Can anyone help me out with proofs/counterexamples? I'm working on an operator-valued multiplicative ergodic theorem and need what may(?) be a well-known fact. This fact (if true) would help me get rid of an annoying asymmetry in the conclusion of a theorem.

I'm assuming that $T$ is an invertible ergodic transformation of a probability space $(X,\mathcal B,\mu)$ and that $f$ is a measurable (but not necessarily integrable) function on $X$.

Is it true that $f(T^nx)/n \to 0$ a.e. if and only if $f(T^{-n}x)/n\to 0$ a.e.?

Comments:

(1) In probability language if you define $X_n=f(T^nx)$ this is a stationary sequence of random variables. Borel-Cantelli 1 shows that if $\mathbb E |X_0|<\infty$ (i.e. $f\in L^1$) then $X_n/n\to 0$ as $n\to\pm\infty$: The probability that $|X_n|/n > 1/k$ is $\mathbb P(|X_0| > n/k)$. The sum of this series is over-estimated by $k\mathbb E|X_0| < \infty$. Hence almost surely $|X_n|/n < 1/k$ for all large $n$. Since this is true for all $k$ you get $X_n/n\to 0$.

If the $X_n$ are i.i.d. random variables, then the converse holds by Borel-Cantelli 2. This shows that for i.i.d. random variables, the boxed question has an affirmative answer.

(2) In the case where the $X_n$ are not i.i.d. I believe there are examples where $X_n/n\to 0$ almost surely even though $\mathbb E|X_0|=\infty$.

In the case that $f\in L^1$, both sides of the implication in the main question are true. The unresolved case is $f\not\in L^1$.