Exercise 1.1.10 (Balls are Jordan measurable)

Proof. We will use the same strategy that we used in the previous exercises, i.e., we will try to represent the figure in some functional way, so that we can easily estimate its measure numerically und then simply apply Exercise 1.1.7. The function describing the shape of an upper half-ball is

In other words, we demonstrate that the area under the graph of the above function is the upper-ball we are looking at

This is easy to verify. Pick an arbitrary $x\in A(f)$. Then,

Since also $x_d\ge 0$, we have $x\in B_{\text{+}}(0,r)$. Conversely, let $x\in B_{\text{+}}(0,r)$. Then from $\sqrt{{\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^d{x}_i^2}<r$ both conditions $\sqrt{{\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{d-1}{x}_i^2}<r$ and $x_d^2<r^2-{\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{d-1}{x}_i^2=f^2(x_1,\dots ,x_d)$ follow. Similarly we can prove that the lower half-ball is the area under the graph of $-f$.
Using this result, we want to demonstrate that $B^d(0,r)$ is Jordan measurable. If we knew that $B^{d-1}(0,r)$ is Jordan measurable, it could be possible to show that $B^d(0,r)$ is Jordan measurable by showing that the area under $f$ is Jordan measurable. This is obviously an induction matter.

• Induction basis. Suppose that $d=2$. Then $B^{d-1}(0,r)$ is simply an interval $[-r,r]$. But it is also a compact box; thus, by Exercise 1.1.7 the area under $f$ must be Jordan measurable.

• Induction step. Now suppose inductively that we have proven this statement for some $d-1>2$; we demonstrate that this implies that the statement is also true for $d$. In other words, assume that $B^{d-1}(0,r)$ is Jordan measurable. We demonstrate the Jordan measurability of $B^d(0,r)$ by the criteria (3) from the Exercise 1.1.5. Thus, pick an arbitrary $𝜖>0$. By Exercise 1.1.5 applied to the previous dimension, we can find an outer cover ${\overline{B^{d-1}}}_{𝜖}$ and an inner cover ${\underline{B^{d-1}}}_{𝜖}$ of $B^{d-1}(0,r)$ such that $m\left({\overline{B^{d-1}}}_{𝜖}∕{\overline{B^{d-1}}}_{𝜖}\right)<\frac{𝜖}{r}$. Note that since ${\underline{B^{d-1}}}_{𝜖}\subset {\overline{B^{d-1}}}_{𝜖}$ we can represent the outer cover as the union of the inner cover with the cover of the "boundary":

  $${\overline{B^{d-1}}}_{𝜖}={\underline{B^{d-1}}}_{𝜖}\cup \left({\overline{B^{d-1}}}_{𝜖}∕{\underline{B^{d-1}}}_{𝜖}\right)$$

  By Exercise 1.1.1 both parts are elementary. At last, by Lemma 1.1.2 we can represent both inner and outer cover as the union of disjoint boxes. However, it will be very helpful later in the proof if we know that the boxes of which this elementary set consists are not longer than ${\left(𝜖∕m_J\left(B^{(d-1)}(0,r)\right)\right)}^2\cdot \frac{1}{2r}$ in each co-ordinate. In other words, select $n\in \mathbb{N}:\frac{2r}{n}<{\left(𝜖∕m_J\left(B^{(d-1)}(0,r)\right)\right)}^2\cdot \frac{1}{2r}$ and using the usual procedure partition the box ${[-r,r]}^{d-1}$ as

  $$G:=\left\{\left[-r+\frac{2r}{n}{k}_1,-r+\frac{2r}{n}{k}_1+\frac{2r}{n}\right]\times \dots \times \left[-r+\frac{2r}{n}{k}_{d-1},-r+\frac{2r}{n}{k}_{d-1}+\frac{2r}{n}\right]:k\in {\mathbb{Z}}^d,0\le k_i<n\right\}$$

  Then considering the collection of disjoint boxes in the inner and outer cover we take a common refinement, and the refined subsets will then be again elementary $⟺$ can be represented as disjoint boxes. We index this disjoint boxes, which have vertices shorter than ${\left(𝜖∕m_J\left(B^{(d-1)}(0,r)\right)\right)}^2\cdot \frac{1}{2r}$, from $1$ to $Z\in \mathbb{N}$:

  $$\begin{array}{llll}\hfill \left\{B\in \underline{B^{d-1}(0,r)}\right\}\mathit{\#\#G}& =\left\{\left[a_i^{(1)},b_i^{(1)}\right]\times \dots \left[a_i^{(d-1)},b_i^{(d-1)}\right]:i=1\dots z\right\}& \hfill & \\ \hfill \left\{B\in \overline{B^{d-1}(0,r)}\right\}\mathit{\#\#G}& =\left\{\left[a_i^{(1)},b_i^{(1)}\right]\times \dots \left[a_i^{(d-1)},b_i^{(d-1)}\right]:i=1,\dots ,z\right\}& \hfill & \\ \hfill & \cup \left\{\left[a_i^{(1)},b_i^{(1)}\right]\times \dots \left[a_i^{(d-1)},b_i^{(d-1)}\right]:i=z+1,\dots ,Z\right\}& \hfill & \end{array}$$

  Using both, we now try to cover the area under the function $f$ on $B^{d-1}(0,r)$.

  Remark 1. Let $E={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{d-1}[a_i,b_i]$ be some box in $B^{d-1}(0,r)$. Then $f$ will achieve its maximum and its minimum on the edges of $E$.

  Proof. First notice that $f$ is monotonically decreasing with respect to the norm, i.e., $\forall <!--\; FUNCTION\; APPLICATION\; -->x,y\in B^{d-1}(0,r):\parallel x\parallel \le \parallel y\parallel ⟹f(x)\ge f(y)$. Thus, we simply need to prove that some of the edges of $E$ have the lowest norm. Pick an arbitrary $x\in E$. Since $x=(x_1,\dots ,x_{d-1})$ we have $a_i\le x_i\le b_i$ in every co-ordinate. This implies, $x_i^2\le \max <!--\; FUNCTION\; APPLICATION\; -->\{\underset{i}{\overset{2}{a}},b_i^2\}=:l_i^2$ and accordingly $\parallel x\parallel =\sqrt{{\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{d-1}{x}_i^2}\le \sqrt{{\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{d-1}{l}_i^2}=\parallel l\parallel$ where $l$ is an edge of $E$. Thus, $l$ has the lowest norm in $E$ (obviously, it may not be unique). Similarly, some edge $h$ of $E$ has the highest norm in $E$. □

  Remark 2. Now consider the maximal difference between $f$ values on $E={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{d-1}[a_i,b_i]$. By monotonicity, we equivalently look at the biggest difference in norms of the points in $E$:

  $$\begin{array}{llll}\hfill \forall <!--\; FUNCTION\; APPLICATION\; -->x,y\in E:\parallel x{\text{∥}}^2-\parallel y{\text{∥}}^2& \le \parallel h{\text{∥}}^2-\parallel l{\text{∥}}^2& \hfill & \\ \hfill & =\sum _{i=1}^{d-1}\max <!--\; FUNCTION\; APPLICATION\; -->\{\underset{i}{\overset{2}{a}},b_i^2\}-\sum _{i=1}^{d-1}\min <!--\; FUNCTION\; APPLICATION\; -->\{\underset{i}{\overset{2}{a}},b_i^2\}& \hfill & \\ \hfill & =\sum _{i=1}^{d-1}|b_i^2-a_i^2|& \hfill & \\ \hfill & =\sum _{i=1}^{d-1}|(b_i-a_i)(b_i+a_i)|.& \hfill & \end{array}$$

  Notice that in our case $b_i-a_i\le 2r∕n\le {\left(𝜖∕m_J\left(B^{(d-1)}(0,r)\right)\right)}^2\cdot \frac{1}{2r}$ and $|b_i+a_i|\le 2r$; thus, the above term will never exceed ${\left(𝜖∕m_J\left(B^{(d-1)}(0,r)\right)\right)}^2$. Furthermore, for $\parallel x\parallel \ge \parallel y\parallel$ we have

  $$\begin{array}{llll}\hfill f(y)-f(x)& =\sqrt{r^2-\parallel y{\text{∥}}^2}-\sqrt{r^2-\parallel x{\text{∥}}^2}& \hfill & \\ \hfill & \le \sqrt{r^2-\parallel y{\text{∥}}^2-r^2+\parallel x{\text{∥}}^2}& \hfill & \\ \hfill & =\sqrt{\parallel x{\text{∥}}^2-\parallel y{\text{∥}}^2}& \hfill & \\ \hfill & \le \frac{𝜖}{m_J\left(B^{(d-1)}(0,r)\right)}.& \hfill & \end{array}$$

  We now use these observations to construct outer and inner covers. By Remark 1, the highest/lowest point on a box $i$ is some $h_i∕l_i$:

  $$\begin{array}{llll}\hfill {\underline{B^d(0,r)}}_{𝜖}& =\bigcup _{i=1}^z\left[a_i^{(1)},b_i^{(1)}\right]\times \dots \left[a_i^{(d-1)},b_i^{(d-1)}\right]\times \left[0,f\left(h_i\right)\right]& \hfill & \\ \hfill {\overline{B^d(0,r)}}_{𝜖}& =\left(\bigcup _{i=1}^z\left[a_i^{(1)},b_i^{(1)}\right]\times \dots \left[a_i^{(d-1)},b_i^{(d-1)}\right]\times \left[0,f(l_i)\right]\right)& \hfill & \\ \hfill & \cup \left(\bigcup _{i=z+1}^Z\left[a_i^{(1)},b_i^{(1)}\right]\times \dots \left[a_i^{(d-1)},b_i^{(d-1)}\times \left[0,f(l_i)\right]\right]\right)& \hfill & \end{array}$$

  Then, by properties of elementary measure, the difference in the measure of the outer and the inner cover can be calculated as follows. By Remark 2, the maximum difference between the values of $f$ on a single box cannot exceed $𝜖$, and $f$ is obviously bounded by $r$. Thus,

  $$\begin{array}{llll}\hfill m\left({\overline{B^d(0,r)}}_{𝜖}-{\underline{B^d(0,r)}}_{𝜖}\right)& =\sum _{i=1}^n\prod _{j=1}^{d-1}\left[b_i^{(j)}-a_i^{(j)}\right]\cdot f\left(l_i\right)& \hfill & \\ \hfill & +\sum _{i=z+1}^Z\prod _{j=1}^{d-1}\left[b_i^{(j)}-a_i^{(j)}\right]\cdot f\left(l_i\right)-\sum _{i=1}^z\prod _{j=1}^{d-1}\left[b_i^{(j)}-a_i^{(j)}\right]\cdot f\left(h_i\right)& \hfill & \\ \hfill & =\sum _{i=1}^z\prod _{j=1}^{d-1}\left[b_i^{(j)}-a_i^{(j)}\right]\cdot \left(f\left(l_i\right)-f\left(h_i\right)\right)& \hfill & \\ \hfill & +\sum _{i=n+1}^Z\prod _{j=1}^{d-1}\left[b_i^{(j)}-a_i^{(j)}\right]\cdot f\left(l_i\right)& \hfill & \\ \hfill & \le \sum _{i=1}^z\prod _{j=1}^{d-1}\left[b_i^{(j)}-a_i^{(j)}\right]\cdot \frac{𝜖}{m_J\left(B^{(d-1)}(0,r)\right)}& \hfill & \\ \hfill & +\sum _{i=n+1}^Z\prod _{j=1}^{d-1}\left[b_i^{(j)}-a_i^{(j)}\right]\cdot r& \hfill & \\ \hfill & =\frac{𝜖}{m_J\left(B^{(d-1)}(0,r)\right)}\cdot m\left(\underline{B^{(d-1)}(0,r)}\right)& \hfill & \\ \hfill & +r\cdot m\left({\overline{B^{d-1}}}_{𝜖}∕{\underline{B^{d-1}}}_{𝜖}\right)& \hfill & \\ \hfill & \le 𝜖+𝜖=2𝜖.& \hfill & \end{array}$$

  Since our choice of $𝜖$ was arbitrary, by Exercise 1.1.5 $B^{(d)}(0,r)$ must be Jordan measurable. This closes the induction.

Proof. To establish the bounds, note the basic relationship

To see why, pick an arbitrary $x\in {\left[-\frac{r}{\sqrt{d}},\frac{r}{\sqrt{d}}\right]}^d$. Then, $\parallel x{\text{∥}}^2={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^d{x}_i^2\le {\sum }_{i=1}^d\frac{r^2}{d}=r^2$; thus $x\in B^{(d)}(0,r)$. Similarly pick an arbitrary $x\in B^{(d)}(0,r)$. Then

By monotonicity of Jordan measure (Exercise 1.1.6):

Evaluating these elementary measures we obtain:

as desired. □