Finite Measure Space integration Let $(\Omega, \mathcal{A}, \mu)$ be a measure space with finite $\mu$. Let $f,f_n:\Omega \to \overline{\mathbb{R}}$ be a $\mathcal{A}-$measurable function $(n \in \mathbb{N})$.
For every $\varepsilon >0$,
$$lim_{n \to \infty} \mu (\bigcup_{m \geq n} x \in \Omega:f_m(x) > f(x)+ \varepsilon))=0$$
it means we may choose an integer $N$ large enough so that 
$$\mu (\bigcup_{m \geq N} \left \{ x \in \Omega:f_m(x) > f(x)+ \varepsilon)) \right \})<\delta?$$  
 A: Suppose $(1)$ is true and let $\delta>0$. Choose an integer $N$ large enough so that 
$$\mu (\bigcup_{m \geq N} \left \{ x \in \Omega:f_m(x) > f(x)+ \varepsilon)) \right \})<\delta.$$
Then, since 
$\Omega \setminus \bigcap_m\bigcup_{m \geq N} \left \{ x \in \Omega:f_m(x) > f(x)+ \varepsilon)) \right \}=(\bigcap_m\bigcup_{m \geq N} \left \{ x \in \Omega:f_m(x) > f(x)+ \varepsilon)) \right \})^c=\bigcup_m(\bigcup_{m \geq N} \left \{ x \in \Omega:f_m(x) > f(x)+ \varepsilon)) \right \})^c=\bigcup_m\bigcap_{m \geq N} \left \{ x \in \Omega:f_m(x) \le f(x)+ \varepsilon) \right \},$ 
it follows that if we take 
$$A_{\delta }= \bigcap_m\bigcup_{m \geq N} \left \{ x \in \Omega:f_m(x) > f(x)+ \varepsilon)) \right \},$$
then 
$$x\in \Omega\setminus A_{\delta }\Rightarrow \exists m_1 \ge N\  \text {such that}\  \forall m\ge m_1,\ f_m(x)\le f(x)+\epsilon.$$
If $(2)$ is true, then take $\delta_n=1/n, $ so that there are sets $A_n$ with $\mu(A_n)<1/n$ and integers $M_n$ with the property that $f_m(x)\le f(x)+\epsilon$ whenever $m\ge M_n$ and $x\in \Omega\setminus A_n.$ Without loss of generality, assume that the $M_n>n$ and that they are increasing, and so $A_{n+1}\subseteq A_n.$
Now, fix $n$ and note that 
$$A_n= \bigcup_{m \geq M_n} \left \{ x \in \Omega:f_m(x) > f(x)+ \varepsilon \right \}$$  
which means that 
$$\mu \left (\bigcup_{m \geq M_n}  \left \{ x \in \Omega:f_m(x) > f(x)+ \varepsilon \right \} \right )<1/n.$$
But then 
$$\lim_{n \to \infty} \mu \left (\bigcup_{m \geq n}  \left \{ x \in \Omega:f_m(x) > f(x)+ \varepsilon \right \} \right )=\lim_{n \to \infty}\mu \left (\bigcup_{m \geq M_n}  \left \{ x \in \Omega:f_m(x) > f(x)+ \varepsilon \right \} \right )=0$$
the first equality being true because if any subsequence of a decreasing sequence of real numbers converges, then so does the sequence itself.
