Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 358 / 3 / a

The map $F$ is a nonsingular transformation with respect to $\omega$ whenEquivalently, the pushforward measure $F_{\ast }\omega (B)=\omega (F^{-1}(B))$ satisfies $F_{\ast }\omega \ll \omega$.

For an essentially bounded observable $g$, define the Koopman operatorNonsingularity makes this well defined on almost-everywhere equivalence classes. By the Radon-Nikodym theorem,Consequently the bounded Koopman operator criterion is

The map $F$ is a measure-preserving transformation whenfor every Borel set $B$, equivalently $F_{\ast }\omega =\omega$. In this case its Koopman operator is an isometry on $L^2(\omega )$.

The map is invertible with respect to $\omega$ when there is a measurable $G:X\to X$ such thatalmost everywhere. For an invertible measure-preserving system, $K_G=K_F^{-1}=K_F^{\ast }$, so its Koopman operator is unitary.