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} andn\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}<p, whence 0<\displaystyle\frac{1}{n}<p-\frac{1}{n}<q_n. So, (\exists n_0\in\mathbb{N})(\forall n\in\mathbb{N}) with n>n_0, it is true that 0<q_n-\displaystyle\frac{1}{n}<p<q_n+\frac{1}{n} 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<q_n-\displaystyle\frac{1}{n}<p<q_n+\frac{1}{n} gives 0<p-\displaystyle\frac{1}{n}<q_n<p+\frac{1}{n}, 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.

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