# 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

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.

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.

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.
• Thanks to the answer in this link, I understood better what a diagonal extraction is. However I still have some questions. Why did you choose to perform the diagonal extraction in $L^2$ instead of $L^1$? Are you using the fact that $L^2$ is reflexive? If yes, how to use this fact to obtain a list of subsequences with which the diagonal extraction procedure is applied? Thanks for your attention! Dec 4, 2023 at 23:28
• "Why did you choose to perform the diagonal extraction in 𝐿2 instead of 𝐿1? " Because we want weak convergence in $L^2$. "Are you using the fact that 𝐿2 is reflexive? " Yes. Extract a sub sequence which works for $k=1$. Then extract a further subsequence that works for $k=2$, and so one. A subsequence that works for each $k$ is given by taking the $j$-th element of the $j$th subsequence. Dec 5, 2023 at 17:27