Remark

— 1) Assume $\mu$ invariant. Then $|\mu|$, $\mathscr{R}(\mu)$, $\mathscr{I}(\mu)$ are invariant. If $\mu$ is real, then $\mu^+$ and $\mu^-$ are invariant. 2) Assume $\mu$ relatively invariant and nonzero. There exists, for every $s \in G$, a unique complex number $\chi(s)$ such that

(7)

and the function $\chi$ on $G$ is a representation of $G$ in $\mathbf{C}^*$, called the multiplier of $\mu$. The formula (5) then gives

and formula (6) gives

With the conventions made above, (7) may also be written

3) Since $|\gamma(s)\mu| = \gamma(s)(|\mu|)$, to say that $\mu$ is quasi-invariant amounts to saying that $|\mu|$ is quasi-invariant.

If $\mu$ is quasi-invariant and $\mu'$ is another measure on $X$ equivalent to $\mu$, then $\gamma(s)\mu'$ is equivalent to $\gamma(s)\mu$, hence to $\mu$, hence to $\mu'$, and so $\mu'$ is quasi-invariant. To say that $\mu$ is quasi-invariant under $G$ therefore means that the class of $\mu$ is invariant under $G$.

For $\mu$ to be quasi-invariant, it is necessary and sufficient that the set of locally $\mu$-negligible subsets of $X$ be invariant under $G$ (Ch. V, §5, No. 5, Th. 2), or again that, for every $\mu$-negligible compact subset $K$ of $X$ and for every $s \in G$, $sK$ be $\mu$-negligible (loc. cit., Remark).

If $\mu$ is quasi-invariant, then the support of $\mu$ is invariant under $G$. In particular if $G$ is transitive in $X$ (A, I, §5, No. 5, Def. 6), this support is either empty (if $\mu = 0$) or equal to $X$ (if $\mu \neq 0$).