Let $(X,\mathcal{A},\mu)$ be a finite measure space. Let $g\in L^p(X)$ where $p\in [1,\infty)$. Define
$$ g_n(x)=\left\{\begin{array}{lll} g(x) & \text{if} & |g(x)|\le n, \\ 0 & \text{if} & |g(x)|> n \, \end{array} \right. $$
We observe that $g_n\in L^\infty$, and $g_n\to g$ pointwise in $X$.
I must prove that $g_n \to g $ in $L^p$
First we observe that
$$\color{blue}{\lim_{n\to \infty}\{x\in X\;:\;\lvert g(x) \rvert>n\}=\bigcap_{n=1}^\infty\{\lvert g \rvert>n\}=\emptyset\implies\lim_{n\to\infty}\chi_{\lvert g \rvert > n}=\chi_{\emptyset}=0.}$$ Then for the Dominated Convergence theorem result
$$\lim_{n\to\infty}\int_X\lvert g_n-g\rvert^p\;d\mu=\lim_{n\to\infty}\int_X\lvert g \rvert^p\chi_{\{\lvert g \rvert>n\}}\;d\mu\color{blue}{=}\int_X\lvert g \rvert^p\left(\lim_{n\to\infty}\chi_{\lvert g \rvert>n}\right)\;d\mu.$$
We could use the Dominated convergence theorem because $$\lvert \chi_{\lvert g \rvert> n}\rvert \le1\in L^p$$
$(1)$ It's correct?
$(2)$ I intuitively understand that $\bigcap_{n=1}^\infty \{\lvert g \rvert>n\}=\emptyset$,but is there a way to see it formally?