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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,


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I suppose one could always filter away any null sets using an equivalence relation. Or not? – Raskolnikov Jul 28 '12 at 10:14
Maybe I'm misunderstanding you but it's pretty much the equivalence relations I wanna get rid off - or at least know when I don't have to use. – Henrik Jul 28 '12 at 10:29
@Raskolnikov: When you do that, you get a measure algebra and not a measure space. – Michael Greinecker Jul 28 '12 at 10:32
up vote 5 down vote accepted

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.

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Is it possible that no singleton is measurable? – Zhen Lin Jul 28 '12 at 10:29
Wow that is of course obvious, wonder why I couldn't make that chain of thought. Thank you very much! – Henrik Jul 28 '12 at 10:31
@Zhen Lin: I see. Yes, I suppose it is possible that no singleton is measurable. I was somehow thinking that we are in a reasonable topological space and at least the Borel sets are measurable. But there is actually no reason for this assumption. You could, for example, take a probability space with all singletons of positive measure and split every singleton into two points. Now the you declare only those sets to be measurable the come from measurable sets in the original space by splitting points. Now the new singletons are not measurable, but there is only the trivial null set. – Stefan Geschke Jul 28 '12 at 10:38
@Stefan: I guess the correct generalization is that the underlying $\sigma$-algebra is atomic and every atom has positive measure. – Michael Greinecker Jul 28 '12 at 10:43
The space I'm working on is defined as to having a 1-1 map onto R where the map and its inverse are both measurable (is this called bimeasurable?). Would that make singletons measurable? (And is it correct such spaces are called standard-borel?) – Henrik Jul 28 '12 at 10:47

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