Measure-Preserving Systems and Mixing Criteria — Examples

The masses of $\mathrm{\varnothing },\{0\},\{1\},X$ are respectively $0,1\mathrm{/}2,1\mathrm{/}2,1$. A disjoint countable family has at most two nonempty members, so adding their cardinalities proves countable additivity of $\mu$. Every inverse image is a subset of X and is measurable.

The inverse images of those four sets are $\mathrm{\varnothing },\{1\},\{0\},X$, respectively. Their masses are unchanged, so $\mu (T^{-1}E)=\mu (E)$ for every measurable E. This is measure preservation.

Take $X=\{0,1\}$ with all subsets measurable, $\mu (E)=\mathrm{\mid }E\mathrm{\mid }\mathrm{/}2$, and $T(x)=x$. The measure is countably additive because a disjoint family has at most two nonempty members. Its total mass is one. For every subset E, $T^{-1}E=E$, so T is measurable and preserves probability.

The set $E=\{0\}$ is strictly invariant and has $\mu (E)=\mu (X\setminus E)=1\mathrm{/}2>0$. It is neither null nor conull, contradicting the ergodicity requirement. The specified T is therefore a counterexample.

The cardinality measure $\mu (E)=\mathrm{\mid }E\mathrm{\mid }\mathrm{/}2$ is countably additive, since a disjoint family has at most two nonempty members. The inverse images of $\mathrm{\varnothing },\{0\},\{1\},X$ are $\mathrm{\varnothing },\{1\},\{0\},X$, with masses $0,1\mathrm{/}2,1\mathrm{/}2,1$. Hence the swap is measurable and preserves this probability measure. For $f(0)=a,f(1)=b$, pullback gives $U_{T}f(0)=b,U_{T}f(1)=a$, the displayed matrix formula.

For finite p, $\mathrm{\parallel }{U}_T(a,b){\mathrm{\parallel }}_p=((\mathrm{\mid }b{\mathrm{\mid }}^p+\mathrm{\mid }a{\mathrm{\mid }}^p)\mathrm{/}2{)}^{1\mathrm{/}p}=\mathrm{\parallel }(a,b){\mathrm{\parallel }}_p$. For infinity, both norms equal $\max \{\mathrm{\mid }a\mathrm{\mid },\mathrm{\mid }b\mathrm{\mid }\}$. Thus the direct calculation, including zero coordinates, agrees with the general Koopman isometry theorem.