Exercise 1.2.21 (Lebesgue measure and linear transformations)

We have to demonstrate that

In case that $m(E)$ is infinite, the proof becomes trivial. Thus, assume that $m(E)<\infty$. We use the fact that we have already proven this theorem for elementary sets in Exercise 1.1.11 (Jordan measure and linear transformations).

• Pick an aribtrary $𝜖>0$. Pick an aribtrary ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }|B_n|$ from the left-hand side corresponding to a box cover ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{B}_n$ of $E$. We give ourselves an epsilon of the room and demonstrate that $\det <!--\; FUNCTION\; APPLICATION\; -->(T)\underset{n=1}{\overset{\infty }{\sum <!--\; FUNCTION\; APPLICATION\; -->}}|B_n|+𝜖$ is contained on the right-hand side. Notice that ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }T(B_n)$ is a (not necessarily box) cover of $T(E)$, which is a countable union of Jordan measurable sets, and is thus Lebesgue measurable. We further have

  $$m\left(\underset{n=1}{\overset{\infty }{\bigcup }}T(B_n)\right)\le \underset{n=1}{\overset{\infty }{\sum }}m\left(T(B_n)\right)=\underset{N\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\underset{n=1}{\overset{N}{\sum }}m\left(T(B_n)\right)=\det <!--\; FUNCTION\; APPLICATION\; -->(T)\times \underset{N\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\underset{n=1}{\overset{N}{\sum }}m\left(B_n\right)=\det <!--\; FUNCTION\; APPLICATION\; -->(T)\times \underset{n=1}{\overset{\infty }{\sum }}m\left(B_n\right)$$

  Since ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }T(B_n)$ is a Lebesgue measurable, we can spend an $𝜖$ to find a box cover ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{C}_n^{(𝜖)}$ such that ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }|C_n^{(𝜖)}|-m({\bigcup }_{n=1}^{\infty }T(B_n))\le 𝜖$. The cover ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{C}_n^{(𝜖)}\supseteq {\bigcup }_{n=1}^{\infty }T(B_n)\supseteq T(E)$ is contained in the right-hand side set, and we have

  $$\underset{n=1}{\overset{\infty }{\sum }}|C_n^{(𝜖)}|-\det <!--\; FUNCTION\; APPLICATION\; -->(T)\underset{n=1}{\overset{\infty }{\sum }}|B_n|\le 𝜖$$

• If $T$ is not invertible, then its has a determinant of $0$, and it sends some cover ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }T(B_n)$ to a low dimensional box cover of Lebesgue measure zero. Thus, consider the case where $T$ is invertible. Then $T^{-1}$ enjoys all of the properties that $T$ does, and the proof of this direction is similar to the first part (e.g. we use that fact that if ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{C}_n$ is a box cover of $T(E)$, then ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{T}^{-1}(C_n)$ is a Lebesgue measurable cover of $E$).