Suppose $\{f_n\}$ is a sequence of nonnegative extended real-valued functions on $X$ and $\lim_{n\to\infty}f_n=f$. Take a simple function $0\leq\varphi\leq f$. If $X_{\infty}=\{x\in X: \varphi(x)=a>0\}$ and $A_n=\{x\in X: f_k(x)\geq \frac{a}{2} ~\text{for all}~ k\geq n\}$. Show that $X_{\infty}\subseteq \bigcup_{n\in \mathbb{N}}A_n$.
I know that if $\lim_{n\to \infty}f_n=f<\infty$. Let $\epsilon=\frac{a}{2}$, so for all $x\in X$ there exists $N\in \mathbb{N}$ such that for each $n\geq N$, \begin{eqnarray*} |f_n(x)-f(x)|&<&\frac{a}{2}\\ -\frac{a}{2}<f_n(x)-f(x)&<&\frac{a}{2}\\ \implies f_n(x)&>&f(x)-\frac{a}{2}\\ \end{eqnarray*} Hence for any $x\in X_\infty$ there exists $N\in \mathbb{N}$ such that for all $n\geq N$, \begin{eqnarray*} f_n(x)&>&f(x)-\frac{a}{2}\\ &\geq&\varphi(x)-\frac{a}{2}~~~~\text{since $\varphi\leq f$ on $X$}\\ &=&a-\frac{a}{2}\\ &=&\frac{a}{2} \end{eqnarray*} implying $x\in A_N$, Therefore, $X_\infty\subseteq \displaystyle\bigcup_{n\in \mathbb{N}}A_n.$
If the limit is infinity, I used this definition provided in the internet "Let $(X,\mathcal{M},\mu)$ be a measure space, $\{f_n\}$ be a sequence of functions with common domain $X$, $f$ be a function on $X$, and $A\subset X$. We say that $\{f_n\}$ converges to $\infty$ on $A$ if $\forall x\in A$ and $\forall c\in \mathbb{R}$, there exists $N\in \mathbb{N}$ such that $\forall n\geq N$ we have that $f_n(x)\geq c$. We call $\infty$ the limit of $\{f_n\}$ and write $\lim_{n\to \infty}f_n=\infty$."
Following this, I will get the result however, I cant find any book that states this one (it has to be a book), and I am having a hard time proving it without considering cases (a proof that would be true regardless). Is there other way to prove this one?
