Setup
Let $A \in \Bbb{R}^{2\times2}$ be an invertible $2\times2$-matrix and $b \in \Bbb{R}^2$.
Let $t:\Bbb{R}^2 \rightarrow \Bbb{R}^2$ be the linear function given by $t(x)=Ax+b, \; x\in\Bbb{R}^2$
Let $f:\Bbb{R}^2\rightarrow\Bbb{R}$ be a non-negative, continuous function, and define a set function $\mu:\Bbb{B}_2\rightarrow[0,\infty ]$ by $$\mu(B)=\int_B f \, dm_2, \;\;B\in\Bbb{B}_2 $$
Show $t(\mu)$ has density w.r.t. the Lebesgue measure $m_2$ $(=m\otimes m)$ given by $$t(\mu)=\int_\mu \hat{f} dm_2$$
Where
$$ \hat{f}(x)= \frac{1}{|det(A)|}f\circ t^{-1}(x) = \frac{1}{|det(A)|}f(A^{-1}(x-b)). $$
My attempt at a solution I'll adopt the notation $f\cdot m_2$ as a short hand for $\int f \,dm_2$, thus e.g., $\mu=f\cdot m_2$
Since $t$ is bijective, we can write $f$ as $f=f\circ t^{-1}\circ t\cdot m_2$. Now define a new function, $g$, to be $g=f\circ t^{-1}$, yielding $f=g\circ t \cdot m_2$. This allows us to express $t(\mu)$ as: $$t(\mu)=g \cdot t(m_2)$$
So far, so good. Given that $t$ itself is a composition of two functions, namely a $\mathit{translation}$ given as $\tau(x)=x+b$, and an $\mathit{isomorphism}$ given as $s(x)=Ax$. Since I know $m_2$ to be invariant under translation, we get $s(\tau(m_2))=s(m_2)=A(m_2)=\frac{1}{|det(A)|}m_2$. Where the last equality is due to $A$ being an isomorphism. Thus we have $$t(\mu)=g\cdot \frac{1}{|det(A)|}m_2$$ Clearly we are almost there, but I'm just not sure, how to move $\frac{1}{|det(A)|}$ to the front? If I write it as an integral, I get $$t(\mu)=\int_\mu g \;\; d\frac{1}{|det(A)|}m_2 $$ But this does not seem helpful.
I know there's a result which states $$\int g \, dt(m_2) = \int g\circ t \, dm_2 $$ but I'm unsure what meaning we can give to $g \circ t$ in our case.
Any help is much appriciated.