Q:

Density Transformation Formulars

From Klenke - "Probability Theory": Let $\mu$ be a measure on $\mathbb{R}^n$, which admits a density $f\colon \mathbb{R}^n\rightarrow [0,\infty)$. Furthermore, let $\varphi\colon A\rightarrow B$ be a continuously differentiable bijection. Then the image measure $\mu\circ\varphi^{-1}$ is given by 
$\displaystyle f_\varphi(x)=\frac{f(\varphi^{-1}(x))}{|\det(\varphi'(\varphi^{-1}(x)))|}$
Now the following is the integral transformation I am used to (from Forster - "Analysis III"):
$\displaystyle \int _{{\varphi (A )}}f(y)\,{\mathrm  {d}}y=\int _{A }f(\varphi (x))\left|\det(\varphi' (x))\right|{\mathrm  {d}}x\;.$
How can I prove equivalence of these two integral transformation formulars?

A:

These are the same formulas, but it is not $\varphi$ but rather $\varphi^{-1}$ which is used in the first one. Since
$$d(\varphi\circ\varphi^{-1})(x)=d\varphi(\varphi^{-1}(x))\circ d\varphi^{-1}(x)=\mathrm{id}$$ 
(I use $d$ for the differential instead of your $'$), you get that $|\det(d\varphi^{-1}(x))|=\frac{1}{|\det(d\varphi(\varphi^{-1}(x)))|}$ and your second formula rewrites as
$$\int_{\varphi^{-1}(B)}f(y)dy=\int_Bf(\varphi^{-1}(x))|\det(d\varphi^{-1}(x))|dx=\int_Bf(\varphi^{-1}(x))\frac{1}{|\det(d\varphi(\varphi^{-1}(x)))|}dx.$$

