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}

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