1. ## Lebesgue Integral

Let $\displaystyle (f_n) \in L^p(X)$ for all $\displaystyle n \in \mathbb{N}$ and $\displaystyle 1\le p <\infty$.Suppose there exist a function $\displaystyle g \in L^p(X)$ such that $\displaystyle |f_n| \le g$for all $\displaystyle n \in \mathbb{N}$.

Prove that for each $\displaystyle \epsilon>0$, there exist a set $\displaystyle E_\epsilon \subseteq X$ with $\displaystyle m(E_\epsilon)< \infty$ such that if $\displaystyle F\subseteq X$ and $\displaystyle F \cap E_\epsilon =\phi$, then $\displaystyle \int_F |f_n|^p dm < \epsilon^p$ for all $\displaystyle n \in \mathbb{N}.$

2. Originally Posted by problem
Let $\displaystyle (f_n) \in L^p(X)$ for all $\displaystyle n \in \mathbb{N}$ and $\displaystyle 1\le p <\infty$.Suppose there exist a function $\displaystyle g \in L^p(X)$ such that $\displaystyle |f_n| \le g$for all $\displaystyle n \in \mathbb{N}$.

Prove that for each $\displaystyle \epsilon>0$, there exist a set $\displaystyle E_\epsilon \subseteq X$ with $\displaystyle m(E_\epsilon)< \infty$ such that if $\displaystyle F\subseteq X$ and $\displaystyle F \cap E_\epsilon =\phi$, then $\displaystyle \int_F |f_n|^p dm < \epsilon^p$ for all $\displaystyle n \in \mathbb{N}.$
Since $\displaystyle |f_n|\leqslant g$ for all n, it suffices to find a set $\displaystyle E_\varepsilon \subseteq X$ with $\displaystyle m(E_\varepsilon)< \infty$ such that if $\displaystyle F\subseteq X$ and $\displaystyle F \cap E_\epsilon =\emptyset$, then $\displaystyle \int_F |g|^p dm < \varepsilon^p$ for all $\displaystyle n \in \mathbb{N}.$ That is a question that you have raised in this forum previously, and you'll find the answer here.