Measurable Partition and Ergodic Decomposition I need some background before asking the question:
Let $\mathcal{P}$ is called a measurable partition if there is a measurable set $M_0\subset M$ with full probability measure such that, restric to $M_0$
$$\mathcal{P}=\bigvee_{n=1}^\infty \mathcal{P}_n$$
for some crescent sequence $\mathcal{P}_1 \prec \dots \prec \mathcal{P}_n \prec \dots$ of enumerable partitions, where $\mathcal{P}_i\prec \mathcal{P}_{i+1}$ means that every element of $\mathcal{P}_{i+1}$ is contained in some element of $\mathcal{P}_i$ (we say that $\mathcal{P}_i$ is less thin than $\mathcal{P}_{n+1}$) and also $\bigvee_{n=1}^\infty \mathcal{P}_n$ is the thinnest such that
$$\mathcal{P}_n \prec \bigvee_{n=1}^\infty \mathcal{P}_n, \forall n$$
It's elements are of the form $\cap_{n=1}^\infty P_n$, where $P_n \in \mathcal{P}_n$.
Questions: i) Show that a partition $\mathcal{P}$ is a measurable partition iff there exists measurable subsets $M_0, E_1, E_2,\dots,E_n,\dots$ such that $\mu(M_0)=1$ and, restrict to $M_0$, 
$$\mathcal{P}= \bigvee_{n=1}^\infty \{E_n, M\setminus E_n\}$$
ii) If $\mu$ is an ergodic probability for a transformation $f$ and $k\geq2$. How is the ergodic decomposition of $f^k$?
 A: For i), we can suppose each partition is finite, i.e., $\mathcal P_n=\{P_1,\cdots,P_{k_n}\}$, then define $E_{k_i}=P_{k_i}$, then one gets $\mathcal P_n\prec \bigvee_{i=1}^{k_n}\{E_{k_i},M\backslash E_{k_i}\}$.Thus, taking all the $E_j$ in this form, we have $\bigvee_n \mathcal P_n\prec \bigvee_j \{E_j,M\backslash E_j\}$.
For ii), Consider $k=2$. There are two cases. Case 1. For any $A\subset M$ such that $f^{-2}(A)=A$, $\mu(A)>0$, we also have $f^{-1}(A)=A$, this implies that $\mu(A)=0$ or $1$. In this case, we have $(f^2,\mu)$ is also ergodic. 
Case two. there exists some $A$ such that $f^{-2}(A)=A$, but $\mu(A\backslash f^{-1}(A))>0$, in this case, 
$f^{-1} (A\backslash f^{-1}(A)\cup f^{-1}(A)\backslash A)=A\backslash f^{-1}(A)\cup f^{-1}(A)\backslash A$ and therefore, $\mu(A\Delta f^{-1}(A))=1$, which implies that $M=A\cup f^{-1}(A)$. 
Now define $\nu_1$ such that $\nu_1(E)=2\mu(E\cap A)$ and $\nu_2(E)=2\mu(E\cap A^c)$, then both $\nu_2$ and $\nu_2$ are ergodic and $\mu=\frac{1}{2} (\nu_1+\nu_2)$.
