TITLE: non-integrable subadditive ergodic theorem
QUESTION [8 upvotes]: Dear MO_World,
I have (another) question about relaxing the assumptions in the sub-additive ergodic theorem. Apologies if this is something I should know already...
There are a number of statements of the Kingman sub-additive ergodic theorem and its extensions. Here is a fairly typical version:
Let $T\colon X\to X$ be a measure-preserving transformation of a probability space. Let 
$(f_n)$ be a sequence of measurable functions from $X$ to $[-\infty,\infty)$ satisfying $f_{n+m}(x)\le f_n(x)+f_m(T^nx)$ for every $m,n\in\mathbb N$ and $x\in X$. Suppose further that $f_1^+$ is integrable. Then $f_n(x)/n$ is convergent for almost every $x$. 
There are also statements about convergence in the mean in certain situations, but I focus on pointwise convergence.
My general question is: When can you remove the integrability assumption? 
In the following example the conclusion holds, even though the hypothesis fails.
Consider a distribution on the positive reals so that if $(X_n)$ is a sequence of iid random variables with the given distribution, then $\mathbb E(\min(X_1,X_2,\ldots,X_n))=\infty$ for all $n$. (As an example, consider a random variable that takes the value $2^{k^3}$ with probability $2^{-k}$ for each $k\ge 1$. Let $M_n$ be $\min(X_1,X_2,\ldots,X_n)$. Then $\mathbb P(M_n=2^{k^3})\approx 2^{-nk}$ so that $\mathbb E M_n=\infty$.
Let $\Omega= \lbrace 2^{k^3}\colon k\ge 1 \rbrace ^\mathbb Z$ equipped with the Bernoulli probability measure arising from the distribution above.
Now define the sequence $f_n(\omega)$ as follows: $f_{2n}(\omega)=0$; $f_{2n+1}(\omega)=\min_{0\le j\le 2n+1}\omega_j$. This sequence of functions is clearly sub-additive. Also $0\le f_n(\omega)/n\le \omega_0/n$, so that $f_n/n\to 0$ everywhere.
My specific question is as follows:

Suppose that $(f_n)$ is a sub-additive sequence of functions taking values in $[-\infty,\infty)$. If we assume that $\limsup f_n(x)/n \lt \infty$ a.e., then does it follow that $f_n(x)/n$ converges almost surely? [ The above example shows that we cannot expect convergence in norm, even if the $f_n$ are non-negative ]

REPLY [5 votes]: Having read Terry's answer, I found a counterexample of my own. It's quite a bit simpler (although less nice in the sense that the liminf is $-\infty$ rather than 0).
It's actually an additive process rather than sub-additive one. Define a transformation $T$ on $X=\mathbb N^\mathbb Z\times\lbrace 0,1\rbrace$ by $T(x,0)=(x,1)$ and $T(x,1)=(Sx,0)$ where $S$ is the shift. Define the function $f(x,0)=-x_0$ and $f(x,1)=x_0$. Let the measure on $X$ be $\mu\times c$ where $c$ is counting measure and $\mu$ is a horrendously non-integrable iid process on $\mathbb N^\mathbb Z$. 
If you look at the values of $f$ along an orbit, you see the values $-x_0,x_0,-x_1,x_1,-x_2,x_2,\ldots$ if you start from an `even' point or $x_0,-x_1,x_1,-x_2,x_2,\ldots$ if you start from an odd point. In the first case the partial sums $S_nf(x)$ are always non-positive and take the value 0 infinitely often, so that $\limsup S_nf(x)/n=0$. In the second case, the partial sums $S_nf(x)$ are bounded above by $x_0$ and take this value infinitely often, so that again $\limsup S_nf(x)/n=0$. 
On the other hand, in the first case, summing $2n+1$ terms, $S_{2n+1}f/(2n+1)=-x_n/(2n+1)$. If the distribution is sufficiently nasty (e.g. the $x_i$'s take the value $3^n$ with probability $2^{-n}$), you get $\liminf S_nf/n=-\infty$. Similarly in the second case, summing $2n$ terms, you see that $\liminf S_nf/n=-\infty$.