When does an action on $X$ makes an well defined action on $L^{\infty}(X)$?

Question

Suppose that $X$ is a compact Hausdroff space. Suppose that $\Gamma$ is a discrete group which acts non-trivially on $X$. Suppose that $\nu$ is a probability measure on $X$. I want to see when does the action of $\Gamma$ on $L^{\infty}(X)$ defined by $s.f(x)=f(s^{-1}x)$ make sense?

I started by taking $\nu=\delta_{x_0}$ for some $x_0 \in X$. Choose $s \in \Gamma$ such that $s.x_0 \ne x_0$. Define $f:X \to \mathbb{C}$ by $$ f(x) = \left\{ \begin{array}{ll} 1 & \mbox{if $x = x_0$}\\ 2 & \mbox{otherwise}\end{array} \right\}$$ Then, $$ s.f(x) = \left\{ \begin{array}{ll} 1 & \mbox{if $x = s.x_0$}\\ 2 & \mbox{otherwise}\end{array} \right\}$$ When $f \in [1_X]$,$s.f \in [2_X]$, where $1_X$ and $2_X$ are the constant functions $1$ and $2$ respectively. Looking at this, I think that for the action to make sense, I have to show that whenever $f_1 \in [f]$, $s.f_1 \in [s.f]$. Looking at the above example, I claim the following:

Action of $\Gamma$ on $L^{\infty}(X)$ defined by $s.f(x)=f(s^{-1}x)$ makes sense iff $\nu$ satisfies the following property: $\nu(E)=0 \iff \nu(s.E)=0, \forall s \in \Gamma$.

Proof: Suppose that $\nu$ satisfies the above mentioned property. Let $f_1 \in [f]$. Then $f_1=f, \nu$-a.e. Let $E=\{x \in X: f_1(x) \ne f(x)\}$. Then $\nu(E)=0$. Let $F=\{x \in X: s.f_1(x) \ne s.f(x)\}$. Then $$F=\{x \in X : f_1(s^{-1}x) \ne f(s^{-1}x)\}=s.E$$ Since $\nu(E)=0 \implies \nu(s.E)=0$, we have that $\nu(F)=0$. Therefore, $s.f_1 \in [s.f]$. By a similar argument we can show that whenever $g \in [s.f] \implies s^{-1}g \in [f]$.

On the other hand suppose that action is well-defined. In addition suppose that $\nu(E)=0$. Let $f=\chi_E$. Then $f \in [0]$. This implies that $s.f \in [0]$. Since $s.\chi_E=\chi_{s.E}$, we have $\chi_{s.E} \in [0]$. Therefore $\nu(s.E)=0$. By a similar argument, replacing $E$ by $s.E$ and multiplying by $s^{-1}$, we can show that $\nu(sE)=0 \implies \nu(E)=0$.

Is this alright?? In particular, I want to know if I framed the well- definedness of the action correctly.

Thanks for the help!!

Answer 1

Yes, your claim is true, and you have a correct proof of it.

Your property is commonly known under the term quasi-invariance; you could say that the action on $L^\infty$ is well-defined iff $\nu$ is quasi-invariant under the action of $\Gamma$.