dynamical systems - Poincaré Recurrence Theorem (measure theory version)

I had a look on the proof of the following Recurrence Theorem of Poincaré:

Let $(\Omega,\Sigma,T,m)$ be a conservative dynamical system in measure theory for which the function $T^{-1}$ preserves null sets. If $f\colon\Omega\to\mathbb{R}$ is measurable, it follows that
$$
\liminf_{n\to\infty}\lvert f(x)-f(T^n(x))\rvert=0~\text{a.s.}
$$

Proof

Let $B\subset\mathbb{R}$ be a measurable set with $m(f^{-1}(B))>0$ and $\text{diam}(B)<\varepsilon$. With the Theorem of Halmos (cited below) it follows that
$$
S_{\infty}1_{f^{-1}(B)}=1_{f^{-1}(B)}+1_{f^{-1}(B)}\circ T+1_{f^{-1}(B)}\circ T^2+\ldots+1_{f^{-1}(B)}\circ T^n+\ldots=\infty
$$
alomost surely on $f^{-1}(B)$.

So it follows that $f(T^n(z))\in B$ for almost all $n\geqslant 0$. So with Halmos it is
$$
\lvert f(z)-f(T^n(z))\rvert\leqslant\text{diam}(B)=\sup_{x,y\in B}\lvert x-y\rvert<\varepsilon.
$$
From this it follows that
$$
\liminf_{n\to\infty}\lvert f(z)-f(T^n(z))\rvert<\varepsilon~~~~~(*)
$$
almost surely on $f^{-1}(B)$.

Now if one covers $\Omega$ by sets $f^{-1}(B)$ with diam smaller than $\varepsilon$ one has (*) almost surely and the Theorem follows with $\varepsilon\to 0$.

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Is that okay?

Why is it possible to cover $\Omega$ by set of the form $f^{-1}(B)$ with $\text{diam}<\varepsilon$? Is $\Omega$ $\sigma$-finite?

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Last but not least

here is the Theorem of Halmos that is used:

Let $(\Omega,\Sigma,T,m)$ a dynamical System in measure theory so that $T^{-1}$ preserves null sets. If $A\in\Sigma$ has positive measure then
$$
A\in\mathcal{C}(T)\Leftrightarrow S_{\infty}1_B=\infty \text{ a.s. on } B \text{ for each }B\subset A\cap\Sigma.
$$

(Here with $\mathcal{C}(T)$ the conservative part of the system is meant.)

With Greetings and best regards

math12