I am reading 'Analysis on Lie groups, an introduction' by Faraud and don't understand the following statement
… the image by the map Ad of the Haar measure $\mu$ of $SU(2)$ is equal to the Haar measure $\nu$ of $SO(3)$.
We know that $Ad:$ $SU(2)\rightarrow SO(3)$ is a covering of order 2. I also know the Haar measure on $SU(2)$, but cannot see why the above statement holds.
I appreciate any hint, comment or answer.
Trying to clarify my brief comment: on a compact (Hausdorff) topological group, up to scalar multiples there is a unique left-and-right translation-invariant (regular, Borel) measure. Thus, any translation-invariant measure on $SO(3)$ is necessarily a multiple of the Haar measure on $SO(3)$. Ok, given a set $E\subset SO(3)$, let $E'$ be its inverse image in $SU(2)$, and say that the measure of $E$ in $SO(3)$ is the measure of $E'$ in $SU(2)$. This measure is invariant: given $k\in SO(3)$, choose any $k'$ in $SU(2)$ mapping down to $k$. The inverse image of $k\cdot E$ is $k'\cdot E'$, which has the same measure as $E'$. And the total mass is $1$ if the measure on $SU(2)$ was normalized in this fashion.