When is there in a probability space no null sets? I remember my lecturer saying that in some cases there will be no other null set than the trivial one (the empty set), but I can't remember exactly the condition.
I've been thinking and convinced my self that for finite sets, equipped with the power set and a probability measure defined through the counting measure the statement would be true pretty obviously, but how about a countable state space?
My reasoning is that it's a big deal in conditional expectations, since we would then not have to work with versions of the conditional expectation.
Hope someone can help give me some insight,
Henrik
 A: Edit:  I am assuming that all singletons are measurable.  See the comments below.  (This is a reasonable assumption.  In many cases, the probabilty measure lives on a topological space in which singletons are closed and all Borel sets are measurable.)
There is only the trivial null set iff every singleton has positive measure.
So in the countable case, if the underlying space is $\mathbb N$, you could assign to each $\{n\}$, $n\in\mathbb N$, the measure $2^{-n-1}$.
This generates a probability space with no non-trivial set of measure $0$.
If your space is uncountable, you will always have a singleton of measure 0, since there cannot be uncountably many pairwise disjoint measurable sets of positive measure.
As for filtering away null sets, yes, you can always consider the $\sigma$-algebra of measurable set and factor out the ideal of sets of measure 0.  This give you the measure algebra of space, and the only measure 0 element of this algebra is the equivalence class of the empty set, but this process doesn't give you a probability space as such, just a complete Boolean algebra.
