Given $\mu(B)=\int_B fdm_2, B \in \Bbb{B}_2$, with $f(x)$ = $1\over|det(A)|$$f(A^{-1}(x-b))$, and $m_2$ the Lebesgue measure on $\Bbb{R}^2$.
Futhermore, we have $t(\mu)=\hat{f}\cdot m_2$, where $\hat{f}(x)=\frac{1}{|det(A)|}f\circ t^{-1}(x)= \frac{1}{|det(A)|}$$f(A^{-1}(x-b)) $
$\mathbf{problem}$:
Assume $f(x)=g(x^Tx)$ for $x \in \mathbb{R}^2$. Show that if $A$ is orthonormal, and $b=0$, then $\mu$ is invariant under $t$, i.e. $t(\mu)=\mu$
I would appreciate some help getting started. I can't figure out how $f(x)=g(x^Tx)$ helps me here. I know, since A is orthonormal, $|det(A)|=1$, and $x^{T}x =(A^{-1}x)^TA^{-1}x$, which reduces both $f$, but I can't don't know how to procede from here. Is it useful to know that $m_2$ is translation invariant?