# Clarification on the proof of Carathéodory’s lemma

Carathéodory’s lemma, as it appears in p. 197 (appendix A1) of the book “Probability with martingales” by David Williams, is stated as follows:

Let $\lambda$ be an outer measure on the measurable space $(S,\mathcal{G})$. Then the λ-sets in $\mathcal{G}$ form a σ-algebra $\mathcal{L}$ on which $\lambda$ is countably additive, so that $(S,\mathcal{L},\lambda)$ is a measure space.

Three aspects of the proof of Carathéodory’s lemma provided in Williams’ book are clarified in this blog post.

Definition of λ-system

The concept of λ-system, which is used implicitly but it is not defined in Williams’ book, is introduced in this post.

A collection $\mathcal{C}$ of subsets of a set $S$ is called a λ-system on $S$ if

1. $S\in\mathcal{C}$,
2. $L\in\mathcal{C}\Rightarrow L^{c}\in\mathcal{C}$ (it is closed under complements),
3. $(\forall n\in\mathbb{N})L_n\in\mathcal{C}$ with $L_i\cap L_j=\emptyset$ for $i\neq j$ it holds that $\underset{n\in\mathbb{N}}{\cup}L_n\in\mathcal{C}$ (it is closed under countable disjoint unions).

Not that the only difference between a λ-system and a σ-algebra is that the former is closed under countable disjoint unions while the latter is closed under countable unions. Moreover, the first condition on the definition of a λ-system could be alternatively set to $\emptyset\in\mathcal{C}$ instead of $S\in\mathcal{C}$ due to closure under complementarity, i.e. due to the second condition of the definition.

Lemma

If a collection of subsets of a set $S$ is a λ-system and a π-system on $S$, it is also a σ-algebra on $S$.

This lemma is used without being proved in Williams’ book for proving Carathéodory’s lemma. In what follows, the lemma will be proved before proceeding with the proof of Carathéodory’s lemma.

Although not relevant to subsequent developments, it is mentioned that a σ-algebra on a set $S$ is also a λ-system on $S$ as it can be trivially seen from the involved definitions.

Proof of the lemma

Let $\mathcal{C}$ be a collection of subsets $S$ that is both a λ-system and a π-system on $S$. To show that $\mathcal{C}$ is a σ-algebra on $S$, it suffices that it is closed under countable unions.

Let $B_n\in\mathcal{C},n\in\mathbb{N}$. The main idea is to express the collection $\{B_n: n\in\mathbb{N}\}$ as a collection $\{L_n: n\in\mathbb{N}\}$ of pairwise disjoint sets ($L_i\cap L_j=\emptyset$ for $i\neq j$) so that $\underset{n\in\mathbb{N}}{\cup}B_n=\underset{n\in\mathbb{N}}{\cup}L_n$. Along these lines, define $L_n:=B_n\setminus \left(\overset{n-1}{\underset{k=1}{\cup}} B_k\right)$.

Obviously, $\underset{n\in\mathbb{N}}{\cup}L_n\subseteq \underset{n\in\mathbb{N}}{\cup}B_n$. To prove the converse set inequality, let $x\in\underset{n\in\mathbb{N}}{\cup}B_n$ and assume that $(\forall n\in\mathbb{N}) x\not\in B_n\setminus \left(\overset{n-1}{\underset{k=1}{\cup}} B_k\right)$. In this case, for each $n\in\mathbb{N}$, either $x\not\in B_n$ or $x\in B_n\cap\left(\overset{n-1}{\underset{k=1}{\cup}} B_k\right)$. There is at least one $n_{*}\in\mathbb{N}$ such that $x\in B_{n_{*}}\cap\left(\overset{n_{*}-1}{\underset{k=1}{\cup}} B_k\right)$, otherwise $(\forall n\in\mathbb{N}) x\not\in B_n$ leads to the contradiction $x\not\in\underset{n\in\mathbb{N}}{\cup}B_n$. Let $n_o\in\mathbb{N}$ be the minimum natural for which $x\in B_{n_{o}}\cap\left(\overset{n_{o}-1}{\underset{k=1}{\cup}} B_k\right)$. In turn, $x\in \overset{n_{o}-1}{\underset{k=1}{\cup}} B_k\Rightarrow (\exists i. Due to $n_o$ being the smallest natural for which $x\in B_{n_{o}}\cap\left(\overset{n_{o}-1}{\underset{k=1}{\cup}} B_k\right)$, it is deduced that $x\not\in B_i\cap\left(\overset{i-1}{\underset{k=1}{\cup}} B_k\right)$, hence $x\not\in\left(\overset{i-1}{\underset{k=1}{\cup}} B_k\right)$. Thus, $(\exists i\in\mathbb{N}) x\in B_i\setminus \left(\overset{i-1}{\underset{k=1}{\cup}} B_k\right)$, which is a contradiction. Thereby, $\underset{n\in\mathbb{N}}{\cup}B_n\subseteq \underset{n\in\mathbb{N}}{\cup}L_n$, and this establishes the equality $\underset{n\in\mathbb{N}}{\cup}L_n= \underset{n\in\mathbb{N}}{\cup}B_n$.

Assume that there are $i,j\in\mathbb{N}$ with $i\neq j$ and $L_i \cap L_j \neq \emptyset$. Let $x\in L_i \cap L_j$. Without loss of generality assume that $i. Then $x\in B_i$ with $i, while $x\in L_j=B_j\setminus \left(\overset{j-1}{\underset{k=1}{\cup}} B_k\right)$, which means that $(\forall k\in\mathbb{N})$ with $k it holds that $x\not\in B_k$, so a contradiction has been reached. Thereby, the sets $L_n, n\in\mathbb{N}$, are pairwise disjoint.

It has thus been shown that the collection $L_n:=B_n\setminus \left(\overset{n-1}{\underset{k=1}{\cup}} B_k\right)$ consists of pairwise disjoints sets that satisfy $\underset{n\in\mathbb{N}}{\cup}L_n=\underset{n\in\mathbb{N}}{\cup}B_n$.

Notice that $L_n:=B_n\setminus \left(\overset{n-1}{\underset{k=1}{\cup}} B_k\right)=B_n \overset{n-1}{\underset{k=1}{\cap}}B_k^c$. Since $\mathcal{C}$ is a λ-system, $B_k^c\in\mathcal{C}$ for the various $k$. Moreover, $\mathcal{C}$ is a π-system, hence the finite intersection $L_n=B_n \overset{n-1}{\underset{k=1}{\cap}}B_k^c$ is also in $\mathcal{C}$. Since the collection $L_n:=B_n\setminus \left(\overset{n-1}{\underset{k=1}{\cup}} B_k\right)$ is a disjoint union of elements $L_n\in\mathcal{C}$ and $\mathcal{C}$ is a λ-system, it follows that the union $\underset{n\in\mathbb{N}}{\cup}L_n= \underset{n\in\mathbb{N}}{\cup}B_n$ is also in $\mathcal{C}$.

Since the countable (but not necessarily disjoint) union $\underset{n\in\mathbb{N}}{\cup}B_n$ of any collection $\{B_n: n\in\mathbb{N}\}$ of sets $B_n\in\mathcal{C}$ is also in $\mathcal{C}$, it follows that $\mathcal{C}$ is a σ-algebra.

First clarification

The above lemma explains why the proof of Carathéodory’s lemma in Williams’ book states that it suffices to show that for a countable collection $\{L_n: n\in\mathbb{N}\}$ of disjoint sets $L_n\in\mathcal{L}$ it holds that $\underset{n\in\mathbb{N}}{\cup}L_n\in\mathcal{L}$. The conclusion then extends to any such countable union of sets, disjoint or not.

Second clarification

It is mentioned in p. 197 of Williams’ book that from

$\lambda (G)\ge\overset{n}{\underset{k=1}{\sum}}\lambda(L_k\cap G)+\lambda(L^c\cap G)$

follows

$\lambda (G)\ge\overset{\infty}{\underset{k=1}{\sum}}\lambda(L_k\cap G)+\lambda(L^c\cap G)$.

To see why this is the case, recall that the outer measure $\lambda:\mathcal{G}\rightarrow [ 0, \infty ]$ takes in values in $\lambda (G)\in[0, \infty]$ for any $G\in\mathcal{G}$.

Distinguish two cases. If $\lambda (G)\in [0,\infty )$ (i.e. if $\lambda (G)$ is finite), then the sequence $a_n=\overset{n}{\underset{k=1}{\sum}}\lambda(L_k\cap G),n\in\mathbb{N}$, is a bounded increasing sequence, therefore it converges, which means that the limit $\overset{\infty}{\underset{k=1}{\sum}}\lambda(L_k\cap G)=\underset{n\rightarrow\infty}{\lim}a_n < \infty$ exists, so taking limits leads from the former to the latter inequality in the book.

If $\lambda (G)=\infty$, then $\infty \ge\overset{\infty}{\underset{k=1}{\sum}}\lambda(L_k\cap G)+\lambda(L^c\cap G)$ holds trivially.

Third clarification

To show that $\lambda (L)=\underset{k\in\mathbb{N}}{\sum}\lambda (L_k)$ for $L=\underset{k\in\mathbb{N}}{\cup}L_k$, notice first that

$\lambda (L)\le\underset{k\in\mathbb{N}}{\sum}\lambda (L_k)$

follows from the countable subadditivity of the outer measure $\lambda$.

Moreover, setting $G=L$ in equation (d) of p. 197 gives

$\lambda (L)\ge\underset{k\in\mathbb{N}}{\sum}\lambda (L_k\cap L)+\lambda (L^c\cap L)=\underset{k\in\mathbb{N}}{\sum}\lambda (L_k)$,

which concludes the argument.

# 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.

# Proof: uniqueness of elementary measure

The following proof is a solution to exercise 1.1.3 of the book “An introduction to measure theory” by Terence Tao.

A box $B\in\mathbb{R}^d$, $d\in\mathbb{N}$, is a Cartesian product $B:={\sf X}_{i=1}^d I_i$, where each interval $I_i$ is $I_i=(a, b)$ or $I_i=(a, b]$ or $I_i=[a, b)$ or $I_i=[a, b]$ for $a,b\in\mathbb{R}$ with $a\le b$. An elementary set $E=\cup_{i=1}^n B_i\subseteq\mathbb{R}^d$ is a finite union of disjoint boxes $B_i\in\mathbb{R}^d$. Let $\mathcal{E}(\mathbb{R}^d)$ denote the collection of elementary sets in $\mathbb{R}^d$. The measure $m:\mathcal{E}(\mathbb{R}^d)\rightarrow R^{+}\cup\left\{0\right\}$ is defined as $m(E)=\displaystyle\lim_{N\rightarrow\infty}\frac{1}{N^d}\#\left(E\cap\frac{1}{N}\mathbb{Z}^d\right)$, where $\#(\cdot)$ denotes set cardinality and $\displaystyle\frac{1}{N}\mathbb{Z}^d:=\left\{\frac{\mathbf{z}}{N}:\mathbf{z}\in\mathbb{Z}^d\right\}$.

Let $m^{'}:\mathcal{E}(\mathbb{R}^d)\rightarrow R^{+}\cup\left\{0\right\}$ be a function satisfying the non-negativity ($m^{'}(E)\ge 0$ for any elementary set $E$), finite additivity ($m^{'}(\displaystyle\cup_{i=1}^n E_i)\le \sum_{i=1}^{n}m^{'}(E_i)$ for disjoint elementary sets $E_i$) and translation invariance ($m^{'}(E+\mathbf{x})=m^{'}(E)$ for any elementary set $E$ and any $\mathbf{x} \in \mathbb{R}^d$) properties.

It will be shown that there exists a positive constant $c\in\mathbb{R}^+$ such that $m^{'}=cm$, i.e. the functions $m^{'}$ and $m$ are equal up to a positive normalization constant $c$.

Observe that $\left[0,1\right)=\displaystyle\cup_{i=0}^{n-1}\left[\frac{i}{n},\frac{i+1}{n}\right)$. Due to translation invariance, $m^{'}\left(\displaystyle\left[\frac{i}{n},\frac{i+1}{n}\right)\right)=m^{'}\left(\displaystyle\left[\frac{i}{n},\frac{i+1}{n}\right)-\frac{i}{n}\right)=m^{'}\left(\left[0,\frac{1}{n}\right)\right)$. Using finite additivity, it follows that $m^{'}\left(\left[0,1\right)^d\right)=n^dm^{'}\left(\left[0,\frac{1}{n}\right)^d\right)$. So, $m^{'}\left(\left[0,\frac{1}{n}\right)^d\right)=\displaystyle\frac{c}{n^d}$ for $c:=m^{'}\left(\left[0,1\right)^d\right)$. Since $m\left(\left[0,\frac{1}{n}\right)^d\right)=\displaystyle\frac{1}{n^d}$, it follows that $m^{'}\left(\left[0,\frac{1}{n}\right)^d\right)=c m\left(\left[0,\frac{1}{n}\right)^d\right)$.

This result generalizes to intervals $\left[0,q\right)$ for any rational $q=\displaystyle\frac{s}{n}$ with $s\in\mathbb{N}$ and$n\in\mathbb{N}$. Since $\left[0,q\right)=\displaystyle\cup_{i=0}^{s-1}\left[\frac{i}{n},\frac{i+1}{n}\right)$, finite additivity and translation invariance lead to $m^{'}\left(\left[0,q\right)^d\right)=c m\left(\left[0,q\right)^d\right)=cq^d$.

It will be shown that the result holds also for intervals $\left[0,p\right)$ for any irrational $p\in\mathbb{P}$.

The set of rationals is dense, which means that $(\forall \epsilon>0)(\forall x\in\mathbb{R})(\exists q\in\mathbb{Q})|x-q|<\epsilon$. For some irrational $x=p$ and for each $n\in\mathbb{N}$, set $\epsilon=\displaystyle\frac{1}{n}$, so $(\forall n\in\mathbb{N})(\exists q_n\in\mathbb{Q})|p-q_n|<\displaystyle\frac{1}{n}$. Pick some $n_0\in\mathbb{N}$ with $n_0>\displaystyle\frac{2}{p}$. For all $n\in\mathbb{N}$ with $n>n_0$, it holds that $\displaystyle \frac{2}{n}<\frac{2}{n_0}, whence $0<\displaystyle\frac{1}{n}. So, $(\exists n_0\in\mathbb{N})(\forall n\in\mathbb{N})$ with $n>n_0$, it is true that $0 and consequently $\left[0,q_n-\displaystyle\frac{1}{n}\right)^d\subseteq\left[0,p\right)^d\subseteq\left[0,q_n+\displaystyle\frac{1}{n}\right)^d$.

For any two elementary sets $E\subseteq F$, it can be shown that $m^{'}(E)\le m^{'}(F)$ via the equality $F=E\cup(F\setminus E)$, non-negativity and finite additivity. Hence, $m^{'}\left(\left[0,q_n-\displaystyle\frac{1}{n}\right)^d\right)\le m^{'}\left(\left[0,p\right)^d\right)\le m^{'}\left(\left[0,q_n+\displaystyle\frac{1}{n}\right)^d\right)$.

Since $q_n\pm\displaystyle\frac{1}{n}$ are rationals, it is deduced that $m^{'}\left(\left[0,q_n\pm\displaystyle\frac{1}{n}\right)^d\right)=c\left(q_n\pm\displaystyle\frac{1}{n}\right)^d$. Thus, $c\left(q_n-\displaystyle\frac{1}{n}\right)^d\le m^{'}\left(\left[0,p\right)^d\right)\le c\left(q_n+\displaystyle\frac{1}{n}\right)^d$.

$0 gives $0, hence the sandwich theorem yields $\displaystyle\lim_{n\rightarrow\infty}q_n=p$.

Combining $c\left(q_n-\displaystyle\frac{1}{n}\right)^d\le m^{'}\left(\left[0,p\right)^d\right)\le c\left(q_n+\displaystyle\frac{1}{n}\right)^d$ and $\displaystyle\lim_{n\rightarrow\infty}q_n=p$ gives $cp^d\le m^{'}\left(\left[0,p\right)^d\right)\le cp^d$, so $m^{'}\left(\left[0,p\right)^d\right)=cp^d$ for any irrational $p$.

This effectively completes the proof. There remains to show that $m^{'}=cm$ is true for Cartesian products of unequal intervals $\left[0,x_i\right)$ in each coordinate $i=1,2,\dots,d$, for any $x_i\in\mathbb{R}$, and for any subinterval of the real line. These are all trivial given the existing foundations.