The following is a lemma from Viana's "Lectures on Lyapunov Exponents".
Lemma 9.9 Let $f:M\to M$ be an invertible transformation and $\mu$ be an invariant aperiodic probability measure. For any measurable set $Y \subseteq M$ with $\mu(Y) > 0$ and any $m \geq 1$ there exists a measurable set $Z \subseteq Y$ such that $\mu(Z) > 0$ and $Z, f(Z), \dots, f^{m-1}(Z)$ are pairwise disjoint.
In the context of the chapter, I think we may assume that $f: M \to M$ is a measurable transformation on a separable complete metric space $M$. And aperiodic means that the set of periodic points has zero measure. Furthermore, they stated somewhere that for the space $(M, \mathcal{B}, \mu)$, separable means that there exists a countable family $\mathcal{E} \subseteq \mathcal{B}$ such that for any $\varepsilon > 0$ and any $B \in \mathcal{B}$ there exists $E \in \mathcal{E}$ such that $\mu(B\triangle E) < \varepsilon$.
The proof starts with the statement:
Since fixed points form a zero measure subset of $Y$, we may find $Y_1 \subseteq Y$ such that $\mu(Y_1 \triangle f(Y_1)) > 0$.
We have no clue why such an $Y_1$ exists. Help is much appreciated.
Does $\mathcal{B}$ mean the Borel $\sigma$-field? If so, let $E_n$ be an enumeration of balls with rational radii and a dense set of centers.
Then $$\{x\in Y:f(x)\neq x\} = \bigcup_{E_n \cap E_m = \emptyset}\{f(x) \in E_n, x \in E_m \cap Y\}$$
The left side has measure $\mu(Y) > 0$, so the right side cannot all be sets of measure zero. Thus there is some $m,n$ such that $E_n \cap E_m = \emptyset$ and $\mu(f(x) \in E_n, x \in E_m \cap Y) > 0$. But $$\{f(x) \in E_n, x \in E_m \cap Y\} \subseteq (Y \cap E_m) \setminus f(Y \cap E_m) \subseteq (Y \cap E_m) \triangle f(Y \cap E_m)$$ and thus the set on the right has positive measure.