Let $U\subset\mathbb R^n$ be a rectifiable, bounded, open subset and $T:\ \mathbb R^n\longrightarrow\mathbb R^n$ be a linear map. Let $f:\ T(U)\longrightarrow\mathbb R$. Suppose $T(U)$ is rectifiable, $f$ is integrable over $T(U)$ and $f\circ T\hspace{.03cm}\big|\det(J_T)\big|$ is integrable over $U$ (in fact, it's known that $f$ is integrable iff $f\circ T\hspace{.03cm}\big|\det(J_T)\big|$ is integrable).
I want to show that $$\intop_{T(U)}f=\intop_{U}f\circ T\hspace{.03cm}\big|\det(J_T)\big| $$ This theorem has been proved in many materials, but in general case where the map $T$ needs not to be linear. But the technique in these proofs are so difficult and complicated. So how can one prove this theorem for the linear maps ? Is there a more simple way to approach compare to the proofs of general case ? I just have no clue where to start, could someone give me hints to solve this ? Thank you
Here is at least an outline of how one can show this:
For $1\leq i \leq n$ let $x_i(\mathbf v) = \mathbf e_i \cdot \mathbf v$, so that $\mathbf v = \sum_{i=1}^n x_i(\mathbf v).\mathbf e_i$, for all $\mathbf v \in \mathbb R^n$. Then if $T\colon \mathbb R^n \to\mathbb R^n$, If $T\colon \mathbb R^n \to \mathbb R^n$ is any linear map, then $T$ may be written as a composition of ``elementary linear maps'' of the following form:
$$ \begin{split} i)& \quad S_{i,j}(\mathbf v) = \sum_{k\notin \{i,j\}} x_k(\mathbf v)e_k + x_j(\mathbf v)\mathbf e_i + x_i(\mathbf v)\mathbf e_j \quad (i,j \in \{1,\ldots, n\}, i\neq j)\\ ii)& \quad E_{ij}(c)(\mathbf v) = \mathbf v + c.x_j(\mathbf v)e_i \quad (i,j \in \{1,\ldots,n\}, i\neq j, c \in \mathbb R)\\ iii)& \quad D_i(c)(\mathbf v) = \sum_{j\neq i} x_j(\mathbf v)\mathbf e_j + cx_i(\mathbf v)\mathbf e_i. \quad (i \in \{1,\ldots,n\}, c \in \mathbb R). \end{split} $$ (The fact that any invertible linear map is a composition of such maps is essentially a rephrasing of the basic results about row and column reductions.)
Now we say that a subset $B\subseteq \mathbb R^n$ is a "box" if it is of the form $B=[a_1,b_1]\times [a_2,b_2]\times \ldots \times [a_n,b_n]$ where each $[a_i,b_i]$ is a closed non-empty interval in $\mathbb R$. The volume of the box $B$ is defined to be $\nu(B) = \prod_{i=1}^n|b_i-a_i|$. One can then show that if $\mathcal S$ is the $\mathbb R$-linear span of the indicator functions $1_B$ of all boxes $B\subseteq \mathbb R^n$, then there is a unique linear map $\int\colon \mathcal S \to \mathbb R$ such that $\int 1_B = \nu(B)$ for any box $B$. Finally, one extends $\int$ to the class of "Riemann integrable functions" by using, for example, Darboux's upper and lower sums definition.
Now it is immediate from the definition of $\nu(B)$ that if $f\in \mathcal S$, then $\int f\circ T = \int f|\det(T)|$ if $T$ is an elementary linear map of the form $S_{ij}$ or $D_i(c)$, and it not too difficult to argue directly that $\int f\circ T = \int f$ if $T=E_{ij}(c)$ -- note in this case $f\circ T$ need not be in $\mathcal S$.
From this one can show that if $f$ is Riemann integrable, then $\int f\circ T = \int f|\det(T)|$, again in the case where $T$ is an elementary linear map, where in particular this shows that $f\circ T$ is Riemann integrable. But then using induction one sees that $\int f\circ T = \int f|\det(T)|$ for any composition of elementary linear maps, and thus for all linear maps $T$.