We are studying measurable and mesaure preserving maps $T : X \to X$ where $(X,\B,\mu)$ is a fixed probability space. At the end of the last section we proved the Poincaré recurrence theorem. It tells us that whenever $\mu(B) \g 0$ one has \[ \mu(B \cap (T^n)^{-1}(B)) \g 0 \] for some $n \in \N$.
It is then reasonable to ask the following question: given disjoint sets $A,B \in \B$ with $\mu(A) \g 0$ and $\mu(B) \g 0$ do we always have \[ \mu(A \cap (T^n)^{-1} B) \g 0 \] for some $n \in \N$?
Let us imagine for the moment that the answer is "no". It must then be the case that the set \[ A \cap \left(\, \bigcup_{n \in \N} (T^n)^{-1} B \right) \] has zero measure. Now the set \[ C = \bigcup_{n \in \N} (T^n)^{-1} B \] satisfies $T^{-1}(C) \subset C$. Since $\mu(T^{-1}(C)) = C$ we see that \[ Y = \bigcap_{j \in \N} (T^j)^{-1} C = \bigcap_{j \in \N} \bigcup_{n \g j} (T^n)^{-1}(B) \] has positive measure equal to $\mu(C)$. Moreover, if $x \in Y$ then $T(x) \in Y$ as well. We therefore have within $X$ a set $Y$ of positive measure but not full measure that the dynamics never escapes. We could therefore study dynamics on $Y$ in isolation.
It is therefore necessary, if the answer to the above question is "yes", for almost all points in $X$ to go "almost everywhere" in the sense that they visit every set of positive measure. This amounts to indecomposability in the sense that $X$ cannot be broken up into pieces that are themselves measure-preserving systems.
Just as one first works, for example, with group representations that are irreducible, we will prefer to avoid the situation where our dynamical system has intermediate invariant subsystems. This leads us to the following definition.
A measure-preserving transformation $T$ on a probability space $(X,\B,\mu)$ is ergodic if \[ \mu(B) \g 0 \Rightarrow \mu \left( \, \bigcup_{n \in \N} (T^n)^{-1} B \right) = 1 \] for every $B \in \B$.
It is convenient to rephrase the definition in the following form.
A measure-preserving transformation $T$ on a probability space $(X,\B,\mu)$ is ergodic if and only if \[ B = T^{-1}(B) \Rightarrow \mu(B) \in \{0,1\} \] for every $B \in \B$.
Fix $\alpha$ irrational. Take $X = [0,1)$ and define \[ T(x) = x + \alpha \bmod 1 \] for all $x \in [0,1)$. Write $\lambda$ for the restriction of Lebesgue measure to $[0,1)$. We will prove that $(X,\B,\lambda,T)$ is ergodic.
The system $(X,\B,\lambda,T)$ is ergodic.
Fix $B \in \B$ with $T^{-1}(B) = B$ and $\lambda(B) > 0$. First we check that \[ \lambda(I \cap B) = \lambda(B) \lambda(I) \] for every interval $I = [a,b)$. But \[ \begin{align} \lambda(I \cap B) &{} = \int 1_I \cdot 1_B \intd \lambda \\ &{} = \dfrac{1}{N} \sum_{n=1}^N \int 1_I \circ T \cdot 1_B \intd \lambda \\ &{} = \lim_{N \to \infty} \int \dfrac{1}{N} \sum_{n=1}^N 1_I(T^n x) \cdot 1_B(x) \intd \lambda(x) \\ &{} = \lambda(I) \lambda(B) \end{align} \] by the dominated convergence theorem and the uniform distribution theorem.
Now define \[ \nu(E) = \dfrac{\lambda(B \cap E)}{\lambda(B)} \] for all $E \in \B$. This is a measure on $\B$ and, by the previous paragraph, we have \[ \nu(I) = \lambda(I) \] for all intervals $I = [a,b)$. But then the π−λ theorem tells us that $\nu = \lambda$. Thus \[ \lambda(B) = \nu(B) = \dfrac{\lambda(B)}{\lambda(B)} = 1 \] as desired. ▮
Take $X = \{0,1\}^\N$ and let $T : X \to X$ be the shift map. Put \[ \mu = \tfrac{1}{2} \delta_a + \tfrac{1}{2} \delta_b \] where \[ \begin{align} a &{} = \mathtt{00000000000000}\cdots \\ b &{} = \mathtt{11111111111111}\cdots \end{align} \] are the constant sequences in $X$. With respect to $\mu$ the map $T$ measure-preserving because $T(a) = a$ and $T(b) = b$. However, it is not ergodic. Indeed, the sets $[0]$ and $[1]$ are $T$ invariant but both have measure $0 \l \tfrac{1}{2} \l 1$.
Next, let us look at the measures $\mu_p$ for $0 \le p \le 1$ fixed. Let $B \in \B$ be a $T$-invariant set. We have \[ \begin{align*} \mu(B \cap [1]) &{} = \lim_{N \to \infty} \dfrac{1}{N} \sum_{n=1}^N \mu(B \cap (T^n)^{-1} [1]) \\ &{} = \lim_{N \to \infty} \int 1_B \cdot \dfrac{1}{N} \sum_{n=1}^N 1_{(T^n)^{-1}[1]} \intd \mu \\ &{} = \int 1_B \cdot \dfrac{1}{N} \lim_{N \to \infty} \sum_{n=1}^N 1_{(T^n)^{-1}[1]} \intd \mu \\ &{} = p \int 1_B \intd \mu = \mu([1]) \mu(B) \end{align*} \] by the strong law of large numbers and the dominated convergence theorem. In fact, by a strengthening of the strong law of large numbers it is the case that \[ \mu(B \cap C) = \mu(B) \mu(C) \] for every cylinder set $C$. It is therefore the case that the measure \[ \nu(C) = \dfrac{\nu(B \cap C)}{\nu(B)} \] agrees with $\mu_p$ on every cylinder set. The π-𝜆 theorem then guarantees $\mu_p = \nu$ so $\mu_p(B) = 1$.