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Let $(\Omega,\Sigma,\mu)$ be a probability space. A $\mu$-atom is an $A\in\Sigma$ such that $\mu(A)>0$ and for all $B\in\Sigma$ such that $B\subseteq A$, either $\mu(B)=\mu(A)$ or $\mu(B)=0$ holds.

Now let $(X,\mathcal{X})$ be a measurable space with the property that for all $x,y\in X$ there is $B\in\mathcal{X}$ such that $x\in B$ and $y\notin B$.

Is every measurable function $f:\Omega\to X$ almost surely constant? That is, is there for every $\mu$-atom $A\in\Sigma$ some point $x\in X$ such that $f^{-1}\big(\{x\}\big)$ has $\mu$-outer measure $\mu(A)$?

It is not hard to prove that the answer is yes if $\mathcal{X}$ is generated by a countable family of sets, but the approach does not generalize.

Motivation: I face the problem in economic modelling. I want the agents to be represented by a probability space, where the atomic part represents "big" players. The behavior is given by a measurable function and want to ensure that the behavior of each big player is well defined.

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No.

Let $\Omega$ be an uncountable set, and $\Sigma$ the $\sigma$-algebra consisting of all countable and co-countable sets. Let $\mu$ be the probability measure assigning measure 0 to all countable sets, and measure 1 to all co-countable sets. Note every co-countable set is an atom. Let $(X, \mathcal{X}) = (\Omega, \Sigma)$ (note that $\Sigma$ separates points), and let $f : \Omega \to \Omega$ be the identity function, which is certainly measurable. Then for each $x$, $f^{-1}(\{x\})$ is a singleton and hence has measure 0.

Indeed, we could replace $f$ with any bijection of $\Omega$.

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  • $\begingroup$ Somehow I managed to oversee the most obvious example of a strange atom. Thank you! $\endgroup$ – Michael Greinecker Nov 25 '13 at 21:44
  • $\begingroup$ I think you overlooked the example, but did not oversee it. $\endgroup$ – Gerald Edgar Nov 26 '13 at 2:40
  • $\begingroup$ @GeraldEdgar Thanks. English is not my native language, so that is quite helpful. $\endgroup$ – Michael Greinecker Nov 26 '13 at 7:28

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