Suppose that $f$ is a non-negative measurable map on a probability space $(\Omega, \lambda)$. Let $T: (\Omega, \lambda) \rightarrow (\Omega, \lambda)$ be measure-preserving ergodic transformation such that $g(\omega) := f(T\omega) - f(\omega)$ is in $L^{1}(\Omega)$. I'm having trouble seeing the following two facts:
- There exists a $C$ such that the set $\{\omega \in \Omega: f(\omega)\leq C \}$ is of positive measure.
- By the ergodic theorem, the set $\{n \in \mathbb{N}: f(T^{n}(\omega)) \leq C \}$ has positive density.
For the first point, I think should follow from the fact that $g$ is integrable and hence bounded almost everywhere but I am having trouble writing a precise proof as to why. For the second point, I'm not quite sure what "density" means here.
Context: These facts came up in a paper I'm reading.
It is understood that $f$ takes only finite values because there is a problem with the definition of $g$ if you allow $f$ to take the value $\infty$. As $C$ increases to $\infty$ the set $\{x:f(x)>C\}$ decreases to the empty set. Hence its measure tends to $0$. 1) follows for this.
Let $A$ be the set in 1) with $C$ chosen so that $\lambda (A) >0$. Let $A_n=T^{-n}(A)$. By the Ergodic Theorem $\frac 1 n \sum\limits_{k=1}^{n} I_{A_n} \to \lambda (A)$ a.s. and in $L^{1}$. This means $\frac 1 n \#\{n: \omega \in A_n\} \to m(A) >0$. This is the definition of $\{n: \omega \in A_n\}$ having positive density.