partial order between sigma algebras Let's call a measurable space $(\Omega_1,\Sigma_1)$ is $\it finer$ than  another measurable space $(\Omega_2,\Sigma_2)$ if there exists an onto measurable mapping $T:\Omega_1 \to \Omega_2$.
Let's call the two measurable spaces equivalent if there exists an one-to-one and onto mapping $T: \Omega_1 \to \Omega_2$ and both $T$ and $T^{-1}$ are measurable.
My question: if $(\Omega_1,\Sigma_1)$ is finer than $(\Omega_2,\Sigma_2)$ and $(\Omega_2,\Sigma_2)$ is finer than $(\Omega_1,\Sigma_1)$, then are they equivalent? In another word, whether the "finer" relationship is a partial order (modulo equivalence)?
Thanks.
 A: No.
Let $\Omega_1 = \Omega_2 = \mathbb{N} = \{0,1,2,\dots\}$.  Consider the $\sigma$-algebras $\Sigma_1 = 2^{\mathbb{N}}$ (all sets are measurable), and $\Sigma_2 = \sigma(\{0,1\},\{2\},\{3\},\dots)$ (0 and 1 have been collapsed into an atom).
The identity map $T_1 : (\Omega_1, \Sigma_1) \to (\Omega_2, \Sigma_2)$ is onto and measurable, so $\Sigma_1$ is finer than $\Sigma_2$.
Let $T_2 : (\Omega_2, \Sigma_2) \to (\Omega_1, \Sigma_1)$ be the map $T_2(0) = 0$, $T_2(1) = 0$, $T_2(n) = n-1$ for $n \ge 2$.  Then $T_2$ is onto and measurable.
But these measurable spaces are not equivalent, since clearly there can be no measurable injection from $\Omega_2$ to $\Omega_1$. (If such an injection $T$ existed, then since every subset of $\Omega_1$ is measurable, the set $T^{-1}(T(\{0\})) = \{0\}$ would have to be measurable in $\Omega_2$, and it isn't.)
Here's a similar example in which the $\sigma$-algebras separate points. Let $\Omega_1 = \Omega_2 = [0,2]$.  Let $\Sigma_1 = 2^{[0,2]}$ be the discrete $\sigma$-algebra.  Let $\Sigma_2$ be the collection of all sets of the form $A \cup B$, where $A \subset [0, 1]$ is Borel and $B \subset (1,2]$ is arbitrary.  Clearly this is a $\sigma$-algebra and separates points (every singleton is measurable).
Let $T_1 : \Omega_1 \to \Omega_2$ be the identity map, which clearly is onto and measurable.  Let $T_2 : \Omega_2 \to \Omega_1$ be
$$T_2(x) = \begin{cases} 0, & 0 \le x \le 1 \\
2(x-1), & 1 < x \le 2 \end{cases}$$
which is clearly onto, and you can check it is measurable.  These spaces are not equivalent because $\Omega_2$ contains non-measurable sets and $\Omega_1$ does not.
If you stick to standard Borel spaces, then this will be true.  Using the axiom of choice you can construct injections from $\Omega_1$ to $\Omega_2$ and vice versa, then Cantor-Schroeder-Bernstein will guarantee that $\Omega_1, \Omega_2$ have the same cardinality.  Standard Borel spaces are completely classified up to equivalence by their cardinality (which is either finite, $\aleph_0$, or $2^{\aleph_0}$).
