Exercise 1.1.3 (Uniqueness of elementary measure)

From the theorem hint we know that $c=m([0,1{\text{)}}^d$; thus, we have to demonstrate that $\forall <!--\; FUNCTION\; APPLICATION\; -->E\in E({ℝ}^d):m^{\prime }(E)=\mathit{cm}(E)$. Notice that $E$ is elementary with the box partition $E=B_1\cup \cdots \cup B_k$; by finite additivity property it suffices to show that for

where $B({ℝ}^d)$ denotes the set of all boxes in ${ℝ}^d$. The key observation here is that for any positive integer $n\in \mathbb{N}$ we have

To see why, notice that we can conveniently rewrite ${\left[0,1\right)}^d$ as the union of squares of the same side length:

This is true, since any point $x\in B$ must satisfy $0\le x_1<p_1,\dots ,0\le x_d<p_d$. We can always find two nearest integers such that for all $i:k_i\le x_i<k_i+1$ (cf. Analysis I). Any combination of $d$ of such $k_i$’s is necessarily contained in the union set. Conversely, the elements of the products of the right-hand side have always the components which lie between $0$ and $p_i$; thus, they must be contained in $B$ by definition. The above representation allows us to use the finite additivity and translation invariance property of $m^{\prime }$:

by translation invariance of $m^{\prime }$. We simply sum up $m^{\prime }\left({\left[0,\frac{1}{n}\right)}^d\right)$ several times; the index set contains $n\cdot \dots \cdot n$ elements; thus,

We now proceed using this result. Let $B$ be an arbitrary box ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^d[a_i,b_i)$ in ${ℝ}^d$. Note that by translation invariance we can equivalently look at the box $B-(a_1,\dots ,a_d)={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^d[0,b_i-a_i)$ (inclusion of the left and right endpoints can be easily generalised at the end of the proof when we have proven that the length of a point is zero). We consider two cases

(i)

Suppose that all $b_i-a_i$ are all rational, i.e., we can represent them in the form $p_i∕n,\dots ,p_d∕n$ with the common denominator $n$. In the following we demonstrate that $$m^{\prime }\left(\prod _{i=1}^d\left[0,\frac{p_i}{n}\right)\right)=c\cdot \prod _{i=1}^d\frac{p_i}{n}$$

To do so, notice that this box can be partitioned into the disjoined boxes of the same form as ${\left[0,\frac{1}{n}\right)}^d$ using a similar argument as before:

$$\begin{array}{llll}\hfill \prod _{i=1}^d\left[0,\frac{p_i}{n}\right)& =\bigcup _{k\in \{(k_1,\dots ,k_d)|k_i\in \mathbb{Z}:0\le k_i<p_i\}}\prod _{i=1}^d\left[\frac{k_i}{n},\frac{k_i+1}{n}\right)& \hfill & \\ \hfill & =\bigcup _{k\in \{(k_1,\dots ,k_d)|k_i\in \mathbb{Z}:0\le k_i<p_i\}}{\left[0,\frac{1}{n}\right)}^d+k& \hfill & \end{array}$$

Since the union index contains $p_1\cdot \dots \cdot p_d$ elements, we obtain

$$\begin{array}{llll}\hfill m^{\prime }\left(\prod _{i=1}^d\left[0,\frac{p_i}{n}\right)\right)& =\sum _{k\in \{(k_1,\dots ,k_d)|k_i\in \mathbb{Z}:0\le k_i<p_i\}}{m}^{\prime }\left({\left[0,\frac{1}{n}\right)}^d+k\right)& \hfill & \\ \hfill & =\sum _{k\in \{(k_1,\dots ,k_d)|k_i\in \mathbb{Z}:0\le k_i<p_i\}}{m}^{\prime }\left({\left[0,\frac{1}{n}\right)}^d\right)& \hfill & \\ \hfill & =c\cdot \prod _{i=1}^d\frac{p_i}{n}& \hfill & \end{array}$$

(ii)

Now we generalise to the case when $b_i-a_i$ are real numbers. We can always find two sequences ${(s_n^{(i)})}_{n\in \mathbb{N}}<b_i-a_i,b_i-a_i<{(q_n^{(i)})}_{n_{\text{∈}}\mathbb{N}}$ or rationals such that $\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{s}_n^i=\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{q}_n^i=b_i-a_i$. In the case before we have shown that $m^{\prime }\left({\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^d[0,s_n^{(i)})\right)=c{\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^d{s}_n^{(i)}$ and similarly with $q$. The key observation in this part is that given the three properties of $m^{\prime }$ in the theorem we can recover the monontinicity, i.e., from $$\prod _{i=1}^d[0,s_n^{(i)})\subseteq \prod _{i=1}^d[0,b_i-a_i)\subseteq \prod _{i=1}^d[0,q_n^{(i)})$$

follows that

$$m^{\prime }\left(\prod _{i=1}^d[0,s_n^{(i)})\right)\le m^{\prime }\left(\prod _{i=1}^d[0,b_i-a_i)\right)\le m^{\prime }\left(\prod _{i=1}^d[0,q_n^{(i)})\right)$$

(Suppose $A\subseteq B$. Then by finite additivity $m^{\prime }(B)=m^{\prime }(B∕A)+m^{\prime }(A)$; thus, $m^{\prime }(B)\ge m^{\prime }(A)$.) In other words, for each $n\in \mathbb{N}$

$$c\cdot s_n^{(1)}\cdot \dots \cdot s_n^{(d)}\le m^{\prime }\left(\prod _{i=1}^d[0,b_i-a_i)\right)\le c\cdot q_n^{(1)}\cdot \dots \cdot q_n^{(d)}$$

from which, by limit laws, we recover

$$m^{\prime }\left(\prod _{i=1}^d[0,b_i-a_i)\right)=c\cdot (b_1-a_1)\cdot \dots \cdot (b_d-a_d)$$

as desired.

By the same limit argument we can repair the fact that $m^{\prime }(\{x\})=0$ for any point $x\in {ℝ}^d$, and generalise to the intervals that include/exclude both of the endpoints.

If $E=B_1\uplus \cdots \uplus B_k$ is an elementary set then $m^{\prime }(E)=m^{\prime }\left(B_1\right)+\cdots +m^{\prime }\left(B_k\right)$. So we only need to show that $m^{\prime }(B)=c\left|B\right|$ for any box $B$. First, note that since

additivity and translation invariance guaranty that

for any $q\in \mathbb{Q}$ (make sure to fill in the details here). By translation invariance, for any $a_j\le b_j\in \mathbb{Q}$ we then have

Set $C=m^{\prime }\left([0,1{\text{)}}^d\right)$. Let $B=I_1\times \cdots \times I_d$. Assume that the endpoints of $I_j$ are $a_j<b_j$. Fix $𝜀>0$ such that for all $j$ we have $b_j-a_j>2𝜀$. Let $a_j^{\text{−}},a_j^{\text{+}},b_j^{\text{−}},b_j^{\text{+}}$be rational numbers such that $a_j^{\text{+}}\in \left(a_j,a_j+𝜀\right),a_j^{\text{−}}\in \left(a_j-𝜀,a_j\right),b_j^{\text{+}}\in \left(b_j,b_j+𝜀\right),b_j^{\text{−}}\in \left(b_j-𝜀,b_j\right)$. Let $I_j^{\text{+}}=\left[a_j^{\text{−}},b_j^{\text{+}}\right)$and $I_j^{\text{−}}=\left[a_j^{\text{+}},b_j^{\text{−}}\right)$. Let $B^{\text{±}}=I_1^{\text{±}}\times \cdots \times I_d^{\text{±}}$, so $B^{\text{−}}\subset B\subset B^{\text{+}}$. Since $m^{\prime }$ is additive and non-negative, it is also monotone. Monotonicity tells us that $m^{\prime }\left(B^{\text{−}}\right)\le m^{\prime }(B)\le m^{\prime }\left(B^{\text{+}}\right).$ Because $a_j^{\text{±}},b_j^{\text{±}}\in \mathbb{Q}$ we get that

Setting $M=\underset{j}{\max <!--\; FUNCTION\; APPLICATION\; -->}\left(b_j-a_j\right)$ we have that

Taking $𝜀\to 0$ we get that

I have no clue what the above solutions are doing...

Simply follow the hints: $m^{\prime }([0,\frac{1}{n}{\text{)}}^d)=m^{\prime }([0,1{\text{)}}^d)m([0,\frac{1}{n}{\text{)}}^d)=m([0,\frac{1}{n}{\text{)}}^d)$ since the the exercise mentions $m^{\prime }([0,1{\text{)}}^d)=1$.

Then simply $m^{\prime }([0,\frac{1}{n}{\text{)}}^d)=m([0,\frac{1}{n}{\text{)}}^d)\forall <!--\; FUNCTION\; APPLICATION\; -->n$ hence m = m’

Done.