## IMT Exercise 1.1.3

(Uniqueness of elementary measure). Let ${d\ge1}$. Let ${m':\mathcal{E}(\mathbb{R}^d)\rightarrow \mathbb{R}^+}$ be a map from the collection ${\mathcal{E}(\mathbb{R}^d)}$ of elementary subsets of ${\mathbb{R}^d}$ to the non-negative reals that obeys the non-negativity, finite additivity and translation invariance properties. Show that there exists a constant ${c\i\mathbb{R}^+}$ such that ${m'(E)=c m(E)}$ for all elementary sets ${E}$. In particular, if we impose the additional normalisation ${m'([0,1)^d)=1}$, then ${m'\equiv m}$. (Hint: Set ${c:=m([0,1)^d)}$, then compute ${m'([0,\frac{1}{n})^d)}$ for any positive integer ${n}$.)

For a box ${B}$ we know that ${m(B)\in\mathbb{R}^+}$ and we are given that ${m'(B)\in\mathbb{R}^+}$ so provided ${B\neq\emptyset}$ we can always write ${m'(B)=c m(B)}$ with ${c\in\mathbb{R}^+}$.

Consider two disjoint boxes ${B_1}$ and ${B_2}$ with ${B_1\neq B_2\neq\emptyset}$ and let

$\displaystyle m'(B_1)=c_1 m(B_1)\mbox{, } m'(B_2)=c_2 m(B_2)\mbox{ and } m(B_1\cup B_2)=c_U m(B_1\cup B_2)$

As ${m'}$ and ${m}$ both obey the finite additivity property we have

$\displaystyle c_U(m(B_1)+m(B_2))=c_1 m(B_1)+c_2 m(B_2)$

Rearranging gives

$\displaystyle (c_U-c_1)m(B_1)=(c_2-c_U)m(B_2)$

This will only be true for all boxes ${B_1}$ and ${B_2}$ if

$\displaystyle c_U=c_1\mbox{ and }c_2=c_U$

and so for all boxes ${B\neq\emptyset}$ we have a constant ${c}$ such that

$\displaystyle m'(B)=c m(B).$

If we consider the box ${[0,\frac{1}{n})^d}$ with ${n}$ a positive integer we have

$\displaystyle m'([0,\frac{1}{n})^d)=cm([0,\frac{1}{n})^d)$

and considering the limit as ${n\rightarrow\infty}$ we obtain the result that ${m'(\emptyset)=0}$.

Using these results and the results of the previous two exercises we can see that for any elementary set

$\displaystyle m'(E)=c m(E).$

If we normalise ${m'}$ so that ${m'([0,\frac{1}{n})^d)=1}$ then ${c=1}$ and we have ${m'\equiv m}$.

${\Box}$