Nonnegative Superharmonic Function is Constant for $d>2$? I have to do the following:

Let $\alpha>0$ be fixed, $(X_i)_{i\geq 1}$ be i.i.d., $\mathbb R^{d}$-valued random variables, uniformly distributed on the ball $B(0,a)$. Set $\displaystyle S_{n}=x+\sum_{i=1}^{n}X_{i}$. Let $f$ be a superharmonic function. Show that $f(S_n)$ defines a supermartingale, and show that if $d\leq 2$, any nonnegative superharmonic function is constant.


I have shown that $f(S_n)$ defines a supermartingale, but how do I address the second part of the question? And what if $d>2$? The definition of superharmonic:
$$f(x)\geq \frac{1}{|B(0,r)|}\int_{B(x,r)}f(y)dy$$
I was looking at Byron Schmuland's answer on A stronger version of discrete "Liouville's theorem":
Since $f$ is superharmonic, the process $M_n:=f(X_n)$ is a supermartingale. Because $f\geq 0$, the process $M_n$ is a nonnegative supermartingale and so must converge a.s. by the martingale convergence them. That is, $M_n\rightarrow M_{\infty}$ a.s. Since superharmonic functions are lower semicontinuous, you can conclude that $f\le M_\infty$ at every point. On the other hand, the supermartingale property implies that $E f(X_n)$ does not increase with $n$. Therefore, $M_\infty\le f(x)$ where $x$ is the starting point. Since the starting point was arbitrary, it follows that $f(x)\equiv M_\infty$. 
But $(X_n)$ is irreducible and recurrent, and so visits every neighborhood of every point infinitely often. Thus with probability $1$, $f(X_n)$ takes on every $f$ value infinitely often.  One can say that the process is "neighborhood recurrent" (for $d\le 2$).
Thus $f$ is a constant function, since $M_n=f(X_n)$ can't take on distinct values infinitely often and still converge. However if $d>2$: In dimensions $d>2$ there are positive superharmonic functions such as $\min(1,|x|^{2-d})$. (Truncated by $1$ just to make it bounded and continuous; $|x|^{2-d}$ is superharmonic too). We can conclude that the process described is not neighborhood recurrent when $d>2$, hence the result does not remain true.

I would change "visits every state infinitely often" to "visits every neighborhood of every point infinitely often". This is because the process you described does not actually hit every point; in fact it has probability zero of hitting any given point. One can say that it's "neighborhood recurrent" (for $d\le 2$) to emphasize this fact.
Since superharmonic functions are lower semicontinuous, you can conclude that $f\le M_\infty$ at every point. On the other hand, the supermartingale property implies that $E f(X_n)$ does not increase with $n$. Therefore, $M_\infty\le f(x)$ where $x$ is the starting point. Since the starting point was arbitrary, it follows that $f(x)\equiv M_\infty$. 
In dimensions $d>2$ there are positive superharmonic functions such as $\min(1,|x|^{2-d})$. (Truncated by $1$ just to make it bounded and continuous; $|x|^{2-d}$ is superharmonic too). You can conclude that the process you described is not neighborhood recurrent when $d>2$.
