Let $(\Omega, A,\mu)$ be a measurable space with a finite measure $\mu$. Further, let $f_{1},f_{2},...$ be measurable numerical functions on $\Omega$. Prove the equivalence of the two assertions:
(i) For all $\varepsilon_{>0}:$ $$\lim_{n \to \infty}\mu(\bigcup_{m\ge n}\{f_{m}>f+\varepsilon\})=0$$
(ii) For every $\delta_{>0}$ there exists $A_{\delta} \in A$ with $\mu(A_{\delta})<\delta$ such that for every $\varepsilon_{>0}$ there exists $N_{\varepsilon} \in \mathbb{N}$ such that: $$f_{n}(x)\le f(x)+\varepsilon \text{ for all }x \in \Omega\setminus A_{\delta} \text{ and } n \ge N_{\varepsilon}$$
So i'm going to prove this with two implications. I got stuck with the first one: (i) $\implies$ (ii).
I got the following hint: Note that (i) is also equivalent to the following statement:
For every $\varepsilon_{>0}$ and $\delta_{>0}$ there exists an $A_{\delta,\varepsilon}\in A$ such that $\mu(A_{\delta,\varepsilon})<\delta$, and an $N_{\delta,\varepsilon} \in \mathbb{N}$ such that $f_{n}(\omega)\le f(\omega)+\varepsilon$ for all $\omega \in \Omega \backslash A_{\delta,\varepsilon}$ and $n\ge N_{\delta,\varepsilon}$. This was the hint, but I don't know how to proceed. Help would be much appreciated!
(Ex. 2 in chapter II in Measure and Integration Theory from Heinz Bauer.)
(ii)$\implies$ (i) is not too difficult: if we take $A_{\delta,\varepsilon}:=A_\delta$ and $N_{\delta,\varepsilon}:=N_\varepsilon$, using the given hint, the implication directly follows.
For (i)$\implies$(ii), define $\delta_n=\delta 2^{-n}$. Then for all $\delta_n$ take $N_{\delta_n,\varepsilon}$ and $A_{\delta_n,\varepsilon}$ as in the hint, and define $A_{\delta,\varepsilon}:=\bigcap_{n\in\mathbb{N}}A_{\delta_n,\varepsilon}$.
I'm sure you're able to write out that $\mu(A_{\delta,\varepsilon})<\delta$ then holds.
Now we have our $A_\delta$ for (ii), and we only need a $N_\varepsilon$. I think the supremum of the $N_{\delta_n,\varepsilon}$ works.