Is every $\sigma$-algebra generated by a partition? I know that every finite $\sigma$-algebra is generated by a finite partition, but is every infinite $\sigma$-algebra also generated by "kind of" partition? Can anyone help provide a explanation or reference?
Thank you!
 A: An atom in a measurable space $(M,\mathcal{M})$ is a nonempty set $A\in\mathcal{M}$ such that $B\in\mathcal{M}$ and $B\subseteq A$ implies that either $B=\emptyset$ or $B=A$. If $\mathcal{M}$ is generated by a countable family of sets, so it is countably generated,  every point in $M$ is in a unique atom and the atoms form a partition of $M$. 
A countably generated measurable space $(M,\mathcal{M})$ is strongly Blackwell if any two countably generated sub-$\sigma$-algebras with the same atoms coincide. They are called that way because David Blackwell has shown that analytic spaces are strongly Blackwell in this paper. This includes for example the unit interval with the Borel sets. So for strongly Blackwell spaces, there is a way to read the question so that the answer is yes.
Not every countably generated measurable space is strongly Blackwell. Just take the unit interval and take the $\sigma$-algebra to be generated by the open sets and one non-Borel set $N$. Then the atoms are the singleton sets but so are the atoms of the proper sub-$\sigma$-algebra consisting of the Borel sets. So the answer to the question is in general no.
A: Borel $\sigma$ algebra $B[0,1])$ of the set $[0,1]$ is infinite but it is not generated  by any partition of the interval $[0.1]$. Assume the contrary and let $(X_i)_{i \in I}$ be such a partition.Since singletons belong to  $B[0,1])$ we deduce that each $X_i$ must be singleton. Then $[0,1/2]$ does not belong to $B[0,1])$ 
because it is not presented as union of countable singletons. The latter relation is the contradiction and we have proved that  $B[0,1])$ is an example of infinite  $\sigma$-algebra which is not generated by any partition of $[0.1]$.
