Suppose there exist an action of group $G$ on $L^{\infty}(X,\mu)$ via measure preserving transformation ( the left translation Koopmans action). $\mu$ is probability measure. Suppose the action is ergodic on the subalgebra of simple functions. Can we say the action is ergodic on $L^{\infty}(X,\mu)$?
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1$\begingroup$ what does "ergodic on the subalgebra of simple functions" mean? $\endgroup$– mathworker21Commented Feb 26, 2019 at 3:44
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$\begingroup$ Only constant functions fixed under the action if the action restricted on the dense subalgebra. $\endgroup$– mathloverCommented Feb 26, 2019 at 3:50
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1 Answer
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Of course. Say $X = X_1\sqcup X_2$ with $X_1,X_2$ both $G$-invariant and $\mu(X_1), \mu(X_2) \not \in \{0,1\}$. Then let $f = 1$ on $X_1$ and $0$ on $X_2$. This is a $G$-invariant simple function.
answered Feb 26, 2019 at 3:54