In Durrett's probability theory and examples, irreducible in the context of Markov chains is defined as follows:
$D$ is irreducible if $x,y \in D$ implies $\rho_{xy} > 0$ where $\rho_{xy} = \mathbb{P}_x(T_y < \infty)$ and $T_y = \inf \{n > 0: X_n = y\}$.
Here, we define a set $D$ to be irreducible. Later in the book, it defines ergodicity as follows:
A set $A \in \mathcal{F}$ is said to be invariant if $\varphi^{-1}A = A$. (Here as usual, two sets are considered to be equal if their symmetric difference has probability 0). Let $\mathcal{I}$ be the collection of invariant events... A measure preserving transformation on $(\Omega, \mathcal{F}, P)$ is said to be ergodic if $\mathcal{I}$ is trivial, i.e., for every $A \in \mathcal{I}, P(A) \in \{0, 1\}$. If $\varphi$ is not ergodic, then the space can be split into two sets $A$ and $A^c$, each having positive measure so that $\varphi(A) = A$ and $\varphi(A^c) = A^c$. In words, $\varphi$ is not irreducible.
Here a measure preserving transformation is not irreducible. So the term irreducible is used for both sets and a mapping which confuses me. When we say $\varphi$ is irreducible, does it mean that $\varphi$ induces sets $A$ and $A^c$ which makes the space $\Omega$ not irreducible?
I think I got what the book is trying to say. Using the rotation of the circle as an example, suppose that $\varphi = m /n$ where $m < n$ are positive integers. If $B$ is a Borel subset of $[0, 1/n)$ and $$A_n = \bigcup_{k=0}^{n-1} (B + k / n)$$ then $A_n$ is invariant to $\varphi$. In other words, the rotation of any point in the set $A_n$ will always be in $A_n$. Hence the measure preserving transformation $\varphi$ on $\Omega$ which is the set of all points on the circle will induce at least two sets $A_n$ and $A_n^c$ which are disjoint, so $\Omega$ won't be irreducible.