# Clarification on the countable additivity of Lebesgue measure

As emphasized in remark 1.2.4, p. 19, of Terence Tao’s book “An introduction to measure theory”, finite additivity doesn’t hold for Lebesgue outer measure $m^{*}(\cdot)$ in general, and therefore it doesn’t hold for Lebesgue measure $m(\cdot)$ either. So, the Lebesgue outer measure $m^{*}(E\cup F)$ of the union of two disjoint sets $E, F$ in the Euclidean metric space $(\mathbb{R}^d, |\cdot|)$ does not necessarily satisfy $m^{*}(E\cup F)=m^{*}(E)+m^{*}(F)$.

If $E,F$ are both Lebesgue measurable, then it holds that $m^{*}(E\cup F)=m^{*}(E)+m^{*}(F)$. Moreover, $m^{*}(E)=m(E)$$m^{*}(F)=m(F)$, which means that the union $E\cup F$ is also Lebesgue measurable and ultimately finite additivity follows as $m(E\cup F)=m(E)+m(F)$.

The main point is that if $E\cup F$ is Lebesgue measurable and $E\cap F=\emptyset$, then finite additivity doesn’t follow. Instead, the disjoint set assumption $E\cap F=\emptyset$ needs to be replaced by the positive distance assumption $\mbox{dist}(E, F)=\inf\{|x-y|: x\in E, y\in F\}>0$ to ensure finite additivity for the outer measure, as explained and proved in lemma 1.2.5, p. 19, of Tao’s book.

Considering this limitation in the applicability of finite additivity, a reader may feel alarmed when reading lemma 1.9.c, p. 21, of David Williams’ book “Probability with martingales”, which states that if $m(S)<\infty$ in a measure space $(S, \Sigma, m)$, then $m(F\cup E) = m(F)+m(E)-m(F\cap E)$ for $F, E\in\Sigma$. If $F\cap E=\emptyset$, then $m(F\cup E) = m(F)+m(E)$, so finite additivity holds.

Recall that a measure $m:\Sigma\rightarrow[0,\infty]$ on $(S,\Sigma)$ has the whole σ-algebra $\Sigma$ on $S$ as its domain. This implies that for any $F, E\in\Sigma$ the measures $m(F),m(E)$ exist by assumption, therefore there is no conflict between the aforementioned statements on finite additivity of measure found in Tao’s and Williams’ book.

It becomes clear that operating on a measure space $(\mathbb{R}^d,\Sigma, m)$ seems to avoid the trouble of having to prove the existence of Lebesgue measure for every set of interest, simply because by definition $m(F)$ exists for every $F\in\Sigma$. However, there is no free lunch. A follow-up question arises, as one then needs to show that $F$ belongs to the defined σ-algebra $\Sigma$, which is not always trivial.

For instance, the experiment of tossing a coin infinitely often is presented in p. 24 of Williams’ book by introducing an associated sample space $\Omega$ and subsequently a probability triple $(\Omega,\mathcal{F},P)$. An event $F$ of possible interest is that the ratio of heads in $n$ tosses tends to $1/2$ as $n\rightarrow\infty$. Even for such a seemingly simple experiment, proving that this event $F$, seen as a set, belongs to the σ-algebra $\mathcal{F}$ on $\Omega$ is already not so trivial to prove.

On a final note, lemma 1.2.15, p. 30, in Tao’s book, proves countable additivity for disjoint Lebesgue measurable sets, which subsumes finite additivity. The proof is easy to follow; the claim is proved first for compact, then for bounded and then for unbounded sets. To conclude the present post, a clarification is made in the proof of the bounded case. In particular, it will be explained why for a bounded Lebesgue measurable set $E_n$ there exists a compact set $K_n$ such that $m(E_n) \le m(K_n)+\epsilon/2^n$.

Since $E_n$ is Lebesgue measurable, it follows from exercise 1.2.7.iv in Tao’s book that for any $\epsilon > 0$ there exists a closed set $K_n$ such that $K_n\subseteq E_n$ and $m^{*}(E_n\setminus K_n) \le \epsilon/2^n$. By applying the countable subadditivity of outer measure (see exercise 1.2.3.iii in Tao’s book), $E_n=K_n\cup (E_n\setminus K_n)$ leads to $m^{*}(E_n)\le m^{*}(K_n)+m^{*}(E_n\setminus K_n) \le m^{*}(K_n)+\epsilon/2^n$. Furthermore, $K_n$ is Lebesgue measurable since it is closed (see lemma 1.2.13.ii in Tao’s book). Thus, Lebesgue outer measure can be replaced by Lebesgue measure to give $m(E_n) \le m(K_n)+\epsilon/2^n$. Finally, $K_n$ is bounded, as a subset of the bounded set $E_n$, and closed, so $K_n$ is compact, which completes the proof.

# Clarification on the proof that an open set is a countable union of almost disjoint boxes

To establish that the Lebesgue outer measure of any open set in the Euclidean metric space of $\mathbb{R}^d$ is equal to the volume of any partitioning of that set into almost disjoint boxes, lemma 1.2.11 in p. 24 of the book “An introduction to measure theory” by Terence Tao first states that any open set $E\subseteq \mathbb{R}^d$ can be expressed as a countable union of almost disjoint boxes (and in fact as a countable union of almost disjoint closed cubes).

Note that this lemma is a generalization of the fact that every open subset of $\mathbb{R}$ can be expressed as a countable union of disjoint open intervals (see theorem 1.3, p. 6, in the book “Real analysis: measure theory, integration and Hilbert spaces” by Elias M. Stein and Rami Shakarchi and also lemma 2 of this blog post of mine).

The purpose of the present blog post is to elaborate on a detail in the proof of lemma 1.2.11 of Tao’s book. In particular, it will be explained why  every closed dyadic cube $Q$ is contained in exactly one maximal cube $Q^{*}$, without reproducing the rest of the proof.

Introduce the notation $Q_{n,\mathbf{i}}=\overset{d}{\underset{k=1}{\times}}\left[\frac{i_k}{2^n},\frac{i_k+1}{2^n}\right]$ to indicate the dependence of closed dyadic cubes on $n\in\mathbb{N}\cup \left\{0\right\}$ and $\mathbf{i}=(i_1,i_2,\cdots, i_d)\in\mathbb{Z}^{d}$. For every $\mathbf{i}=(i_1,i_2,\cdots, i_d)\in\mathbb{Z}^{d}$ for which there exists a closed dyadic cube $Q_{n,\mathbf{i}}$ contained in $E$, choose the biggest closed dyadic cube $Q_{n,\mathbf{i}}$ in $E$, i.e. choose $n_0=\underset{n}{\min}\left\{n\in\mathbb{N}\cup \left\{0\right\}:Q_{n,\mathbf{i}}\in E\right\}$. It is now obvious that by having capped the cubes by a sidelength of 1, there exists a maximum cube $Q_{n_0,\mathbf{i}}$ among $\left\{Q_{n,\mathbf{i}}\in E:n\in\mathbb{N}\cup \left\{0\right\}\right\}$, which is a maximal cube among those contained in $E$.

What is left to show is that every closed dyadic cube $Q$ can’t be contained in two or more maximal cubes. Assume that $Q$ is contained in two distinct maximal cubes $Q^*$ and $Q^{**}$, i.e. $Q\subseteq Q^*$ and $Q\subseteq Q^{**}$ with $Q^*\neq Q^{**}$. The dyadic nesting property then leads to the contradiction $Q^*\subseteq Q^{**}$ or $Q^{**}\subseteq Q^*$ for the maximal cubes $Q^*$ and $Q^{**}$, since $Q\subseteq Q^*$ and $Q\subseteq Q^{**}$ excludes the possibility of $Q^*$ and $Q^{**}$ being almost disjoint.

# 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.