Koopman operators are linear isometries

Composition with measurable $T$ is measurable. If $f=g$ outside a measurable null set $N$, then $f\circ T=g\circ T$ outside $T^{-1}N$, which is measurable and null. Thus pullback respects the a.e. equivalence relation used to define $U_T$.

For $p<\mathrm{\infty }$, integral invariance applied to the nonnegative function $\mathrm{\mid }f{\mathrm{\mid }}^p$ gives $\mathrm{\parallel }f\circ T{\mathrm{\parallel }}_p^p=\int \mathrm{\mid }f{\mathrm{\mid }}^p\circ T=\int \mathrm{\mid }f{\mathrm{\mid }}^p=\mathrm{\parallel }f{\mathrm{\parallel }}_p^p$. Thus the pullback belongs to $L^p$ and preserves the norm, including at $p=1$.

For every $M\ge 0$, $\{\mathrm{\mid }f\circ T\mathrm{\mid }>M\}=T^{-1}\{\mathrm{\mid }f\mathrm{\mid }>M\}$ has the same measure as $\{\mathrm{\mid }f\mathrm{\mid }>M\}$. The sets of finite essential bounds therefore coincide, so their infima coincide. The least-essential-bound result applies to the real modulus, proving membership and equality of the infinity norms.

Pointwise, $(\alpha f+\beta g)\circ T=\alpha (f\circ T)+\beta (g\circ T)$. The real and complex quotient norm theorems make these the quotient vector operations. Together with steps 1.1, 2.1 and 2.2 this proves linear isometry.

For an actual measurable inverse $S=T^{-1}$, $TE=S^{-1}E$ is measurable and $\mu (TE)=\mu (T^{-1}(TE))=\mu (E)$. Thus $S$ preserves measure. Both compositions $U_T{U}_S$ and $U_S{U}_T$ are the identity, so $U_T$ is onto.

In the modulo-null case restrict to the measurable conull invariant $X_0$ of the definition. For measurable $E\subseteq X_0$, $T^{-1}E$ differs from its restricted inverse image only within $X\setminus X_0$, so the restricted map preserves restricted measure. Step 4.1 applies there. For any measurable $f$ on $X$, compose $f{\mathrm{\mid }}_{X_0}$ with the restricted inverse and extend by zero on $X\setminus X_0$. This extension is measurable, has the same Lp norm as $f$, and pulls back to $f$ on $X_0$. It supplies a preimage class.