Let $U, V \in \mathbb{R}^n$ be open sets and $\varphi: U \rightarrow V$ a ($C^1$)-diffeomorphism. So we have the identity for every measurable $A$ in $U$ that $\mu(\varphi(A))=\int_{A}^{} |\det(d\varphi)| d\mu$.
Suppose further that f is a function $V \rightarrow \mathbb{R}$ then f is integrable is equivalent to the fact that the function $U \in x \mapsto f(\varphi(x)) *|\det(d\varphi_x)|\in \mathbb{R}$ is integrable. In this case we get $\int_{V}^{} f d\mu = \int_{U}^{} (f \circ \varphi)*|\det(d\varphi)| d\mu$ (what I want to show).
In a first step I want to use the first fact about the n-dimensional Lebesgue-measure for step functions but I do not really get along with this. I come to a point where I have the sum $\sum_{i = 1}^{n} a_k \mu(B_k)$ where $B_k$ are the measurable sets on which we get the value $a_k$ for f(x).
Now I want to use the first fact for the sum to get the second fact but I do not see how to apply the first fact because I do not know whether there is a measurable A so that $\varphi(A)=B_k$. How can I show that?