Let $(\Omega,A,\mu)$ be a measure space. For a mapping $T: \Omega \to \Omega$ which together with its inverse is an measurable bijection. I want to show that for every $f$ measurable non negative holds the following \begin{equation*} T(f\mu) = (f \circ T^{-1})T(\mu). \end{equation*}
So, this is what i have tried, but i think that I'm not getting anywhere.
Let $g$ a measurable non negative function, then trying to prove the equality using integrals i start with the left side \begin{align*} \int g dT(f\mu) &= \int g \circ T d(f\mu) \\ &= \int(g \circ T)\cdot fd\mu. \end{align*} And this is the part where i tried writing like $T\circ T^{-1}$ or keep using some ideas of integration of image measure but does not seem to reach something. So i know it probably seems like i have not done too much work, but i have already tried a lot, so any hints or any help would be really helpfull.
Thank you in advance.
It's just a matter of unfolding the definitions correctly. On the one hand we have
$$T(f\mu)(A):=(f\mu)(T^{-1}(A)):=\int_{T^{-1}(A)} f d\mu$$
On the other hand the transformation theorem implies that
$$(f \circ T^{-1})T(\mu)(A):=\int_A f \circ T^{-1} dT(\mu)=\int 1_A(f \circ T^{-1}) dT(\mu)$$ $$=\int 1_A(f \circ T^{-1}) \circ T d\mu=\int_{T^{-1}(A)} f \circ T^{-1} \circ T d\mu$$ $$=\int_{T^{-1}(A)} fd\mu$$