# Proof: characterisation of Jordan measurability

The following proof is a solution to exercise 1.1.5 of the book “An introduction to measure theory” by Terence Tao.

Some notation will be established before proceeding with the exercise. Let $\mathcal{E}(\mathbb{R}^d)$ be the set of elementary sets on $\mathbb{R}^d$. Denote by $\mathcal{L}(E):=\left\{m(A):A\in\mathcal{E}(\mathbb{R}^d),A\subseteq E\right\}$ and by $\mathcal{U}(E):=\left\{m(B):B\in\mathcal{E}(\mathbb{R}^d),E\subseteq B\right\}$ the sets of elementary measures of all elementary subsets and supersets of a bounded set $E\subseteq\mathbb{R}^d$, respectively. Let $m_{*}(E)=\sup{\mathcal{L}(E)}$ and $m^{*}(E)=\inf{\mathcal{U}(E)}$ be the Jordan inner and Jordan outer measures of $E$, respectively.

Exercise 1.1.5 requires to prove the equivalence of the following three statements, as a way of characterising Jordan measurability:

1. $E$ is Jordan measurable, which means that $m_{*}(E)=m^{*}(E)$.
2. For every $\epsilon>0$ there exist elementary sets $A\subseteq E\subseteq B$ such that $m(B\setminus A)\le\epsilon$.
3. For every $\epsilon>0$ there exists an elementary set $A$ such that $m^{*}(A\triangle E)\le\epsilon$.

It suffices to prove that $[1]\Rightarrow [2]\Rightarrow [3]\Rightarrow [1]$. To provide further practice and familiarity with Jordan inner and outer measures, it will be additionally shown how to prove $[1]\Rightarrow [3]\Rightarrow [2]\Rightarrow [1]$.

$\boxed{[1]\Rightarrow [2]}$

A reductio ad absurdum argument will be used. Assume that there exists $\epsilon_0>0$ such that for all elementary sets $A,B\in\mathcal{E}(\mathbb{R}^d)$ with $A\subseteq E\subseteq B$ the inequality $m(B\setminus A)>\epsilon_0$ holds.

Considering the set equality $B=A\cup (B\setminus A)$, it follows from $A\subseteq B$ that $m(B\setminus A)=m(B)-m(A)$. Hence, $m(B\setminus A)=m(B)-m(A)>\epsilon_0$.

So, $m(A)+\epsilon_0\le m^{*}(E)$, since $m(A)+\epsilon$ is a lower bound of $\mathcal{U}(E)$ and $m^{*}(E)=\inf{\mathcal{U}(E)}$. In turn, $m^{*}(E)-\epsilon_0$ is an upper bound of $\mathcal{L}(E)$ and $m_{*}(E)=\sup{\mathcal{L}(E)}$, therefore $m_{*}(E)\le m^{*}(E)-\epsilon_0$.

Thus, $m_{*}(E), which contradicts the assumption $m_{*}(E)=m^{*}(E)$.

$\boxed{[2]\Rightarrow [3]}$

Assume that there exists $\epsilon_0>0$ such that $\forall C\in\mathcal{E}(\mathbb{R}^d)$ holds $m^{*}(C\triangle E)>\epsilon_0$.

According to the assumed statement [2], for $\epsilon=\epsilon_0$, $\exists A, B\in\mathcal{E}(\mathbb{R}^d)$ with $A\subseteq E\subseteq B$ such that $m(B\setminus A)\le\epsilon_0$.

Pick $C=B$, so $m^{*}(B\triangle E)=m^{*}(C\triangle E)>\epsilon_0$. It follows from $E\subseteq B$ and $B\triangle E=(B\setminus E)\cup(E\setminus B)$ that $B\triangle E=B\setminus E$, so $m^{*}(B\setminus E)=m^{*}(B\triangle E)>\epsilon_0$.

$B\setminus E\subseteq B\setminus A$, since $A\subseteq E$. By also taking into account that $B\setminus A$ is elementary and the inequality $m(B\setminus A)\le\epsilon_0$, the conclusion is $m^{*}(B\setminus E)\le m(B\setminus A)\le\epsilon_0$.

A contradiction has been reached, as it has been deduced $m^{*}(B\setminus E)>\epsilon_0$ and $m^{*}(B\setminus E)\le\epsilon_0$ on the basis of the negation of statement [3].

$\boxed{[3]\Rightarrow [1]}$

Before proceeding with the proof of $[3]\Rightarrow [1]$, four lemmas will be proved.

Lemma 1

The Jordan inner measure $m_{*}(E)$ of any bounded set $E\subseteq\mathbb{R}^d$ is less than or equal to its Jordan outer measure $m^{*}(E)$, i.e. $m_{*}(E)\le m^{*}(E)$.

Proof of lemma 1

For any elementary sets $A,B$ with $A\subseteq E\subseteq B$, the set relation $A\subseteq B$ yields $m(A)\le m(B)$. This means that any $m(A)\in\mathcal{L}(E)$ is a lower bound of $\mathcal{U}(E)$, therefore $m(A)\le m^{*}(E)$. Since the last inequality holds for any $m(A)\in\mathcal{L}(E)$, it follows that $m^{*}(E)$ is an upper bound of $\mathcal{L}(E)$, thus $m_{*}(E)\le m^{*}(E)$.

Lemma 2

The elementary and Jordan measures of any elementary set $X$ coincide, that is $m_{*}(X)=m^{*}(X)=m(X)$.

Proof of lemma 2

It suffices to notice that $X\subseteq X$, whence $m^{*}(X)\le m(X)$ and $m(X)\le m_{*}(X)$, thereby $m^{*}(X)\le m_{*}(X)$. It is also known from lemma 1 that $m_{*}(X)\le m^{*}(X)$, so $m_{*}(X)=m^{*}(X)$. Finally, $m^{*}(X)\le m(X)$, $m(X)\le m_{*}(X)$ and $m_{*}(X)=m^{*}(X)$ yield $m(X)=m_{*}(X)=m^{*}(X)$.

Lemma 3

Let $E\subseteq\mathbb{R}^d$ be a bounded set and $B\in\mathcal{E}(\mathbb{R}^d)$ with $E\subseteq B$. It then holds that $m(B)-m_{*}(E)\le m^{*}(B\setminus E)$.

Proof of lemma 3

Recall that for a set $X\subseteq\mathbb{R}$ the following equivalences hold:

• $\ell=\inf{X}\Leftrightarrow(\forall\epsilon>0)(\exists x\in X)x<\ell+\epsilon$.
• $u=\sup{X}\Leftrightarrow(\forall\epsilon>0)(\exists x\in X)u-\epsilon < x$.

According to the former equivalence, for any $\epsilon>0$ there exists an elementary set $B\setminus E\subseteq C$ such that

$m(C).

By using the latter equivalence, there exists an elementary set $A\subseteq B$ such that

$m_{*}(B)-\epsilon /2.

It follows from lemma 2 and inequality (2) that $m(B)-m_{*}(E). $B\setminus E\subseteq C$ and $A\subseteq B$ lead to $A\setminus C\subseteq E$, so $m(A\setminus C)\le m_{*}(E)$, which in turn gives

$m(B)-m_{*}(E).

Inequality (3) connects the statement of lemma 3 to an analogous simpler statement for elementary sets; in particular, it will be shown that

$m(A)-m(C)\le m(A\setminus C)\ \ \ \ (4)$.

Indeed, $(A\setminus C)\cup (A\cap C)=A$, so $m(A)-m(A\cap C)=m(A\setminus C)$. Moreover, $A\cap C\subseteq C$, which means $m(A\cap C)\le m(C)$, therefore $m(A)-m(C)\le m(A)-m(A\cap C)$. Inequality (4) has thus been reached.

(3) and (4) yield $m(B)-m_{*}(E), which is combined with (1) to give

$m(B)-m_{*}(E).

For two reals $x, y$, if $(\forall\epsilon>0) x\le y+\epsilon$, then $x\le y$. This can be shown by assuming $x>y$, whence $0, contradiction. As inequality (5) holds for any $\epsilon>0$, the conclusion $m(B)-m_{*}(E)\le m^{*}(B\setminus E)$ follows.

Lemma 4

The Jordan outer measure of the union of two bounded sets $X\subseteq\mathbb{R}^d$ and $Y\subseteq\mathbb{R}^d$ is less than or equal to the sum of the Jordan outer measures of the two sets, i.e. $m^{*}(X\cup Y)\le m^{*}(X)+m^{*}(Y)$.

Proof of lemma 4

Let $V\subseteq\mathbb{R}^d,W\subseteq\mathbb{R}^d$ be elementary sets with $X\subseteq V,Y\subseteq W$.

Start by noticing the set equality $V\cup W=V\cup(W\setminus V)$. So, $m(V\cup W)=m(V)+m(W\setminus V)$. Moreover, $W\setminus V\subseteq W$ implies that $m(W\setminus V)\le m(W)$. Thus,

$m(V\cup W)\le m(V)+m(W)\ \ \ \ (6)$.

Inequality (6) holds for elementary sets, thereby it is a special case of lemma 4.

$m^{*}(X\cup Y)\le m(V\cup W)$, since $X\cup Y \subseteq V\cup W$. By taking into account inequality (6), $m^{*}(X\cup Y)\le m(V)+m(W)$. So, $m^{*}(X\cup Y)-m(V)$ is a lower bound of $\mathcal{U}(Y)$ and $m^{*}(X\cup Y)-m(V)\le m^{*}(Y)$. Furthermore, $m^{*}(X\cup Y)-m^{*}(Y)$ is a lower bound of $\mathcal{U}(X)$, so $m^{*}(X\cup Y)-m^{*}(Y)\le m^{*}(X)$, quod erat demonstrandum.

Proof of $\boldsymbol{[3]\Rightarrow [1]}$ using the lemmas

The main idea of the proof of $\boldsymbol{[3]\Rightarrow [1]}$ is to show that for any $\epsilon>0$ the inequality $m^{*}(E)-m_{*}(E)\le \epsilon$ holds.

It is known from statement [3] that for any $(\epsilon>0)(\exists X\in\mathcal{E}(\mathbb{R}^d))$ such that

$m^{*}(X\triangle E)\le\epsilon/3\ \ \ \ (7)$.

Using the relevant property of infimum, for any $\epsilon>0$ there exists an elementary superset $E\setminus X\subseteq Y$ of $E\setminus X$ such that

$m(Y).

Introduce the set $B:=X\cup Y$. Obviously $B$ is an elementary set and $E\subseteq B$, so $m^{*}(E)\le m(B)$. It thus follows from lemma 3 that

$m^{*}(E)-m_{*}(E)\le m^{*}(B\setminus E)\ \ \ \ (9)$.

$B\setminus E=(X\cup Y)\setminus E=(X\setminus E)\cup (Y\setminus E)$, so by lemma 4

$m^{*}(B\setminus E)\le m^{*}(X\setminus E)+m^{*}(Y\setminus E)\ \ \ \ (10)$.

$X\setminus E\subseteq X\triangle E\Rightarrow \mathcal{U}(X\triangle E)\subseteq\mathcal{U}(X\setminus E)$, so $m^{*}(X\setminus E)=\inf{\mathcal{U}(X\setminus E)}\le \inf{\mathcal{U}(X\triangle E)}=m^{*}(X\triangle E)$. Consequently, inequality (10) gives

$m^{*}(B\setminus E)\le m^{*}(X\triangle E)+m^{*}(Y\setminus E)\ \ \ \ (11)$.

In a similar way, $E\setminus X\subseteq X\triangle E$ implies

$m^{*}(E\setminus X)\le m^{*}(X\triangle E)\ \ \ \ (12)$.

Finally, $Y\setminus E\subseteq Y$ means that

$m^{*}(Y\setminus E)\le m(Y)\ \ \ \ (13)$.

All the components of the proof have been established. More concretely, inequalities (9), (11), (7), (13), (8) and (12) produce $m^{*}(E)-m_{*}(E)\le \epsilon$ for any $\epsilon>0$, so $m_{*}(E)=m^{*}(E)$.

$\boxed{[1]\Rightarrow [3]}$

Assume that $(\exists\epsilon_0>0)(\forall X\in\mathcal{E}(\mathbb{R}^d))m^{*}(X\triangle E)>\epsilon_0$.

As $m^{*}(E)=\inf{\mathcal{U}(E)}$, it is deduced that there exists an elementary set $E\subseteq B$ such that

$m(B).

Since $m_{*}(E)=\sup{\mathcal{L}(E)}$, there exists an elementary set $A\subseteq E$ such that

$m_{*}(E)-\epsilon_0/2.

$A\subseteq E\subseteq B\Rightarrow E\setminus A\subseteq B\setminus A$, so $m^{*}(E\setminus A)\le m(B\setminus A)$. Moreover, $A\subseteq B$ means $m(B\setminus A)=m(B)-m(A)$, hence

$m_{*}(E\setminus A)\le m(B)-m(A)\ \ \ \ (16)$.

Combining (14), (15) and (16) gives $m_{*}(E\setminus A). By the assumed statement [1], $m_{*}(E)=m^{*}(E)$, thus $m_{*}(E\setminus A)<\epsilon_0$. This is a contradiction, as the assumed negation of statement [3] gives $m_{*}(A\triangle E)=m_{*}(E\setminus A)>\epsilon_0$ for $A\subseteq E$.

$\boxed{[3]\Rightarrow [2]}$

The proof of $[3]\Rightarrow [2]$ is similar in spirit to the proof of $[3]\Rightarrow [1]$. It will be shown that for any $\epsilon >0$ there exist $A,B\in\mathcal{E}(\mathbb{R}^d)$ with $A\subseteq E\subseteq B$ such that $m(B\setminus A)\le \epsilon$.

It is known from statement [3] that for any $(\epsilon>0)(\exists X\in\mathcal{E}(\mathbb{R}^d))$ such that

$m^{*}(X\triangle E)\le\epsilon/4\ \ \ \ (17)$.

According to the relevant property of infimum, for any $\epsilon>0$ there exists an elementary set $Y$ with $E\setminus X\subseteq Y$ such that

$m(Y).

Consider the set $B:=X\cup Y$, which is an elementary set and $E\subseteq B$, so $m^{*}(E)\le m(B)$. By application of lemma 3,

$m(B)-m_{*}(E)\le m^{*}(B\setminus E)\ \ \ \ (19)$.

It is noted that $B\setminus E=(X\cup Y)\setminus E=(X\setminus E)\cup (Y\setminus E)$, so by lemma 4

$m^{*}(B\setminus E)\le m^{*}(X\setminus E)+m^{*}(Y\setminus E)\ \ \ \ (20)$.

$X\setminus E\subseteq X\triangle E$, so $m^{*}(X\setminus E)\le m^{*}(X\triangle E)$. Thus, inequality (20) leads to

$m^{*}(B\setminus E)\le m^{*}(X\triangle E)+m^{*}(Y\setminus E)\ \ \ \ (21)$.

Similarly, $E\setminus X\subseteq X\triangle E$ implies

$m^{*}(E\setminus X)\le m^{*}(X\triangle E)\ \ \ \ (22)$.

Moreover, $Y\setminus E\subseteq Y$ gives

$m^{*}(Y\setminus E)\le m(Y)\ \ \ \ (23)$.

Finally, there exists an elementary set $A$ with $A\subseteq E$ such that

$m_{*}(E)-\epsilon /4.

Inequalities (19), (21), (17), (23), (18) and (22) produce

$m(B)-m_{*}(E)\le 3\epsilon /4 \ \ \ \ (25)$.

Combining inequalities (24) and (25) confirms that for any $\epsilon > 0$ there exist elementary sets $A,B$ with $A\subseteq E\subseteq B$ such that $m(B)-m(A)\le \epsilon$, which completes the proof of $[3]\Rightarrow [2]$.

$\boxed{[2]\Rightarrow [1]}$

It is known from lemma 1 that $m_{*}(E)\le m^{*}(E)$. Assume the negation of statement [1], that is assume $m_{*}(E)\neq m^{*}(E)$. So, $m_{*}(E)< m^{*}(E)$.

Set $\epsilon_0=(m^{*}(E)-m_{*}(E))/2>0$. From the assumed statement [2], it is known that there exist $A,B\in\mathcal{E}(\mathbb{R}^d)$ with $A\subseteq E\subseteq B$ such that

$m(B\setminus A)=m(B)-m(A)\le \epsilon_0=(m^{*}(E)-m_{*}(E))/2 \ \ \ \ (26)$.

Note that $A\subseteq E\Rightarrow m(A)\le m_{*}(E)$ and $E\subseteq B\Rightarrow m^{*}(E)\le m(B)$, which leads to

$m^{*}(E)-m_{*}(E)\le m(B)-m(A) \ \ \ \ (27)$.

It follows from equation (26) that $2(m(B)-m(A))\le m^{*}(E)-m_{*}(E)$, which is combined with equation (27) to give $2(m(B)-m(A))\le m(B)-m(A)$, and finally $m(B)-m(A)\le 0$. Moreover, equation (27) yields $0 < 2\epsilon_0=m^{*}(E)-m_{*}(E)\le m(B)-m(A)$, i.e. $m(B)-m(A)>0$. Thus, a contradiction has been reached.