Suppose that $X$ and $Y$ are measurable spaces with the property: there are measurable bijections $f:X \to Y$ and $g:Y \to X$. Is it possible to find non-isomorphic spaces $X,Y$ with this property? One can state similar question in any catgeory and for example in the topological category it is possible.

  • 1
    $\begingroup$ I would bet the answer can be found somewhere in Fremlin... $\endgroup$ – Nate Eldredge Apr 1 '17 at 18:44

(This is an edited example, simpler than the first version.) Yes, there are such spaces $X$ and $Y$.

For $n=0,1,2$, let $A_n$ be a set of cardinality $\aleph_2$ with the $\sigma$-algebra of subsets of cardinality $\leq\aleph_n$ and their complements. Let $X$ be the sum (=disjoint union) of the spaces $A_n$, $n=0,2$. Let $Y$ be the sum of $X$ and $A_1$. Let $f\colon X\to Y$ be a bijection that is identity on $A_0$ and maps $A_2$ onto $A_1 \cup A_2$. Let $g\colon Y\to X$ be a bijection that maps $A_0 \cup A_1$ onto $A_0$ and is identity on $A_2$.

To prove that $X$ and $Y$ are not isomorphic, let $h\colon Y\to X$ be any bijection. Since $h(A_1)=(h(A_1)\cap A_0)\cup(h(A_1)\cap A_2)$, there is $k\in\{0,2\}$ such that the cardinality of $h(A_1)\cap A_k$ is $\aleph_2$. If $k=0$ then $h^{-1}$ is not measurable, and if $k=2$ then $h$ is not measurable. That proves that $X$ and $Y$ are not isomorphic.

  • $\begingroup$ Very nice example! Nitpick: you mean the $\sigma$-algebra of subsets of cardinality at most $\aleph_n$. And at the end, maybe it would help to explain how we know this must hold for $k=0$ or $k=2$? $\endgroup$ – Nate Eldredge Apr 1 '17 at 21:46
  • $\begingroup$ @NateEldredge Good points, thank you. I have updated the answer. $\endgroup$ – user95282 Apr 2 '17 at 2:46

Recall that a Borel standard space is a space that is isomorphic to the unit interval together with its Borel $\sigma$-algebra. Any Borel subset of a complete metric space is either countable or the union of a Borel standard space and a countable space.

The image of a Borel set by an injective Borel map between two Borel standard spaces is a Borel set.

As a corollary, a bijective Borel map between two Borel spaces is a Borel isomorphism and this answers your question in that specific setting.

I don't expect that to be true for all measurable spaces though, but many measurable spaces are Borel standard, so one must go a bit further to find a counterexample. A space may fail to be Borel standard either by being too big (the $\sigma$-algebra is not separable) or by being non Borel e.g. one of these famous non Borel Lebesgue-measurable subsets of $[0,1]$ (or even non measurable but then this is probably harder to deal with these).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.