I would like to construct a non-invariant measure on a compact Lie group but I'm not sure what is allowed and what the consequences are. Take the simplest example of $SO(2)$. The unnormalized invariant measure is $$ d\mu=d\phi. $$ for $0\leq\phi\leq2\pi$. So the integral of a function on a circle with this measure remains the same as one moves around. So far so good.
QUESTIONS
How much can I modify $d\mu$ to still have a legitimate (non-invariant) measure on $S^1$? How about $$ d\eta=\phi d\phi $$ The invariance is broken $d\eta\to d\eta'=(\phi-a)d\phi$ so the integral of a function on $S^1$ will depend on $a$ as well as its normalization. Something else bad happened I can't see? Is it a legitimate measure for $SO(2)$?
Suppose that I am interested in the probabilistic interpretation of the non-invariant measure. It does not preserve the normalization of a function I'm integrating over (the normalization depends on $a$). Can I still use this measure to randomly and uniformly choose points on the circle? For example $d\eta$ `favors' some point over the others but it looks like I can counter the bias just by writing $$ d\eta=1/2d(\phi^2), $$ set $x=\phi^2$ so $\sqrt{x}=\pm\phi$ and choose $x$ uniformly and randomly. So it looks like a complicated it way of undoing non-invariance but other than that the question is: Can I in this way use a non-invariant measure to randomly and uniformly pick points on $S^1$?
If nothing is really wrong, how about more complicated compact Lie groups. For $SU(2)$ the invariant measure is $$d\mu=\sin{\beta}\cos{\beta}\ d\beta\ d\psi\ d\phi.$$ Can I modify it to $$d\eta=\sin^6{\beta}\cos^3{\beta}\ d\beta\ d\psi\ d\phi$$ and still choose the group elements uniformly and at random?