Fix $a_1,\dots,a_n\in\mathbb{R}\setminus\{0\}$ and let $L:\mathbb{R}^n\to\mathbb{R}^n$ given by $L(x_1,\dots,x_n)=(x_1/a_1,\dots,x_n/a_n)$.
Let $f:\mathbb{R}^n\to\mathbb{R}$ be a funtion such that there exsists a sequence of step functions, $(h_k)_{k\in\mathbb{N}}$, such that $\|f-h_k\|_1\to 0$ (i.e. $f$ is integrable).
Now I consider $f\circ L$. If we write the step functions as follows $h_k=\sum_ic_i^{(k)} 1_{Q_i^{(k)}}$ with cuboids $Q_i^{(k)}$ in $\mathbb{R}^n$, then $h_k\circ L=\sum_ic_i^{(k)} 1_{L(Q_i^{(k)})}$, is that right?
Now, let $M\subset\mathbb{R}^n$ measurable, i.e. the characteristic function $1_M$ is Lebesgue integrable. Very elementary, and without using that $L$ and $L^{-1}$ are continuous, how to prove that $1_{L^{-1}(M)}$ is Lebesgue integrable? Furthermore, I have to show/conclude that $|(det(L)|\int 1_{L^{-1}(M)}dx=\int 1_{M}dx$. But before I do the second part, I need a hint for the first part. I appreciate any hints for this and a correction or a confirmation of $h_k\circ L$.
I think you do need to note that $L$ is a homeomorphism; otherwise, $L(M)$ might not be Borel for every Borel set $M$. And you need to know the basics of how the Lebesgue measure is developed. At least, you need this in the proof I know, (adapted from Rudin's Real and Complex Analysis):
Let $Q$ be any box in $R^n$, of side length $2^{-k}$. Then,
$m(L(Q))= \frac{2^{-kn}}{\prod a_i} =\frac{1}{\prod a_i}\cdot 2^{-kn}=(\det L)m(Q).$
Thus,
$\int 1_Q= \frac{1}{\det L} \int 1_{L(Q)}=(\det L^{-1})\int 1_{L(Q)}.$
This formula is true for $any$ invertible linear transformation, so we may substitute $L^{-1}$ for $L$, to obtain
$\int 1_Q=(\det L)\int 1_{L^{-1}(Q)}.$
This shows that the result is true for boxes. But then, since every open set is a countable disjoint union of such boxes, the result is also true for any open set (use the countable additivity of $m.)$ This in turn shows that the result is true for all Borel sets, and finally, since every Lebesgue meaurable set is a union of a Borel set and a set of measure zero which of course does not affect the integral, you have the formula on the whole of $\frak M.$