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I have a question regarding the following definition of an ergodic map from my lecture notes.

Let $(X,\mathcal{M},\mu)$ be a probability space. A function $f\in L^p(X,\mu)$ is invariant if and only if $f(x)=f(\varphi(x)),$ $\mu$ almost everywhere. We note that the following conditions are equivalent.

  1. $f\in L^1(X,\mu)$ invariant $\Longrightarrow$ $f$ constant $\mu$-a.e.
  2. $f\in L^2(X,\mu)$ invariant $\Longrightarrow$ $f$ constant $\mu$-a.e.
  3. $S\in \mathcal{M}$ invariant $\Longrightarrow$ $\mu(S)=0\text{ or }\mu(S)=1.$

Here we say that $S\in \mathcal{M}$ is invariant iff $$\mu(\varphi^{-1}(S)\Delta S)=0$$ where $A\Delta B:=(A\setminus B)\cup (B\setminus A).$

To see the equivalence, note that if $f\in L^1(X,\mu)$ is invariant, then all the sets $S_{\lambda}:=\{x\in X\mid f(x)>\lambda\}$ are invariant, so $(3)\Longrightarrow (1).$ It is clear that $(1)\Longrightarrow (2)\Longrightarrow (3).$ A measure-preserving map satisfying the above equivalent conditions is said to be ergodic.

My question: I can see why $(1)\Longrightarrow (2)\Longrightarrow (3),$ but I am a bit confused on the argument for $(3)\Longrightarrow (1).$ In particular, how does one show that $S_{\lambda}:=\{x\in X\mid f(x)>\lambda\}$ are invariant and why does it imply $(3)\Longrightarrow (1)?$

I wasn't able to come up with a satisfactory explanation to convince myself, so any help will be useful.

neophyte
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1 Answers1

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$x \in S_\lambda$ means $f(x) > \lambda$. If $f$ is invariant, then except on a set of$\mu$-measure $0$, this is equivalent to $f(\phi(x)) > \lambda$, i.e. $\phi(x) \in S_\lambda$. So $S_\lambda$ is invariant. Thus (3) says for every $\lambda$, either $f(x) > \lambda$ $\mu$-almost everywhere or $f(x) \le \lambda$ $\mu$-almost everywhere. Let $y$ be the supremum of those $\lambda$ for which $f(x) > \lambda$ $\mu$-almost everywhere, then show that $f(x) = y$ $\mu$-almost everywhere.

Robert Israel
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