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?