Superharmonic function and super martingale. The definition (from Durrett's "Probability: Theory and Examples"):

Superharmonic functions. The name (super martingale)
  comes from the fact that if $f$ is superharmonic (i.e., f has continuous derivatives of
  order $\le 2$ and $\partial^2 f /\partial^2 x_1^2 + · · · + \partial^2 f /\partial^2 x_d^2)$, then
  $$
f (x) \ge \frac 1 {|B(0, r)|} \int_{B(x,r)} f(y) dy
$$
  where $B(x, r) = \{y : |x − y| \le r\}$ is the ball of radius $r$, and $|B(0, r)|$ is the volume of the ball of radius $r$.

The question is

Suppose $f$ is superharmonic on $R^d$. Let $\xi_1 , \xi_2 , ...$ be i.i.d. uniform
  on $B(0, 1)$, and define $S_n$ by $S_n = S_{n−1} + \xi_n$ for $n \ge 1$ and $S_0 = x$. Show that $X_n = f (S_n)$ is a supermartingale.

Here the filtration should be $\mathcal{F}_n = \sigma \{X_n, X_{n-1}...\}$.
I know to prove $X_n, n \ge 0$ is a super martingale, we only need to show
$$
E\{X_{n+1} ~|~ \mathcal{F}_n\} \leq X_n.
$$
This is easy when $n=0$. But for $n > 0$, I've no idea how to, or if it is possible, to derive the formula of the conditional expectation.

$f(S_{n+1})=f(S_n+\xi_{n+1})$
Using Did's Hint: If $g(x)=E(f(x,Y))$ then $g(X)=E(f(X,Y)|X)$
$\Rightarrow E(f(S_n+\xi_{n+1})|\mathcal F_n)=g(S_n)$
but $\displaystyle g(x)=\int f(x+y)\nu(dy)$
and in our case $\nu(dy)=\mathbf 1_{B(0,1)}\frac{1}{|B(0,1)|}dy$ (because $\xi_i$ are uniform on $B(0,1)$)
$\displaystyle\Rightarrow g(S_n)=\int  f(S_n+y)\mathbf 1_{B(0,1)}\frac{1}{|B(0,1)|}dy=\frac{1}{|B(0,1)|}\int_{B(S_n,1)}f(y)dy\le f(S_n)$
