Assume we have a $2$-dimensional sample space $(\Omega, B, P_\Omega)$, with $\Omega =\mathbb R^2$ with borel measure and probability measure $P$, where the axes are simply equal to random variables $X$ and $Y$. That is, if $\omega=(x,y)$ then $X(\omega)=x$ and $Y(\omega)=y$. Now assume we have the probability density function given by the Radon-Nikodym derivative w.r.t. Lebesgue measue: $$\forall A\in B: \quad P_\Omega(A)=\int_Af_\Omega(\omega)d\lambda(\omega)$$ Now also assume that we have a transformation from this sample space to another sample space $(\Gamma, B,P_\Gamma)$, which also is equal to $\mathbb R^2$, and where the axes are again equal to random variables $W$ and $V$. And now assume we have a mapping $g:\Omega \to \Gamma$, such that $P_\Gamma$ is the probability measure induced by $g$ and $P_\Omega$.
Then we must have that $$\forall A\in B: \quad \int_Af_\Omega(\omega)d\lambda(\omega)=\int_{g(A)}f_\Gamma(\gamma)d\lambda(\gamma)$$
My question is, how do we find $f_\Gamma$? I know what the result has to be, I am just not sure how to derive it using the setup I have here. I want to get rid of the $g(A)$, so that we can get rid of the integral.
I am tempted to do the following:
$$\int_{g(A)}f_\Gamma(\gamma)d\lambda(\gamma)=\sup\left\{\sum_{a\in P_{g(A)}}\inf_{\gamma \in a}(f_\Gamma(\gamma))\cdot \lambda (a) : \text{partition } P_{g(A)}\right\}=
\sup\left\{\sum_{b\in P_{A}}\inf_{\omega\in b}(f_\Gamma(g(\omega)))\cdot \lambda (g(b)) : \text{partition } P_{A}\right\}=\int_{A}f_\Gamma(g(\omega))d\lambda(g(\omega))$$
But I am not sure if that last expression is rigorously defined. I know that the result we should get to is $|Dg|$, but I'm not sure how to get there.
