left\{d(x,y):y\in B\right\}$ between $x$ and the outer cover $B$. Obviously, there are only two possible cases: $d(x,B)=𝜖>0$ and $d(x,B)=0$. In the first case, we can build a ball $B(x,𝜖∕2)$, which, by definition of $x$ must satisfy $B(x,𝜖∕2)\cap E\ne \varnothing$, and by definition of $𝜖$ must satisfy $B(x,𝜖∕2)\cap B=\varnothing$. But these both conditions create a contradiction since we then have $x\in E$ and not $x\in B$. Thus, this case is impossible. The only left case is that $d(x,B)=0$. In this case, $x$ is an adherent point of $B$. Since $B$ is closed, $x\in B$, as desired.

To sum up, let $B$ be any cover of $E$. Then $\overline{B}$ covers both $E$ and $\overline{E}$, thus $m(B)=m(\overline{B})$ is contained in the Jordan outer covers set of both $E$ and $\overline{E}$. The vice versa is also true; thus, the infimums coincide.

(ii)

Here we have to demonstrate that $$\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{m(A):\begin{array}{c}E\supset A\hfill \\ A\text{ is elementary}\hfill \end{array}\right\}=\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{m(A^{\prime }):\begin{array}{c}\overline{E}\supset A^{\prime }\hfill \\ A^{\prime }\text{ is elementary}\hfill \end{array}\right\}$$

Using the same strategy as before, we consider the set of open elementary inner covers of both $E$ and $E^{\text{∘}}$.

$$\left\{A\subset {ℝ}^d:\begin{array}{c}{E}^{\text{∘}}\supset A\hfill \\ A\text{ is elementary and open}\hfill \end{array}\right\}=\left\{A^{\prime }\subset {ℝ}^d:\begin{array}{c}E\supset A^{\prime }\hfill \\ A^{\prime }\text{ is elementary and open}\hfill \end{array}\right\}$$

• $\supseteq$ Let $A^{\prime }$ be an open elementary inner cover of $E$. We can represent $E=E^{\text{∘}}\cup (E\cap \mathit{\partial E})$. If we manage to demonstrate that $A^{\prime }\subset E^{\text{∘}}$ by showing that $A^{\prime }\cap \mathit{\partial E}=\varnothing$ then we are done. Suppose for the sake of contradiction that we can find a $x\in A^{\prime }\cap \mathit{\partial E}$. On one hand, since $x\in A^{\prime }$ it must be an inner point of $A^{\prime }$ (i.e., $B(x,𝜖)\subseteq A^{\prime }$ for some $𝜖>0$). On the other hand, since $x\in \mathit{\partial E}$ we must have $B(x,𝜖)\cap E^c\ne \varnothing$ - a contradiction. Hence, $A^{\prime }\subseteq E^{\text{∘}}$.
• $\subseteq$ Let $A$ be an open elementary inner cover of $E^{\text{∘}}$. Since $E^{\text{∘}}\subseteq E$, $A$ must be an open elementary inner cover of $E$ as well.

(iii)

We demonstrate the both directions of equivalence.

• $⟹$ Suppose that $E$ is Jordan measurable. Note that by part (1) and (2) we have $m(E)=m_{\text{∗}}(E^{\text{∘}})=m^{\text{∗}}(\overline{E})$. Thus, for any $𝜖>0$ we can find an elementary outer cover $B$ of $\overline{E}$ and an elementary inner cover $A$ of $E^{\text{∘}}$ such that $m(B∕A)\le 𝜖$. But $\mathit{\partial E}=\overline{E}∕E^{\text{∘}}\subset B∕A$; hence, $B∕A$ is an elementary cover of $\mathit{\partial E}$, and we have $m^{\text{∗}}(\mathit{\partial E})\le 𝜖$ for any $𝜖>0$.

• $⟸$ Suppose that $m^{\text{∗}}(\mathit{\partial E})=0$. Notice how we can reformulate the theorem assertion. We have to prove that

  $$E\text{ is Jordan measurable}$$

  By Exercise 1.1.14 this assertion is equivalent to

  $$\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{2}^{-\mathit{dn}}\cdot \left(E^{\text{∗}}\left(E,2^{-n}\right)-E_{\ast }\left(E,2^{-n}\right)\right)=0$$

  A key observation at this point is that $E^{\text{∗}}\left(E,2^{-n}\right)-E_{\text{∗}}\left(E,2^{-n}\right)$ is the number of the sets that intersect $E$ minus the number of sets that are contained in $E$. But this is the number of sets that intersect $E$ and are not contained in $E$ - in other words, the sets that outer cover the boundary of $E$ (why?). Thus, the theorem asks us to prove that

  $$\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{2}^{-\mathit{dn}}{E}^{\text{∗}}\left(\mathit{\partial E},2^{-n}\right)=0$$

  But $\mathit{\partial E}$ is a Jordan measuarble set, since $m^{\text{∗}}(\mathit{\partial E})=0$ and $m_{\text{∗}}(\mathit{\partial E})\le m^{\text{∗}}(\mathit{\partial E})=0⟹m_{\text{∗}}(\mathit{\partial E})=0$. Thus, the above limit is indeed zero in our case, and the theorem assertion must be true.

(iv)

Let $A$ be an arbitrary elemetary inner cover of ${[0,1]}^2∕{\mathbb{Q}}^2$. Suppose that $A$ is non-empty, i.e., it contains a box $B=[a_1,b_1]\times [a_2,b_2]$. Between $0\le a_1<b_1\le 1$ we can find a rational number $q_1$ and similarly $q_2:a_2\le q_2\le b_2$ (why?). Then $(q_1,q_2)\in B$ - a contradiction. Thus, $A$ is empty, and the inner cover of ${[0,1]}^2∕{\mathbb{Q}}^2$ must be zero. Now let $B$ be an elementary outer cover of ${[0,1]}^2∕{\mathbb{Q}}^2$. The closure of this set is ${[0,1]}^2$ by the topology of real line. By part (1) the outer measure is then $m({[0,1]}^2∕{\mathbb{Q}}^2)=m({[0,1]}^2)=1$.
A similar argument applies to the set of bullets.

First, $A\subset Ā$ so $J^{\text{∗}}(A)\le J^{\text{∗}}(A)$, and we only need to prove $J^{\text{∗}}(A)\le J^{\text{∗}}(A).$ Let $𝜀>0.$ Then choose disjoint boxes ${\left(B_n\right)}_{n=1}^N$ such that $A\subset {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_n{B}_n$ and ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n}m\left(B_n\right)\le J^{\text{∗}}(A)+𝜀$. Note that

and since ${\overline{B}}_n$ are also boxes,

Taking $𝜀\to 0$ completes the proof of the first item. For the second item, since $A^{\text{∘}}\subset A$ we only need to show that $J_{\text{∗}}(A)\le J_{\text{∗}}\left(A^{\text{∘}}\right).\mathrm{Fix}$ $\xi >0$ and let ${\left(B_n\right)}_{n=1}^N$ be a finite number of disjoint boxes such that ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_n{B}_n\subset A$ and ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n}m\left(B_n\right)\ge J_{\text{+}}(A)-𝜀.$ Then, since $B_n^{\infty }$ are disjoint boxes, and since $m\left(B_n^{\text{∘}}\right)=m\left(B_n\right)$, from ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_n{B}_n^{\text{∘}}\subset A^{\text{∘}}$ we deduce that

Taking $𝜀\to 0$ completes the second item. For the final item: First, note that ${(\mathit{\partial A})}^{\text{∘}}=\varnothing$. If $D_n(z)\subset \mathit{\partial A}$ for some dyadic cube $D_n(z)$, then ${(\mathit{\partial A})}^{\text{∘}}\supset {\left(D_n(z)\right)}^{\text{∘}}\ne \varnothing$, a contradiction. So $\mathit{\partial A}$ contains no dyadic cubes, which is to say that $E_{\text{∗}}(\mathit{\partial A},n)=0$ for all $n$. Thes, $\mathit{\partial A}$ is Jordan measurable with $J(\mathit{\partial A})=0$ if and only if $2^{-\mathit{dn}}{E}^{\text{∗}}(\mathit{\partial A},n)\to 0.$

Note that if $D_n(z)$ is a dyadic cube, then it intersects $A$ and is not contained in $A$, if and only if it intersects $\mathit{\partial A}$. That is, $E^{\text{∗}}(\mathit{\partial A},n)=E^{\text{∗}}(A,n)-E_{\text{∗}}(A,n)$.

Now, by Exercise 1.1.14, $A$ is Jordan measurable, if and only if $2^{-\mathit{dn}}(E^{\text{∗}}(A,n)-$ $E_{\text{∗}}(A,n))\to 0$ if and only if $2^{-\mathit{dn}}{E}^{\text{∗}}(\mathit{\partial A},n)\to 0$, if and only if $\mathit{\partial A}$ is Jordan measurable with $J(\mathit{\partial A})=0$, if and only if $J^{\text{∗}}(\mathit{\partial A})=0$.

The last step follows from the fact that if $J^{\text{∗}}(\mathit{\partial A})=0$ then $J_{\text{∗}}(\mathit{\partial A})\le J^{\text{∗}}(\mathit{\partial A})=0$ and so $\mathit{\partial A}$ is Jordan measurable with $J(\mathit{\partial A})=0$.