I want to prove the following statement:
Let $(X,\mu)$ be a probability space and $T:X\to X$ be a measure-preserving map, then $T$ is ergodic iff for all measurable functions $f:X\to\mathbb{R}$, $f\circ T= f$ a.e. implies $f$ is constant a.e.
The 'only if' direction is no problem. For the 'if' direction i consulted a book by M. Bekka and M. Mayer on ergodic theory of group actions, and their proof (adapted for my situation) is as follows:
Let $f:X\to\mathbb{R}$ be measurable and $T$-invariant a.e., then for each $k\in\mathbb{Z}$ and $n\in\mathbb{N}$, the set $$X(k,n):=\left\{x\in X:\frac{k}{2^n}\leq f(x)<\frac{k+1}{2^n}\right\}$$ is $T$ invariant, i.e. $T^{-1}(X(k,n))=X(k,n)$, hence $\mu(X(k,n))=0$ or $\mu(X(k,n))=1$ by ergodicity of $T$. Now fix $n\in\mathbb{N}$, then $X=\dot\bigcup_{k\in\mathbb{Z}}X(k,n)$ (disjoint union). Hence there exists exactly one $k=k_n$ with $\mu(X(k_n,n))=1$. Then $$Y:=\bigcap_{n\in\mathbb{N}}X(k_n,n)$$ has full measure and $f$ is constant on $Y$.
I have no problem understanding the proof until the last statement. Why is $f$ constant on $Y$? If the intervals $[k_n/2^n,(k_{n+1}+1)/2^n)$ where nested i would understand, but as far as i can see, they are not necessarily so since $k_n$ could be any integer. How could i go about proving that $Y$ is the preimage of one point without using the interval nesting principle? Thanks in advance for any help.
If $x \in Y$ then $\frac{k_n}{2^n}\leq f(x)<\frac{k_n+1}{2^n}$ for all $n$. This gives $f(x)-\frac {1} {2^{n}}\leq \frac{k_n}{2^n}\leq f(x)$. This proves that $\frac{k_n}{2^n}$ converges to $f(x)$. [Note that $k_n$ does not depend on $x$]. It also proves that $f(x)$ has the same value (viz. $\lim \frac{k_n}{2^n}$) for all $x \in Y$. Since $P(Y)=1$ it follows that $f$ has the constant value $\lim \frac{k_n}{2^n}$ a.e. on $Y$.