Let $(\Omega,\mathcal{A},\mu,T)$ be a dynamic system in measure theory. Let this system be ergodic. Show then then this implies $$ \forall f\colon\Omega\to\mathbb{R}\mbox{measurable}: f=f\circ T\text{ a.s. }\Rightarrow f=\text{const a.s.} $$
The idea of the proof which we got from the professor is to define the set $$ \Omega_{kn}:=\left\{\omega\in\Omega| k2^{-n}\leq f(\omega)< (k+1)2^{-n}\right\}. $$ Because of the measurability of $f$ it is $\Omega_{kn}\in\mathcal{A}$ and because $f\circ T=f\text{ a.s. }$ it is $T^{-1}(\Omega_{kn})\subset\Omega_{kn}\text{ a.s. }$. So, because the system is ergodic, it is $\mu(\Omega_{kn})=0\text{ or }\mu(\Omega_{kn}^C)=0$.
Additionally $\Omega_{kn}\cap\Omega_{jn}=\emptyset$ for $j,k\in\mathcal{Z}, j\neq k$.
The professor said, that from all this it follows that $\mu(\Omega_{kn})=1$.
Then he consideres $n\to\infty$ getting that $f=\text{ const a.s.}$.
Could you please explain me why it follows that $\mu(\Omega_{kn})=1$ and why it is necessary to show this?
Let $j,k\in\mathbb Z$. Since $\Omega_{kn}\cap\Omega_{jn} = \emptyset \qquad \forall j\neq k$, it follows $$\Omega_{jn} \subset \Omega_{kn}^C \qquad \forall j \neq k$$ However, for fixed $k$ we know $$\mu(\Omega_{kn}) = 0 \text{ or } \mu(\Omega_{kn}^C)=0$$ Let $k$ now be such that $\mu(\Omega_{kn}^C)=0$ (and thus by ergodicity and $T^{-1}(\Omega_{kn}) \subset \Omega_{kn}: \mu(\Omega_{kn})=1$). Then we have for all $j\neq k$: $$\mu(\Omega_{jn}) \stackrel{\Omega_{jn} \text{ measurable}}\leq \mu(\Omega_{kn}^C) = 0$$ This means if we have one such $k$, all other $\Omega_{jn}$ must be nullsets, as claimed.