Let $G$ be a finitely generated group acting on a set $S$ (on the right). Define the heirarchy of "marginal sets" as follows:

The emptyset is 0-marginal.
A set E is $(k+1)$-marginal if $E$ can be covered by finitely many sets $A$ such that for some $k$-marginal set $B$ and some $g$ in $G$, if $a$ is in $A$ and $a \cdot g^i$ is in $A$ for some $i > 0$, then there is a $j < i$ such that $a \cdot g^j$ is in $B$.
a set is marginal if it is $k$-marginal for some $k$.
So if we recall that $A$ is wandering if, for some $g$ in $G$, the sets $A \cdot g^i$ $(i < \infty)$ are pairwise disjoint, then 1-marginal just means that the set is a finite union of wandering sets.

The point is that marginal sets are assigned measure 0 by any invariant measure (or even by any invariant exhaustive submeasure --- see my other recent question).

Now for the question(s):

Are there non-amenable actions which are aperiodic but not marginal? (Notice that if $G$ is a non-amenable torsion group acting on itself then the emptyset is the only marginal set.)
Can $S$ be $(k+1)$-marginal but not $k$-marginal for some $k > 0$?
Has the notion of marginality been considered (and given a name)? (I used the notion to give estimates in my proof of a lower bound for the Folner function for Thompson's $F$, but it seems this "must" have been considered before.)
thenon-wandering set? There are sets which are wandering with respect to the action of the group (i.e. there is some element of the group g such that $E \cdot g^k$ $(k < \infty)$ is a pairwise disjoint family, where $E$ is the wandering set) and then every other set is non wandering. – Justin Moore Nov 12 '10 at 14:57