# Non-isomorphic measurable spaces

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.

• I would bet the answer can be found somewhere in Fremlin... – 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.

• 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 $k=2$? – Nate Eldredge Apr 1 '17 at 21:46
• @NateEldredge Good points, thank you. I have updated the answer. – 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.

Theorem
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).