Schröder–Bernstein theorem for measurable spaces

The phrase "${\displaystyle f}$ is bimeasurable" means that, first, ${\displaystyle f}$ is measurable (that is, the preimage ${\displaystyle f^{-1}(B)}$ is measurable for every measurable ${\displaystyle B\subset Y}$), and second, the image ${\displaystyle f(A)}$ is measurable for every measurable ${\displaystyle A\subset X}$. (Thus, ${\displaystyle f(X)}$ must be a measurable subset of ${\displaystyle Y,}$ not necessarily the whole ${\displaystyle Y.}$)

An isomorphism (between two measurable spaces) is, by definition, a bimeasurable bijection. If it exists, these measurable spaces are called isomorphic.

First, one constructs a bijection ${\displaystyle h:X\to Y}$ out of ${\displaystyle f}$ and ${\displaystyle g}$ exactly as in the proof of the Cantor–Bernstein–Schroeder theorem. Second, ${\displaystyle h}$ is measurable, since it coincides with ${\displaystyle f}$ on a measurable set and with ${\displaystyle g^{-1}}$ on its complement. Similarly, ${\displaystyle h^{-1}}$ is measurable.

The open interval (0, 1) and the closed interval [0, 1] are evidently non-isomorphic as topological spaces (that is, not homeomorphic). However, they are isomorphic as measurable spaces. Indeed, the closed interval is evidently isomorphic to a shorter closed subinterval of the open interval. Also the open interval is evidently isomorphic to a part of the closed interval (just itself, for instance).

The real line ${\displaystyle \mathbb {R} }$ and the plane ${\displaystyle \mathbb {R} ^{2}}$ are isomorphic as measurable spaces. It is immediate to embed ${\displaystyle \mathbb {R} }$ into ${\displaystyle \mathbb {R} ^{2}.}$ The converse, embedding of ${\displaystyle \mathbb {R} ^{2}.}$ into ${\displaystyle \mathbb {R} }$ (as measurable spaces, of course, not as topological spaces) can be made by a well-known trick with interspersed digits; for example,

g(π,100e) = g(3.14159 265…, 271.82818 28…) = 20731.184218 51982 2685….

The map ${\displaystyle g:\mathbb {R} ^{2}\to \mathbb {R} }$ is clearly injective. It is easy to check that it is bimeasurable. (However, it is not bijective; for example, the number ${\displaystyle 1/11=0.090909\dots }$ is not of the form ${\displaystyle g(x,y)}$).