Schroeder-Bernstein theorem for measurable spaces

The Cantor-Bernstein-Schroeder theorem of set theory has a counterpart for measurable spaces, sometimes called "the Borel Schroeder-Bernstein theorem", since measurable spaces are also called Borel spaces. This theorem, whose proof is quite easy, is instrumental when proving that two measurable spaces are isomorphic. The general theory of standard Borel spaces contains very strong results about isomorphic measurable spaces, see Kuratowski's theorem. However, (a) the latter theorem is very difficult to prove, (b) the former theorem is satisfactory in many important cases (see Examples), and (c) the former theorem is used in the proof of the latter theorem.

The theorem

Let $X,$ and $Y,$ be measurable spaces. If there exist one-to-one, bimeasurable maps $f\; :\; X\; o\; Y,\; ,$ $g\; :\; Y\; o\; X,\; ,$ then $X,$ and $Y,$ are isomorphic.

Comments

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

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

Proof

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

Examples

Example 1

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).

Example 2

The real line $mathbb\{R\},$ and the plane $mathbb\{R\}^2,$ are isomorphic as measurable spaces. It is immediate to embed $mathbb\{R\},$ into $mathbb\{R\}^2.,$ The converse, embedding of $mathbb\{R\}^2.,$ into $mathbb\{R\},$ (as measurable spaces, of course, not as topological spaces) can be made by a well-known trick with interspersed digits; for example,: $g(pi,100mathrm\{e\})\; =\; g(3.14159,26535,89793dots,\; 271.82818,28459,04523dots)\; =$::: $=\; 20731.18421,85198,\; 22685dots$The map $g\; :\; mathbb\{R\}^2\; o\; mathbb\{R\},$ is clearly one-to-one. It is easy to check that it is bimeasurable. (However, it is not bijective; for example, the number $1/11\; =\; 0.090909dots\; ,$ is not of the form $g(x,y)\; ,$).

References

* S.M. Srivastava, "A Course on Borel Sets", Springer, 1998.:: See Proposition 3.3.6 (on page 96), and the first paragraph of Section 3.3 (on page 94).