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. $x_n$ takes 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 $\liminf S_nf/n=-\infty$.

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 $\liminf S_nf/n=-\infty$.

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. $x_n$ takes the value $2^{n^2}$ with probability $2^{-n}$), you get $\liminf S_nf/n=-\infty$. Similarly in the second case, summing $2n$ terms, you see that $\liminf S_nf/n=-\infty$.

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. $x_n$ takes 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 $\liminf S_nf/n=-\infty$.