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