Lebesgue Integral

**problem** · April 3rd 2010, 05:55 AM

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

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

**Opalg** · April 3rd 2010, 11:52 AM

Originally Posted by problem

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

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

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

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