I'm looking for a proof, in English, of the following theorem due to von Neumann (which apparently originates in the paper *Einige Sätze über Messbare Abbildungen*, Ann. of Math, 1932):



> 
> Every automorphism of the Boolean algebra of (Lebesgue) measurable subsets of $[0,1]$ modulo null sets is realized by a Borel measurable bijection on $[0,1]$.
> 
> 
> 


Obviously, $[0,1]$ can be substituted with any standard finite measure space.


I know that this is generalized by results of Mackey, Maharam, and others, but I'm really just interested in the original result and its proof. Alas, I cannot read German.



Von Neumann's theorem on realizing automorphisms of the measure algebra
Here is a hint for a proof, in English.
"Obviously, $[0,1]$ can be substituted with any standard finite measure space",
so let's substitute it with the cantor set $C=\{0,1\}^\mathbb{N}$ endowed with the standard measure $\mu=(1/2\delta\_0+1/2\delta\_1)^\mathbb{N}$
and let $T$ be an automorphism of the corresponding measured algebra.
For every $n$ let $A\_n=\{(x)\mid x\_n=1\} \subset C$ and fix $B\_n\subset C$, representatives for $T^{-1}A\_n$.
Define $\hat{T}:C\to C$ by $(\hat{T}x)\_n=\chi\_{B\_n}(x)$ and check that the corresponding operator on the measure algebra coincides with $T$.


You can check it using the auxiliary space $D=\{0,1\}^{\mathbb{N}\times\mathbb{Z}}$. Define a measurable map
$\phi:C\to D$ by $(\phi(x))\_{n,m}=\chi\_{T^{-m}A\_n}(x)$ (for this to make sense, you need here representatives for $T^{-m}A\_n$)
and set $\nu=\phi\_\*(\mu)$.
Identifying $C$ with $\{0,1\}^{\mathbb{N}\times\{0\}}$ we obtain a continuous factor map $\psi:D\to C$.
Check that $\phi:(C,\mu)\to (D,\nu)$ and $\psi:(D,\nu)\to (C,\mu)$ form mutatual inverses (up to a.e identifications) and they intertwine the $T$ action on $C$ with the shift action on $D$ with respect to the $\mathbb{Z}$ indexes.


