I have the product space $(\mathbb{R}^2,\mathbb{B}_2)$ and a function $t:\mathbb{R}^2\to\mathbb{R}^2$ with $t(x)=Ax+b$. I have to show that $t$ is invariant for $b=0$ and $A$ is a $\mathbb{R}^{2\times 2}$ invertible matrix.
The density of $t$ regarding the Lebesgue measure is, with b=0: $$t(\mu)=\hat{f}\cdot m_2=\int_B\frac{1}{|\det(A)|}f(A^{-1}x)dm_2,\quad B\in\mathbb{B}_2$$ and $\mu(B)$ is: $$\mu(B)=f\cdot m_2=\int_B f(x)dm_2,\quad B\in\mathbb{B}_2.$$
I have shown that $|\det(A)|=1$ (I am not going to write the proof here), so I get: $$t(\mu)=\int_Bf(A^{-1}x)dm_2=\int_B f(x) dm_2$$ I am unsure about whether the last step is correct, however, I think so, as when the determinant is 1 of $A$, then it also 1 of $A^{-1}$ so the area should not change?