Group action on a topological space and Haar integral (A question on a proof in a Palais' paper)

Question

The paper is: R. Palais, "On the existence of slices for actions of non-compact Lie groups", Annals of Mathematics, vol. 73, 1961. In the proof of Proposision 1.2.6, p. 301, $X$ is a Tychonoff space and a locally compact group $G$ acts on it, $f$ is a map from $X$ to a linear space $V$ on which $G$ acts too and $μ$ is right invariant Haar measure on $G$. I would like an explanation of the following equality: $\int{g^{-1}f(gγx)dμ(g)}=\int{γg^{-1}f(gx)dμ(g)}$, where $g,γ \in G$, $x\in X$. Because of the invariance of Haar measure shouldnt $γ$ be vanished completely?

Answer 1

You have\begin{align}\int g^{-1}f(g\gamma x)\,\mathrm d\mu(g)&=\int\gamma\gamma^{-1}g^{-1}f(g\gamma x)\,\mathrm d\mu(g)\\&=\int\gamma(\color{red}{g\gamma})^{-1}f(\color{red}{g\gamma}x)\,\mathrm d\mu(g)\\&=\int\gamma g^{-1}f(gx)\,\mathrm d\mu(g),\end{align}since $\mu$ is right invariant. In fact, if $F(g)=\gamma g^{-1}f(gx)$, then$$\int\gamma g^{-1}f(gx)\,\mathrm d\mu(g)=\int F(g)\,\mathrm d\mu(g),$$whereas$$\int\gamma(g\gamma)^{-1}f(g\gamma x)\,\mathrm d\mu(g)=\int F(g\gamma)\,\mathrm d\mu(g).$$