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Consider description of the model solution in this mathexchange post: Proving that an ergodic and invariant map is constant a.e. We have that $(E, F, \mu)$ is a probability space, $\theta:E\to E$ is an ergodic transformation and $f:E\to \mathbb{R}$ is measurable function such that $f\circ \theta = f$ everywhere.

What isn't entirely clear to me from the model solution or from @user160629's answer https://math.stackexchange.com/a/852384/877219 is that how do we argue that the defined $c \equiv \inf \{a \in \mathbb{R}: \mu(f > a) = 0\}$ is finite? I understand it has something to do with $f \circ T = f$ on $E$ but I am not sure what exactly.

Cartesian Bear
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1 Answers1

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Here is one way to think about it: for $a\in\mathbb{R}$, put $E_a=\{x\in E\,|\, f(x)\in ]a,\infty[\}$. $f$ being measurable, any $E_a$ is measurable; $f$ being $\theta$-invariant, any $E_a$ is $\theta$-invariant; $(\mu,\theta)$ being ergodic, any $E_a$ has either zero xor full measure.

Note that if $a<b$, then $E_{a}\supseteq E_{b}$. Consequently for any sequences $a_\bullet,b_\bullet:\mathbb{Z}_{\geq1}\to \mathbb{R}$ with $\lim_{n\to\infty} a_n=-\infty$ and $\lim_{n\to\infty} b_n=\infty$, we have $E = \bigcup_{n\in\mathbb{Z}_{\geq1}}E_{a_n}$ and $\emptyset = \bigcap_{n\in\mathbb{Z}_{\geq1}}E_{b_n}$. Thus there are $a^\ast,b^\ast\in\mathbb{R}$ such that $\mu(E_{a^\ast})=1$ and $\mu(E_{b^\ast})=0$, whence the set

$$\{a\in \mathbb{R}\,|\, \mu(E_a)=0\}$$

is nonempty and bounded from below, and as such, its infimum is a real number.

Alp Uzman
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