# 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.