Exercise 1.1.11 (Jordan measure and linear transformations)

Pick an arbitrary Jordan measurable set $E\subseteq {ℝ}^d$. By Exercise 1.1.5 this is equivalent to the fact that for any $𝜖>0$ we can find two elementary sets $E_1,E_2\subseteq {ℝ}^d$ such that

This implies

If we manage to demonstrate that $T(E_1)$ and $T(E_2)$ are elementary as well, with $m(T(E_2))-m(T(E_1))=\det <!--\; FUNCTION\; APPLICATION\; -->(T)\cdot (m(E_1)-m(E_2))\le 𝜖$, then we are done. Thus, it suffices to prove the theorem assertion only for elementary sets. By Lemma 1.1.2 we can write represent an elementary set $E$ as a disjoint union of boxes $E={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^N{E}_n$. Knowing that a finite union of Jordan measurable sets $T({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^N{B}_n)={\bigcup }_{n=1}^{N}T(B_n)$ is again Jordan measurable by Exercise 1.1.6, it is sufficient to demonstrate that $T(B_n)$ is Jordan measurable for any box $B_n$.

We use the strategy of expanding the proof to an always more general case with respect to the type of $T$.

(i)

$T$ is an elementary linear transformation
We have further three cases (we only consider the cases with respect to matrix rows; the cases with columns follow analogously):

• $T$ is a Type I linear transformation
  In other words, $T$ exchanges any two rows $i,j$ of a point, and it can be represented by a matrix which is an identity matrix with rows $i$ and $j$ switched:

  $$[T]=\left[\begin{array}{ccccccc}\hfill 1\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \ddots \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill 0\hfill & \hfill \hfill & \hfill 1\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill 1\hfill & \hfill \hfill & \hfill 0\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill 1\hfill \end{array}\right]$$

  Notice that $T$ transforms any box $B=[a_1,b_1]\times \dots \times [a_d,b_d]$ into another box, which is Jordan measurable; namely, it is easy to verify pointwise that

  $$\begin{array}{llll}\hfill & T\left([a_1,b_1]\times \dots \times [a_i,b_i]\times \dots \times [a_j,b_j]\times \dots \times [a_d,b_d]\right)& \hfill & \\ \hfill & =[a_1,b_1]\times \dots \times [a_j,b_j]\times \dots \times [a_i,b_i]\times \dots \times [a_d,b_d].& \hfill & \end{array}$$

  It is only left to demonstrate how $T$ changes the magnitude of the Jordan measure:

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->B\subseteq \mathrm{\text{boxes}}({ℝ}^d):m(T(B))=|\det <!--\; FUNCTION\; APPLICATION\; -->T|\times m(B).$$

  But this follows directly from the fact that $T$ only switches two intervals of $B$; thus, leaving the measure of $T(B)$ unchanged. Since $\det <!--\; FUNCTION\; APPLICATION\; -->(T)=-1$ (Linear Algebra, Theorem 4.5), and thus $m(T(B))=m(B)\cdot 1$.

• $T$ is a Type II linear transformation
  Now suppose that $T$ multiplies the $i$th row of the input with a non-zero scalar $k$, i.e.,

  $$\left[\begin{array}{ccccccc}\hfill 1\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \ddots \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill 1\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill k\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill 1\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill 1\hfill \end{array}\right]$$

  Then it is easy to verify pointwise that

  $$T\left([a_1,b_1]\times \dots \times [a_i,b_i]\times \dots \times [a_d,b_d]\right)=[a_1,b_1]\times \dots \times [k{a}_i,k{b}_i]\times \dots \times [a_d,b_d].$$

  As it was in the previous case, $T$ maps a box to a box, which is Jordan measurable. We need to demonstrate the change in the magnitude of the Jordan measure:

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->B\subseteq \mathrm{\text{boxes}}({ℝ}^d):m(T(B))=|\det <!--\; FUNCTION\; APPLICATION\; -->T|\times m(B).$$

  But this follows directly from the fact that $T$ only enlarges the $I_i$ interval of $B$; thus, leaving the measure of $m(T(B))=(b_1-a_1)\cdot \dots \cdot (k{b}_i-k{a}_i)\cdot \dots \cdot (b_d-a_d)=k\cdot m(B)$. Since $\det <!--\; FUNCTION\; APPLICATION\; -->(T)=m$ (Linear Algebra, Theorem 4.6), the assertion follows.

• $T$ is a Type III linear transformation
  In other words, $T$ adds a row $i$ of the input to its another row $j$.

  $$\left[\begin{array}{ccccccc}\hfill 1\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \ddots \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill 1\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill 1\hfill & \hfill \hfill & \hfill 1\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill 1\hfill \end{array}\right]$$

  In this case, $T$ does not map a box into another box; we have to improvise here. The trick is to look closer at what $T$ maps a box into (the answer is parallelepiped). If the image is Jordan measurable, then we can simply spend another $𝜖$ and approximate this set by another elementary set from below (in case the elementary set is $E_1$) or from above (in case the elementary set is $E_2$). If we denote by $T^1(i,j)$ an elementary linear transformation that interchanges $i$th and $j$th columns, then:

  $$\begin{array}{llll}\hfill T(B)& =\left\{\left(x_1,x_2,\dots ,x_i,\dots ,x_i+x_j,\dots ,x_d\right):x_l\in [a_l,b_l]\right\}& \hfill & \\ \hfill & =T_{i,1}^1{T}_{j,2}^1\left\{\left(x_i,x_i+x_j\dots ,x_1,\dots ,x_2,\dots ,x_d\right):x_l\in [a_l,b_l]\right\}& \hfill & \\ \hfill & =T_{i,1}^1{T}_{j,2}^1\left\{\left(x,x+y\right):x\in I_i,y\in I_j\right\}\times I_3\times \dots \times I_{i-1}\times & \hfill & \\ \hfill & \times I_1\times I_{i+1}\dots \times I_{j-1}\times I_2\times I_{j+1}\times \dots \times I_d& \hfill & \end{array}$$

  This representation has nice properties. Since $T^1$ is proven to map a Jordan measurable set to a Jordan measurable set, while preserving the original volume, it suffices to demonstrate that the set

  $$\left\{\left(x,x+y\right):x\in I_i,y\in I_j\right\}\times I_3\times \dots \times I_{i-1}\times I_1\times I_{i+1}\dots \times I_{j-1}\times I_2\times I_{j+1}\times \dots \times I_d$$

  is Jordan measurable, and has a measure equal to the original set. Knowing how Jordan measure interacts with Cartesian products from Exercise 1.1.16, it suffices to demonstrate that the parallelepiped in ${ℝ}^2$ given by

  $$\left\{\left(x,x+y\right):x\in I_i,y\in I_j\right\}$$

  is Jordan measurable.

       

  Assuming $b_1<b_1$ without loss of generality, it can be tediously verified (exercise) that a parallelepiped can be partitioned into two triangles and a box.

  $$\begin{array}{llll}\hfill \left\{\left(x,x+y\right):x\in I_i,y\in I_j\right\}=& △\left((a_1,a_1+a_2)--(a_1,a_2+b_1)--(b_1,a_2+b_1)\right)& \hfill & \\ \hfill & \cup [a_1,b_1]\times [a_2+b_1,b_2+a_1]& \hfill & \\ \hfill & \cup △\left((a_1,b_2+a_1)--(b_1,b_2+b_1)--(b_1,b_2+b_1)\right)& \hfill & \end{array}$$

  Since all of these figures are Jordan measurable by Exercise 1.1.8, their union must be Jordan measurable as well, and its measure is

  $$m\left(\left\{\left(x,x+y\right):x\in I_i,y\in I_j\right\}\right)=(b_1-a_1)\cdot (b_2-a_2).$$

  Hence, the set

  $$\left\{\left(x,x+y\right):x\in I_i,y\in I_j\right\}\times I_3\times \dots \times I_{i-1}\times I_1\times I_{i+1}\dots \times I_{j-1}\times I_2\times I_{j+1}\times \dots \times I_d$$

  is Jordan measurable, with a measure of $\det <!--\; FUNCTION\; APPLICATION\; -->(T)m(B)=m(B)$ by Exercise 1.1.16. The rest of the proof can be finished by choosing an elementary set with covers from inner and from outer, and by adjusting $𝜖$ accordingly.

(ii)

$T$ is an invertible linear transformation.
By Linear Algebra, Corollary 3.6 (3), $T$ can be represented as a sequence of elementary transformations $T=T_1\dots T_n$. On one hand, by Theorem 4.7 we have $\det <!--\; FUNCTION\; APPLICATION\; -->(T)=\det <!--\; FUNCTION\; APPLICATION\; -->(T_1)\cdot \dots <!--\; FUNCTION\; APPLICATION\; -->\cdot \det <!--\; FUNCTION\; APPLICATION\; -->(T_n)$. On the other hand, by the previous part of this exercise $m(T(B))=m(T_1(\dots (T_n(B))))=\det <!--\; FUNCTION\; APPLICATION\; -->(T_1)(\dots <!--\; FUNCTION\; APPLICATION\; -->(\det <!--\; FUNCTION\; APPLICATION\; -->(T_n)m(B)))=\det <!--\; FUNCTION\; APPLICATION\; -->(T_1)\cdot \dots <!--\; FUNCTION\; APPLICATION\; -->\cdot \det <!--\; FUNCTION\; APPLICATION\; -->(T_n)\cdot m(B)$, and we are done.

(iii)

$T$ is a linear transformation.only class of linear transformation for which the theorem assertion is left to verify are non-invertible linear transformations. Recall from Linear Algebra, Theorem 3.6., that using elementary linear transformations $T_1,\dots ,T_n$ we can transform $T$ into a block matrix of the form $$T=T_1\dots T_n\left[\begin{array}{cc}\hfill I_{r\times r}\hfill & \hfill 0_{r\times (d-r)}\hfill \\ \hfill 0_{(d-r)\times r}\hfill & \hfill 0_{(d-r)\times (d-r)}\hfill \end{array}\right],$$

where $I_{r\times r}$ is an identity matrix of size $r$, and the rest are zero matrices. The last block matrix $T_{n+1}$ will send any box into a lower dimensional box

$$T_{n+1}(B)=T_{n+1}\left([a_1,b_1]\times \dots \times [a_r,b_r]\times \dots \times [a_d,b_d]\right)=[a_1,b_1]\times \dots \times [a_r,b_r]\times \{0\}\times \dots \times \{0\}$$

and the Jordan measure of the above box is obviously zero. On one hand, applying the elementary transformations $T_1\dots T_n$ to obtain the end result will only multiply the sets Jordan measure by scaled determinants, so the result will remain zero. Hence, $m(T(B))=0$. On the other hand, the determinant of the non-invertible linear transformation is zero $\det <!--\; FUNCTION\; APPLICATION\; -->(T)=0$ (Linear Algebra, Corollary 4.7), and we are done.