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I found this question very interesting and gave it much thought this week. I believe I have a proof now. I think it would be interesting to see what generalizations one can get from this argument. Below I use probabilistic notation, which I'm more used to.
Let $\{X_n\}_{n\in \Z}$ be a stationary stochastic process, taking values in $\R$. Let $L=\limsup_{n\to -\infty} \frac{X_n}{|n|}$ and $R=\limsup_{n\to \infty} \frac{X_n}{|n|}$.
Theorem: $L=R$, almost surely.
Proof:
It is enough to prove the theorem for ergodic processes. We may also assume, WLOG, that all the values are nonnegative integers.
Let $A$ be the event that for some $n<0$ we have $X_n\ge |n|$ and let $B$ be the event that for some $n>0$ we have $X_n\ge |n|$.
Lemma: $\P(A) \le 2 \P(B) .$
Proof of lemma:
For a given realization of $X_n$, let $I$ be all indices $i$ for which $T^i X$ is in $A$ and let $J$ be all the indices for which $T^i X$ is in $B$. We claim that the density of $J$ is no more than twice the density of $I$, which then implies the conclusion.
$I$ can be written as $\cup_{n \in \Z} \{n,\ldots,n+X_n\}$, while $J=\cup_{n \in \Z} \{n-X_n,\ldots,n\}$. In particular, $J$ is contained in $\bar{J}=\cup_{n \in \Z} \{n-X_n,\ldots,n+X_n\}$. But if we write $I$ as a union of disjoint intervals, then in $\bar{J}$ each of these intervals is extended to the left by at most the length of the interval. Hence, the density of $\bar{J}\supset J$ is at most twice that of $I$. $\blacksquare$
Of course, by symmetry, we also have $\P(B) \le 2 \P(A)$. Let $A_K$ be the event that for some $n<0$ we have $X_n \ge \max(|n|,K)$ and define $B_K$ analogously. By applying the lemma to the process $\{Y_n\}_{n\in \Z}$ defined by $Y_n=X_n$ if $X_n\ge K$ and $Y_n=0$ otherwise, we get that $\P(A_K) \le 2 \P(B_K)$ and vice verse.
In particular, $\lim_{K\to \infty} \P(A_K) =0$ if and only if $\lim_{K\to \infty} \P(B_K) = 0$. These limits necessary exists, since these are monotone decreasing events.
For ergodic processes, we have that $L$ and $R$ are a.s. constant. Now, if $L<1$ a.s. then we have $\P(A_K)\to 0$. In the other direction, if $\P(A_K)\to 0$ then a.s. $L\le 1$. Similar implications hold for $R$ and $B_K$. Hence, we get that if $L<1$ then $R\le 1$ and vice verse. By applying these to a rescaled process $X_n / \alpha$, we get that for any $\alpha$ we have $L<\alpha$ implies $R\le \alpha$ (and vice verse), so we must have $L=R$. $\blacksquare$
Best Answer
You could look at the paper
Mackey, George W. Ergodic theory and virtual groups. Math. Ann. 166 1966 187–207.
which ends up by discussing the notion of ergodic groupoid, andfollow this up with the citations of this paper. The intuitive idea is that while a transitive action of a group corresponds to a subgroup, then what does an ergodic action, correspond to? His theory went through various stages, and ended up with the notion of ergodic groupoid. This introduction of groupoids into analysis is part of the historical background to Noncommutative geometry!
Mackey told me of this work in 1967, and made me realise that there was more in groupoids than I had then thought; the idea did not come just from algebraic topology.
Of course groupoid theory is not the same as category theory, but is in that direction. At least, people who liked category theory found it easy to be happy with groupoids.