Are the measurable spaces $(\mathbb{R}^n, Bor(\mathbb{R}^n))$ and $(\mathbb{R}^m, Bor(\mathbb{R}^m))$ isomorphic for $n\neq m$

Solution 1:

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.