Let $T$ be a measure-preserving transformation on a probability space $(X,\B,\mu)$. In particular $T$ is a map from $X$ to itself, and we can use composition to induce a map on spaces of functions on $X$. For example, given $f : X \to \C$ we can define a new function $Tf$ by \[ (Tf)(x) = f(T(x)) \] for all $x \in X$. Thus $Tf$ is just $f \circ T$. One then has the option of studying $T$ not by its direct effect on points in $X$ but by its indirect effect on functions. For any two functions $f,g$ and any $\alpha$ in $\C$ we have \[ \begin{gather*} (T(f+g))(x) = (Tf)(x) + (Tg)(x) \\ (T(\alpha f))(x) = (\alpha (Tf))(x) \end{gather*} \] for all $x$ in $X$. Thus \[ \begin{gather*} T(f+g) = Tf + Tg \\ T(\alpha f) = \alpha (Tf) \end{gather*} \] and $T$ acts linearly on functions. We can therefore study $T$ using the tools of linear algebra.
In our context, where $T$ is measurable and measure-preserving, it makes sense to consider the effect $T$ has on measurable and integrable functions. The relevant spaces - Lebesgue spaces - are not just vector spaces. They also have topological structure coming from the norm we have put on them. We can therefore use functional analysis - which is essentially the study of topological vectors spaces and the continuous, linear maps between then - to study our transformation.
In this course we will mainly look at how $T$ affects functions in the Lebesgue space $\Lp[2](X,\B,\mu)$. Our first duty is to check that composition with $T$ gives us a well-defined map from $\Lp[2](X,\B,\mu)$ to $\Lp[2](X,\B,\mu)$.
Fix a measurable space $(X,\B)$ and a measurable map $T : X \to X$. If $f : X \to \R$ is $(\B,\borel(\R))$ measurable then $f \circ T$ is also $(\B,\borel(\R))$ measurable.
We check the definition of measurability. Fix a set $B$ in $\borel(\R)$. From \[ (f \circ T)^{-1}(B) = T^{-1}(f^{-1}(B)) \] and the fact that $f^{-1}(B) \in \B$ we get $(f \circ T)^{-1}(B) \in \B$ as desired. ▮
Fix a probability space $(X,\B,\mu)$ and a measure-preserving map $T : X \to X$. If $f$ belongs to $\mathscr{L}^{\mathsf{2}}(X,\B,\mu)$ then $Tf$ does as well.
Fix $f : X \to \R$ measurable such that \[ \int |f|^2 \l \infty \] holds. By the previous lemma the function $Tf$ is measurable and therefore \[ \int |Tf|^2 \intd \mu \] exists in $[0,\infty]$. Let $n \mapsto \phi_n$ be a non-decreasing sequence of non-negative and simple functions with $\phi_n \to |f|^2$ pointwise. It is immediate that \[ T\phi_n \to T(|f|^2) = |Tf|^2 \] and therefore \[ \begin{align*} \int |Tf|^2 \intd \mu &{} = \lim_{n \to \infty} \int T\phi_n \intd \mu \\ \int |f|^2 \intd \mu &{} = \lim_{n \to \infty} \int \phi_n \intd \mu \end{align*} \] by the monotone convergence theorem. It therefore suffices to check that \[ \int T\psi \intd \mu = \int \psi \intd \mu \] for all simple functions $\psi : X \to \R$.
Fix a simple and measurable function \[ \psi = a(1) 1_{B(1)} + \cdots + a(s) 1_{B(s)} \] from $X$ to $\R$ with $B(1),\dots,B(s)$ all in $\B$. One calculates \[ \begin{align*} T\psi &{} = a(1) 1_{B(1)} \circ T + \cdots + a(s) 1_{B(s)} \circ T \\ &{} = a(1) 1_{T^{-1}B(1)} + \cdots + a(s) 1_{T^{-1} B(s)} \end{align*} \] so that \[ \begin{align*} \int T\psi \intd \mu &{} = a(1) \mu(T^{-1} B(1)) + \cdots + a(s) \mu(T^{-1} B(s)) \\ &{} = a(1) \mu(B(1)) + \cdots + a(s) \mu(B(s)) = \int \psi \intd \mu \end{align*} \] because $\mu$ is $T$ invariant. ▮
Fix a measure-preserving transformation $T$ of a probability space $(X,\B,\mu)$. For every $f$ in $\Lp[2](X,\B,\mu)$ the composition $Tf$ also belongs to $\Lp[2](X,\B,\mu)$.
Fix $f$ in $\Lp[2](X,\B,\mu)$. Recall that $f$ is in fact an equivalence class of measurable functions \[ f = F + \mathscr{Z}^{\mathsf{2}}(X,\B,\mu) = \{ g \in \mathscr{L}^{\mathsf{2}}(X,\B,\mu) : \| g - F \|_\mathsf{2} = 0 \} \] where $F : X \to \R$ is a fixed representative. We would like to define \[ Tf = TF + \mathscr{Z}^{\mathsf{2}}(X,\B,\mu) \] and must check this is well-defined. Thus fix $g : X \to \R$ measurable with $\| F - g \|_\mathsf{2} = 0$. By the two lemmas above $|T(F-g)|^2$ is measurable and \[ \int |T(F-g)|^2 \intd \mu = \int |F-g|^2 \intd \mu \] so $\| (TF) - (Tg) \|_\mathsf{2} = 0$ and $Tf$ is well-defined. ▮
The above theorem gives us a well-defined map \[ T : \Lp[2](X,\B,\mu) \to \Lp[2](X,\B,\mu) \] whenever $T$ is a measure-preserving transformation of the probability space $(X,\B,\mu)$. This map on Lebesgue space is called the Koopman operator associated with $T$.
Our goal is to understand dynamical properties of $T$ in terms of its Koopman operator, and to use the Koopman operator to deduce dynamical properties of $T$. Generally, this involves replacing statements about sets with statements about functions, the connection being that every set $E \subset X$ gives rise to a function $1_E$. We begin by rephrasing ergodicity of $T$ in terms of its Koopman operator.
We defined $T$ to be ergodic if and only if all invariant sets have measure 0 or measure 1. Since \[ T1_E = 1_{T^{-1} E} = 1_E \] whenever $E$ is invariant, it seems reasonable to look at those functions in $\Lp[2](X,\B,\mu)$ that are fixed by the Koopman operator.
A measure-preserving transformation $T$ on $(X,\B,\mu)$ is ergodic if and only if all $T$ invariant functions in $\Lp[2](X,\B,\mu)$ are equal almost-everywhere to a constant function.
First suppose that every invariant function $f$ in $\Lp[2](X,\B,\mu)$ is equal almost-everywhere to a constant function. This means the following: if $f$ is invariant then there is $c \in \C$ such that \[ \mu( \{ x \in X : f(x) = c \} ) = 1 \] holds. Fix $E \subset X$ with $T^{-1} E = E$. Then $T 1_E = 1_E$ and $1_E$ as a member of $\Lp[2](X,\B,\mu)$ is fixed by the Koopman operator. By hypothesis there is then a constant $c \in \C$ with $1_E = c$ almost everywhere As $1_E$ only takes the values $0$ and $1$ it must be the case that $c \in \{0,1\}$ and therefore $\mu(E) \in \{0,1\}$.
Conversely, suppose that $T$ is ergodic and that $f$ from $\Lp[2](X,\B,\mu)$ is fixed by the Koopman operator. Suppose first that $f$ is real-valued. Fix $k \in \N$. Put \[ E(k,N) = \{ x \in X : \tfrac{N}{2^k} \le f(x) \l \tfrac{N+1}{2^k} \} \] for each $N \in \Z$. We have \[ \mu(E(k,N) \triangle T^{-1} E(k,N)) = 0 \] and therefore $\mu(E(k,N)) \in \{0,1\}$. Exactly one of these sets can have full measure. As we increase $k$ we converge to a specific value $c \in \R$ with \[ \mu( \{ x \in X : f(x) = c \}) = 1 \] and therefore $f$ is constant almost-everywhere. ▮
In analogy with linear algebra, we next look for eigenfunctions of $T$. Eigenfunctions of the Koopman operator indicate that $T$ has some component of regularity.
By an eigenfunction of a measure-preserving transformation $T$ we mean a member $f$ of $\Lp[2](X,\B,\mu)$ such that $Tf = \eta f$ for some $\eta \in \C$.
For the the irrational rotation $T(x) = x + \alpha \bmod 1$ there are lots of eigenfunctions. Indeed \[ \psi_k(x + \alpha) = e^{2 \pi i k (x + \alpha)} = e^{2 \pi i k \alpha} e^{2 \pi i k x} = e^{2 \pi i k \alpha} \psi_k(x) \] so \[ T \psi_k = e^{2 \pi i k \alpha} \psi_k \] and $\psi_k$ is an eigenfunction of $T$ with eigenvalue $e^{2 \pi i k \alpha}$.
In contrast with the existence of eigenfunctions, we have the property of mixing.
A measure-preserving transformation $T$ on a probability space $(X,\B,\mu)$ is mixing if \[ \lim_{n \to \infty} \langle f, T^n g \rangle = \langle f,1 \rangle \langle 1,g \rangle \] for all $f,g$ in $\Lp[2](X,\B,\mu)$.