I am currently reading an article in which the author goes to certain lengths which could be avoided if the following result were true:



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> > *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 $\lim\_{n \to \infty} \frac{1}{n}\int f\_n d\mu <0$.
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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?



Non-oscillatory behaviour in the subadditive ergodic theorem
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.




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*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.


