If $\{f_n\}$ is uniformly integrable, then there's a subsequence $(f_{k_n})$ of $(f_n)$ such that $\big(\int _Ef_{k_n}d\mu \big)$ is a Cauchy sequence

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The page 20 of the book "Topological Fixed Point Theory for Singlevalued and Multivalued Mappings and Applications" (written by Ben Amar and O'Regan) has the following lemma.

Lemma: Let $(X,\Sigma,\mu )$ be a finite measure space and $\{f_n\}_{n\in\mathbb{N}}\subseteq \mathcal{L}^1_\mathbb{R}(\mu )$. Suppose that

  1. $\sup_{n\in\mathbb{N}}\Vert f_n\Vert _{L^1}<\infty$
  2. $(\forall \varepsilon >0)(\exists \delta >0)(\forall E\in \Sigma )\left(\mu (E)<\delta\Rightarrow \sup_{n\in\mathbb{N}}\int _E|f_n|d\mu <\varepsilon \right)$.

Then there's a subsequence $(f_{k_n})_{n\in\mathbb{N}}$ of $(f_n)_{n\in\mathbb{N}}$ such that $\left(\int _Ef_{k_n}d\mu \right)_{n\in\mathbb{N}}$ is a Cauchy sequence for all $E\in \Sigma$.


Unfortunately, the book doesn't give any tips on how to prove this lemma.

My question is: how can I prove that lemma?


I couldn't do anything worth mentioning, but I know that the conclusion of that lemma is true if and only if there's a subsequence $(f_{k_n})_{n\in\mathbb{N}}$ of $(f_n)_{n\in\mathbb{N}}$ that converges weakly.

Thank for your attention!


EDIT: Please don't use the Dunford-Pettis Theorem because I want to use that lemma to prove this theorem.

Please don’t use either the Eberlein-Smulian Theorem, because I want to avoid advanced theorems of functional analysis.

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For each fixed $k$, the sequence $\left(f_n\mathbf{1}_{\{\lvert f_n\rvert\leqslant k\}}\right)$ is bounded in $L^2$ hence using the diagonal extraction, we can find an increasing sequence of integers $(n_j)$ such that for each $k$, $\left(f_{n_j}\mathbf{1}_{\{\lvert f_{n_j}\rvert\leqslant k\}}\right)$ converges weakly to some $g_k$ in $L^2$.

For each $E\in \Sigma$, the sequence $(\int_E f_{n_j})$ is Cauchy. Indeed, for a fixed $\varepsilon$, pick $k$ such that $\sup_n \int \lvert f_n\rvert\mathbf{1}_{\{\lvert f_n\rvert> k\}}<\varepsilon$. Then use the fact that $(\int_E f_{n_j}\mathbf{1}_{\{\lvert f_{n_j}\rvert\leqslant k\}})$ converges to reach the conclusion.