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Consider a measurable space $(\Omega, \mathcal{F})$. Given a partition $\mathcal{P}$ of $\Omega$ into measurable sets we consider the map $\pi\colon \Omega\to \mathcal{P}$ associating to each $\omega\in \Omega$ the atom $P\in \mathcal{P}$ that contains it. In the space $\mathcal{P}$ we consider the largest $\sigma$-algebra in which $\pi$ is measurable, that is, $A$ is a measurable subset of $\mathcal{P}$ if and only if $\pi^{-1}(A)$ is measurable.

I was reading Rokhlin's book "On the fundamental ideas of measure theory", and he says that ( see chapter 1, No. 2, page 5) the map $\pi$ constructed above satisfies the following property: If $B\subset \Omega$ is measurable then $\pi(B)$ is measurable.

If I understood right this is equivalent to say that $$ \bigcup_{ P\in \mathcal{P},P\cap B\neq\emptyset,} P $$ is measurable. Since the partition $\mathcal{P}$ and the set $B$ may not be countable, I don't see why he assertion is true.

Can someone help me to clarify this fact or suggest some reference on elementary properties of factor spaces?

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    $\begingroup$ The smallest $\sigma$-algebra on $\mathcal{P}$ that makes $\pi$ measurable is $\{\mathcal{P},\emptyset\}$. You are talking about the largest $\sigma$-algebra that makes $\pi$ measurable. $\endgroup$ – Michael Greinecker Dec 6 '16 at 17:33
  • $\begingroup$ Yes, you are right!!! $\endgroup$ – Didi Dec 6 '16 at 20:13
  • $\begingroup$ I have edited... $\endgroup$ – Didi Dec 6 '16 at 20:38
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The general statement as written in the question is false. For example, consider the projection $\pi\colon[0,1]\times[0,1]\to[0,1]$, both $[0,1]\times[0,1]$ and $[0,1]$ with their Borel $\sigma$-algebras. There is a Borel subset of $[0,1]\times[0,1]$ whose $\pi$-image in $[0,1]$ is not Borel. To make the statement true, some assumptions must be added. For example, if $(\Omega,\mathcal{F})$ is a standard Borel space (i.e. isomorphic to $[0,1]$ with its Borel $\sigma$-algebra) and $B\in\mathcal{F}$ then $\pi(B)$ is measurable with respect to every measure on $\mathcal{P}$. A good reference is Fremlin's Measure Theory, chapters 42 and 43.

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    $\begingroup$ Rokhlin does not work with Borel spaces. This is the paper where he introduces the notion of Lebesgue spaces. Here, the setting is abstract measure spaces and the $\sigma$-algebras are complete. $\endgroup$ – coudy Dec 3 '16 at 18:54
  • $\begingroup$ @coudy Your comment is about the question, isn't it, not about the answer. $\endgroup$ – user95282 Dec 6 '16 at 19:58
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    $\begingroup$ Your answer does not provide a counterexample to the statement in Rokhlin's paper because the Borel σ-algebra is not complete. Rokhlin considered complete σ-algebras. As you mention, the image of a Borel set is in the completion of the Borel σ-algebra. Arguably, one must look at the paper in order to really understand what is the question here. And preferably at the russian paper, because I think the english translation is not clear wrt to the question at hand. $\endgroup$ – coudy Dec 6 '16 at 20:21
  • $\begingroup$ @coudy Thank you for your clarification. My answer is for the question as asked. It seems that OP should have asked a different question, in view of your comment about Rokhlin's paper. And likely that different question would be answered by the second part of my answer above. $\endgroup$ – user95282 Dec 6 '16 at 21:17

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