Non-oscillatory behaviour in the subadditive ergodic theorem I am currently reading an article in which the author goes to certain lengths which could be avoided if the following result were true:


Lemma (proposed): Let $T$ be an ergodic measure-preserving transformation of a probability space $(X,\mathcal{F},\mu)$, and let $(f_n)$ be a sequence of integrable functions from $X$ to $\mathbb{R}$ which satisfy the subadditivity relation $f_{n+m} \leq f_n \circ T^m + f_m$ a.e. for all integers $n,m \geq 1$. Suppose that $f_n(x) \to -\infty$ in the limit as $n \to \infty$ for $\mu$-a.e. $x \in X$. Then $\lim_{n \to \infty} \frac{1}{n}\int f_n d\mu <0$.


Via the subadditive ergodic theorem, this effectively states that if $f_n(x) \to -\infty$ almost everywhere then it must do so at an asymptotically linear rate. The supposed lemma would also be equivalent to the statement that if $\frac{1}{n} f_n(x) \to 0$ almost everywhere, then for almost every $x$ the sequence $(f_n(x))$ must return infinitely often to some neighbourhood of $0$ which is not a neighbourhood of $-\infty$. If the sequence $(f_n)$ is additive rather than just subadditive then this last formulation of the result follows from a well-known theorem of G. Atkinson, but the more general subadditive case is less clear.
If the lemma were true then several parts of the paper I am reading would be redundant, which makes me wonder whether it is in fact false. Yet it seems rather plausible. Does anyone know whether this result is true or not?
 A: I believe the Lemma you propose is true, via a relatively straightforward adaptation of the proof given for the additive case in Giles Atkinson, Recurrence of co-cycles and random walks, J. Lond. Math. Soc. (2) 13 (1976), 486-488.  (I assume this is the result you were referencing.)  This may already be known and written somewhere -- I can't speak to that. In case it's not, here's a proof.
Proof of Lemma.  By the subadditive ergodic theorem, there exists a measurable function $f\colon X\to \mathbb{R}\cup\{-\infty\}$ such that $\lim \frac 1n f_n(x) = f(x)$ for $\mu$-a.e. $x\in X$ and $\lim \frac 1n \int f_n\,d\mu = \int f\,d\mu$.  So our task is to show that $\int f\,d\mu < 0$.
To this end, given $x\in X$, let $M_x = \{n \mid f_n(x) \geq -1 \}$, and observe that $M_x$ is finite $\mu$-a.e.  Thus writing $A_n = \{x\in X \mid \# M_x < n \}$, we see that there exists $N$ such that $\mu(A_N) > \frac 12$.
Furthermore, given $x\in X$, write $L_x = \{ k \mid T^k(x) \in A_N \}$.  Since $\mu(A_N) > \frac 12$, we see that $\mu$-a.e. $x$ has
$$
(*)\qquad\qquad\qquad\# L_x \cap [1,n] \geq \frac n2\qquad\qquad\qquad\qquad\qquad\quad
$$
for all sufficiently large $n$.  We fix such an $x$ and show that $f(x) < 0$.
Let $k_0$ be the smallest element of $L_x$.  We define $k_i\in L_x$ recursively with the property that
$$
f_{k_i}(x) < f_{k_0}(x)-i,
$$
as follows.  Let $J_i$ be the $N$ smallest elements of $L_x \cap (k_i,\infty)$.  Because $T^{k_i}(x)\in A_N$, there exists $k_{i+1}\in J_i$ such that $k_{i+1} - k_i \notin M_{T^{k_i}(x)}$.  In particular, we have
$$
f_{k_{i+1} - k_i}(T^{k_i}(x)) < -1.
$$
Now subadditivity gives
$$
f_{k_{i+1}}(x) \leq f_{k_i}(x) + f_{k_{i+1} - k_i}(T^{k_i}(x)) < f_{k_0(x)} - i-1.
$$
The next observation to make is that by $(*)$, we have $k_i \leq 2Ni$ for all sufficiently large $i$.  Thus we have
$$
\frac 1{k_i} f_{k_i}(x) \leq \frac 1{2Ni}(f_{k_0}(x) - i),
$$
and sending $i\to\infty$, we obtain $f(x) \leq -\frac 1{2N}$.  Since this holds for $\mu$-a.e. $x$, we have $\int f\,d\mu \leq -\frac 1{2N} < 0$, which completes the proof.

Note.  It's quite important in this proof that we're dealing with negative values of $f_n$; the proof would fail if we tried to show that $f_n(x)\to+\infty$ for $\mu$-a.e. $x$ implies that $\int f\,d\mu > 0$.  I'm not sure if the result is true in this case.
