The map $F$ is a <nonsingular transformation> with respect to $\omega$ when
$$
\omega(N)=0\quad\Longrightarrow\quad
\omega(F^{-1}(N))=0.
$$
Equivalently, the <pushforward measure> $F_*\omega(B)=\omega(F^{-1}(B))$ satisfies $F_*\omega\ll\omega$.

For an essentially bounded observable $g$, define the <Koopman operator>
$$
\boxed{K_Fg=g\circ F.}
$$
Nonsingularity makes this well defined on almost-everywhere equivalence classes. By the <Radon-Nikodym theorem>,
$$
\|K_Fg\|_2^2
=\int|g|^2\,d(F_*\omega)
=\int|g|^2\frac{d(F_*\omega)}{d\omega}\,d\omega.
$$
Consequently the <bounded Koopman operator criterion> is
$$
\boxed{
K_F:L^2(\omega)\to L^2(\omega)\text{ is bounded}
\iff
\frac{d(F_*\omega)}{d\omega}\in L^\infty(\omega).}
$$

Solution

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

{title2=$T$}
{wiki=Nonsingular_transformation}

A measurable transformation $T:X\to X$ is nonsingular with respect to $\mu$ when $\mu(T^{-1}N)=0$ for every $\mu$-null set $N$, equivalently when the <pushforward measure> $T_*\mu$ is <absolute continuity of measures>[absolutely continuous] with respect to $\mu$.

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 Incoming links: Nonsingular transformation