Kolmogorov automorphism

Let ${\displaystyle (X,{\mathcal {B}},\mu )}$ be a standard probability space, and let ${\displaystyle T}$ be an invertible, measure-preserving transformation. Then ${\displaystyle T}$ is called a K-automorphism, K-transform or K-shift, if there exists a sub-sigma algebra ${\displaystyle {\mathcal {K}}\subset {\mathcal {B}}}$ such that the following three properties hold:

${\displaystyle {\mbox{(1) }}{\mathcal {K}}\subset T{\mathcal {K}}}$

${\displaystyle {\mbox{(2) }}\bigvee _{n=0}^{\infty }T^{n}{\mathcal {K}}={\mathcal {B}}}$

${\displaystyle {\mbox{(3) }}\bigcap _{n=0}^{\infty }T^{-n}{\mathcal {K}}=\{X,\varnothing \}}$

Here, the symbol ${\displaystyle \vee }$ is the join of sigma algebras, while ${\displaystyle \cap }$ is set intersection. The equality should be understood as holding almost everywhere, that is, differing at most on a set of measure zero.

Assuming that the sigma algebra is not trivial, that is, if ${\displaystyle {\mathcal {B}}\neq \{X,\varnothing \}}$, then ${\displaystyle {\mathcal {K}}\neq T{\mathcal {K}}.}$ It follows that K-automorphisms are strong mixing.

All Bernoulli automorphisms are K-automorphisms, but not vice versa.

Kolmogorov automorphisms are precisely the natural extensions of exact endomorphisms,[2] i.e. mappings ${\displaystyle T}$ for which ${\displaystyle \bigcap _{n=0}^{\infty }T^{-n}{\mathcal {M}}}$ consists of measure-zero sets or their complements, where ${\displaystyle {\mathcal {M}}}$ is the sigma-algebra of measureable sets,.

• Christopher Hoffman, "A K counterexample machine", Trans. Amer. Math. Soc. 351 (1999), pp 4263–4280.