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It is well known, that the topological spaces $\mathbb{R^n}$ and $\mathbb{R}^m$ are non-homeomorphic for $n\neq m$. However for a formal proof of this one usually needs strong methods like local homology.

Now one can ask the same question (are there bi-measurable bijections) for the measurable space $(\mathbb{R}^n, Bor(\mathbb{R}^n))$, where $Bor(\mathbb{R}^n)$ denotes the Borel $\sigma$-algbra. Of course $(\mathbb{R}^n, Bor(\mathbb{R}^n))$ and $(\mathbb{R}^m, Bor(\mathbb{R}^m))$ are generated by the non-homeomorphic topologies, but I don't see how this would imply that the measurable spaces are non-isomorphic. Is there any sophisticated theory on measurable spaces comparable to the one for topological spaces one can use for this?

If the measurable spaces are isomorphic, how about the measure spaces $(\mathbb{R}^n, Bor(\mathbb{R}^n), \mu^n)$ and $(\mathbb{R}^m, Bor(\mathbb{R}^m), \mu^m)$ where $\mu^n$ is the n-dimensional Lebesgue measure?

If they are non-isomorphic: can there be measurable bijections between $(\mathbb{R}^n, Bor(\mathbb{R}^n))$ and $(\mathbb{R}^m, Bor(\mathbb{R}^m))$ for $n\neq m$?

Takirion
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  • Obviously, if you take the completion wrt the Lebesgue measure, they are isomorphic, so if there's a difference its in the sets of lebesgue measure 0. One possible path forward: we may decompose $\mathbb{R}^n$ into $\mathbb{R}$ indexed sheets of dimension $\mathbb{R}^{n-1}$, induce the relevant measures, and you have that $\mathbb{R}^n$ contains an uncountable infinity of copies of $\mathbb{R}^{n-1}$. Can a borel sigma algebra do that with itself?? – user24142 Oct 09 '17 at 22:30
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    Why is it obvious, that they are isomorphic if we consider the Lebesgue algebra? – Takirion Oct 09 '17 at 22:33
  • I intended that to be read as "the spaces are isomorphic modulo 0", as described by https://en.wikipedia.org/wiki/Standard_probability_space – user24142 Oct 09 '17 at 23:14

1 Answers1

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All the measure spaces $(\mathbb{R}^n, Bor(\mathbb{R}^n), \mu^n)$ are isomorphic. One quick way to prove this is to observe that $(\mathbb{R}, Bor(\mathbb{R}), \mu)$ is isomorphic to the product measure space $\mathbb{Z}\times\{0,1\}^\mathbb{N}$, where $\mathbb{Z}$ has counting measure and $\{0,1\}$ has the uniform probability measure. It is clear that a product of any finite number of copies of this measure space is isomorphic to itself, since you can choose a bijection $\mathbb{Z}^n\to\mathbb{Z}$ and $(\{0,1\}^\mathbb{N})^n$ is again just a countably infinite product of copies of $\{0,1\}$.

So, why is $\mathbb{R}$ isomorphic to $\mathbb{Z}\times\{0,1\}^\mathbb{N}$? Morally, this is because you can take the map $f:\mathbb{Z}\times\{0,1\}^\mathbb{N}\to\mathbb{R}$ given by $f(n,s)=n+b(s)$, where $b(s)\in[0,1]$ is the number with binary expansion $s$. This map is easily seen to be measurable and measure-preserving. Unfortunately, it is not quite a bijection, since dyadic rationals have two different binary expansions. To fix this, let $S\subset\{0,1\}^\mathbb{N}$ be the set of sequences which are eventually constant. Since both $\mathbb{Z}\times S$ and $f(\mathbb{Z}\times S)$ are countably infinite, we can choose some bijection $g$ between them. Then define $f':\mathbb{Z}\times\{0,1\}^\mathbb{N}\to\mathbb{R}$ by $f'(n,s)=g(n,s)$ if $(n,s)\in \mathbb{Z}\times S$ and $f'(n,s)=f(n,s)$ otherwise. This $f'$ is now a bijection, and it is easy to see it is in fact an isomorphism of measure spaces.

Eric Wofsey
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    I did not know of this result! Does it mean that I can use $\mathbb{R}^n$ to integrate over $\mathbb{R}^m$ via an isomorphism?! That looks so odd! – Ivan Di Liberti Aug 09 '18 at 10:59
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    Why is it easy to show that it is an isomorphism of measure spaces? I agree that it's a measure space isomorphism mod sets of measure 0, but to the OP's question, why are $(\mathbb R, \mathcal B(\mathbb R))$ and $(\mathbb R^2, \mathcal B(\mathbb R^2))$ isomorphic as measurable spaces (i.e. without ignoring sets of Lebesgue measure 0)? – D Ford Feb 14 '19 at 03:51
  • @DFord: It's a Borel bijection with Borel inverse that preserves the measure. Which part of that do you have trouble with? – Eric Wofsey Feb 14 '19 at 03:53
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    Well the part I was having trouble with was showing that $f'$ is measurable in the first place. Clearly it is on the complement of $\mathbb Z \times S \cup T$, but I suppose it's obvious since countable subsets of $\mathbb Z \times {0,1}^\mathbb N$ are measurable. But why do you need the disjoint countable set $T$? – D Ford Feb 14 '19 at 03:58
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    Hmmm, I think you're right that $T$ is not necessary, since $f(\mathbb{Z}\times S)$ is already countably infinite. I'll get rid of it. I think I originally included $T$ because I had in mind a similar argument where the "bad set" $\mathbb{Z}\times S$ is finite, and so you need an extra set $T$ to make the function bijective on the bad set. – Eric Wofsey Feb 14 '19 at 04:00