Proposition 1

— Assume that G is locally compact. Let $\mu$ be a nonzero relatively invariant measure on X. Then its multiplier $\chi$ is a continuous function on G.

For, let $f \in \mathscr{K}(X)$, S the support of $f$, $s_0$ a point of G, and V a compact neighborhood of $s_0$ in G; then, the set

is compact in X; by Lemma 1 and formula (8), $\chi(s^{-1}) \langle \mu, f \rangle$ depends continuously on $s$; if $f$ is chosen so that $\langle \mu, f \rangle \neq 0$, one sees that $\chi$ is continuous.

Now let G be a topological group operating continuously on the right in a locally compact space X; for $s \in G$ and $x \in X$, let $xs$ be the transform of $x$ by $s$. We denote by $\delta_X(s)$, or $\delta(s)$, the homeomorphism of X defined by

(1')

We have

(2')

By transport of structure, one defines the action of $\delta(s)$ on functions and measures on X:

(3')

(4') (5') (6')

We agree to write $d\mu(xs)$ in place of $d(\delta(s)\mu)(x)$, and (5') then takes the form

One defines in an analogous manner the measures on X invariant, relatively invariant and quasi-invariant under G. If $\mu$ is relatively invariant, its multiplier $\chi$ is defined by the formulas

If one regards the group $G^0$ opposite to G as operating in X by $(x,s) \mapsto xs$, then $\mu$ is relatively invariant under $G^0$ with the same multiplier $\chi$.

Finally, let G be a locally compact group. It operates on itself by left and right translations, according to the formulas $\gamma(s)x = sx$, $\delta(s)x = xs^{-1}$. Then

All of the foregoing is applicable here, thus we have on G the concepts of measures that are left-invariant, right-invariant, relatively left-invariant, relatively right-invariant, left quasi-invariant, right quasi-invariant (however, see Nos. 8 and 9).

The mapping $x \mapsto x^{-1}$ is a homeomorphism of G onto G. For every function $f$ on G, we define the function $\check{f}$ on G by

For every measure $\mu$ on G, we define the measure $\check{\mu}$ by

In other words,

If A is a $\check{\mu}$-integrable set, then $A^{-1}$ is $\mu$-integrable and

We agree to write $d\mu(x^{-1})$ in place of $d\check{\mu}(x)$, and (14) then takes the form