Measure-theoretic analog of homeomorphism and isometry If $(X,\tau_X)$ and $(Y,\tau_Y)$ are topological spaces and $f:X\to Y$ is a continuous bijective function between them such that $f^{-1}$ is also continuous, then the two topological spaces are said to be homeomorphic and they can be essentially identified through $f$, because their topological properties are the same.
Similarly, if $(X,d_X)$ and $(Y,d_Y)$ are metric spaces and $f:X\to Y$ is a bijective isometry between them that preserves distance, then the two metric spaces can be essentially identified through $f$, because their metric properties are the same.
Now, suppose that $(X,\Sigma_X)$ and $(Y,\Sigma_Y)$ are measurable spaces and $f:X\to Y$ is a measurable bijective function between them such that $f^{-1}$ is also measurable. Can one say that the two spaces can be essentially identified from a measure-theoretic point of view? Is there a distinguished name for this phenomenon (analogous to “homemorphic” and “isometric” in the previous two cases)?
(Please refrain from using category-theoretic arguments in your answer, I'm not familiar with it. Thank you!)
 A: 
Two measurable spaces $(X,\mathcal {A})$ and $(Y,\mathcal{B})$ are called isomorphic  if there exists a bijection $f:X\to Y$ such that $f$ and $f^{-1}$ are measurable (such $f$ is called an isomorphism).

This is from the EOM article Measurable space. Same definition is found, e.g., in Real Analysis by Royden.
There is a weaker notion of isomorphism between $\sigma$-algebras: a bijection $\Phi:\mathcal A\to\mathcal B$ that commutes with algebra operations. (Using $A\subseteq B\iff A\cup B=B$ etc, one can show that preservation of binary operations implies that countable unions and intersections are preserved as well.) In general, an isomorphism of $\sigma$-algebras does not have to be induced by any point mapping between spaces. 
A: As pointed out in the comments and answers, there is a disctinction between isomorpshisms of measurable spaces and measure spaces. The former one is defined in the answer by Rafflesia adnoldii, and is very much similar to homeomorpshim in spirit: we just compare two structures on sets being $\sigma$-algebras (rather than topologies). There is a famous Borel isomorphism theorem stating that two Borel subsets of Polish spaces are isomorphic as long as they are equipotent, that is have the same cardinality (clearly such a strong statement does not hold for homeomorphisms). 
Yet, I would not say that isomoprhisms between measurable and measure spaces differ in the same way as they do for topological and metric spaces. When you endow measurable spaces with measures, you make the structure finer (similar to fixing a metric over a topological space) and so you expect the definition of isomoprhism to be more restrictive. Indeed, this is the case if we are talking about measure-preserving isomoprhisms between measure space: they have to be isomoprhisms between measurable spaces, and in addition they must preserve the measures. However, there also exists a notion of being isomoprhic modulo null sets, which on the one hand requires the measures to be preserve, but on the other hand only cares about non-null sets. For example, $X = \{1\}$ and $Y = X\cup \Bbb R$ endowed with Borel $\sigma$-algebras are not isomoprhic as measurable spaces, whereas $(X,\delta_1)$ and $(Y,\delta_1)$ are isomoprhic modulo zero as measure spaces. Thus, in contrast to the fact that isomorphic metric spaces are always isomoprhic as topological spaces, isomoprhic measure spaces may not be isomoprhic as measurable spaces: it depends on the notion of isomoprhism we use for measure spaces - and notice that both (strong and modulo zero) version of isomoprhism are natural in relevant applications.
