Please check whether the proof is correct or not.

Question

Please check my solving. I want to know where to be wrong or illogical, or where logical jumps are.

Problem

Let $y=Tx$ be a nonsingular linear transformation of $\mathbb{R}^n$.

If $\displaystyle\int_E f(y) dy$ exists, then $\displaystyle\int_E f(y) dy = |\det T| \int_{T^{-1}E}f(Tx)dx$.

(Since $E$ is a subset of $\mathbb{R}^n$, I have to write $\mathbf{x}$ and $\mathbf{y}$ instead of $x$ and $y$, respectively. However, I will use just $x$ and $y$ since it is hard to type \mathbf for every $x$ and $y$.)

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Solve

1. $f=f^+ - f^-$
2. Since $f^+ \ge 0$, There exists $\{f_k\}_{k=0}^{\infty}$ such that $f_k$ are nonnegative simple functions and $f_k\nearrow f^+$. That is $$f_k=\sum_{i=1}^Na_i^{(k)}\chi_{E_i^{(k)}}\quad\text{and}\quad f_k\le f_{k+1}\quad\text{and}\quad\lim_{k\to\infty}f_k=f^+$$
3. $\displaystyle\int_E f^+(y)dy = \int_E \lim_{k\to\infty}f_k(y)dy$
4. Since $f_k\nearrow f^+$ and $f_k\ge 0$, by M.C.T., \begin{align} \int_E f^+(y)dy=\int_E \lim_{k\to\infty}f_k(y)dy &= \lim_{k\to\infty} \int_E f_k(y)dy \\ &= \lim_{k\to\infty} \int_E \sum_{i=1}^{N} a^{(k)}_i\chi_{E^{(k)}_i}(y)dy \\ &= \lim_{k\to\infty} \sum_{i=1}^{N} \int_E a^{(k)}_i\chi_{E^{(k)}_i}(y)dy \\ &= \lim_{k\to\infty} \sum_{i=1}^{N} a^{(k)}_i|E^{(k)}_i| \\ &= \lim_{k\to\infty} \sum_{i=1}^{N} a^{(k)}_i\frac{|T^{-1}E^{(k)}_i|}{|\det T^{-1}|} \\ &= \lim_{k\to\infty} \sum_{i=1}^{N} a^{(k)}_i |\det T| |T^{-1}E^{(k)}_i| \\ &= \lim_{k\to\infty} \sum_{i=1}^{N} \int_{T^{-1}E} a^{(k)}_i|\det T|\chi_{T^{-1}E^{(k)}_i}(x)dx \\ &= \lim_{k\to\infty} \sum_{i=1}^{N} \int_{T^{-1}E} a^{(k)}_i|\det T|\chi_{E^{(k)}_i}(Tx)dx \\ &= |\det T|\lim_{k\to\infty} \int_{T^{-1}E} \sum_{i=1}^{N} a^{(k)}_i\chi_{E^{(k)}_i}(Tx)dx \\ &= |\det T|\lim_{k\to\infty} \int_{T^{-1}E} f_k(Tx)dx \\ &= |\det T| \int_{T^{-1}E} \lim_{k\to\infty}f_k(Tx)dx \\ &= |\det T| \int_{T^{-1}E} f^+(Tx)dx \\ \end{align}
5. For the same way for $f^-$, $$\int_E f^-(y)dy = |\det T| \int_{T^{-1}E} f^-(Tx)dx$$
6. From the process 1, \begin{align} \int_E f(y)dy &= \int_E \left(f^+-f^-\right)(y)dy \\ &= \int_E f^+(y)dy - \int_E f^-(y)dy \\ &= |\det T| \int_{T^{-1}E} f^+(Tx)dx - |\det T| \int_{T^{-1}E} f^-(Tx)dx \\ &= |\det T| \int_{T^{-1}E} \left(f^+ - f^-\right)(Tx)dx \\ &= |\det T| \int_{T^{-1}E} f(Tx)dx \end{align}

Answer 1

Looks ok to me, if you justify that $$ |TE|=|\det T|\,|E|. $$

The result you are trying to prove is the change of variable formula. Just assuming that $f$ is measurable you get directly from the substitution formula that $$ \int_Ef(y)\,dy=\int_{T^{-1}E} f(Tx)\,|\det T|\,dx, $$ where the only verification to be made is that the Jacobian of $T$ is $T$ itself.