I'd would like to confirm if the following proposition is indeed true in the case of an arbitrary measure space.

Theorem: Let $(X,\Sigma,\mu)$ be a measure space and $\{f_n\}_{n\in\mathbb{N}}\subseteq \mathcal{L}^1_\mathbb{R}(\mu )$. If the limit $\lim_{n\to\infty}\int _Ef_nd\mu\in \mathbb{R} $ exists for all $E\in \Sigma$, then $\{f_n\}_{n\in\mathbb{N}}$ uniformly integrable.

The definition of uniformly integrable I'm using is: $\{f_n\}_{n\in\mathbb{N}}\subseteq \mathcal{L}^1_\mathbb{R}(\mu )$ is uniformly integrable if

$$(\forall \varepsilon >0)(\exists w\in \mathcal{L}_\mathbb{R}^1(\mu ))\Big(\sup _{n\in\mathbb{N}}\int _{\{|f_n|>|w|\}}|f_n|d\mu <\varepsilon \Big).$$

We can show that if $\mu$ is finite, then $\{f_n\}_{n\in\mathbb{N}}$ is uniformly integrable if and only if

$$\lim_{M\to \infty }\sup _{n\in\mathbb{N}}\int _{\{|f_n|>M\}}|f_n|d\mu =0$$

According to the 4.5.6. Theorem of the book "Measure Theory" written by V.I. Bogachev (see this link), the previous theorem is true when $\mu$ is finite. The author also shows in the proof of this theorem a version of the following lemma:

Lemma: Let $(X,\Sigma,\mu)$ be a measure space and $\{f_n\}_{n\in\mathbb{N}}\subseteq \mathcal{L}^1_\mathbb{R}(\mu )$. Then there're a subset $\{X_k\}_{k\in\mathbb{N}}\subseteq \Sigma$ of pairwise disjoint sets with $\mu(X_k)<\infty$ for all $k\in\mathbb{N}$, a sequence $(a_k)_{k\in\mathbb{N}}$ of $\mathbb{R}$ and a finite measure $\nu:\Sigma\to \mathbb{R}$ such that $\forall n\in\mathbb{N}$ the function $g_n:X\to \mathbb{K}$ given by $g_n:=\Sigma _{k=0}^\infty a_k^{-1}\mathbf{1}_{X_k}f_n$ is $\nu$-integrable and satisfies $\int _E|g_n|d\nu =\int _E|f_n|d\mu $ and $\int _Eg_nd\nu =\int _Ef_nd\mu $ for all $E\in \Sigma$.

In my opinion, the previous lemma allow us to reduce that theorem to the case in which $\mu$ is a finite measure and, therefore, obtain the desired result. However, maybe I'm doing some mistakes (since I didn't see that theorem in any book) and that theorem is in fact false.

Thank you for your attention!

I'd would like to confirm if the following proposition is indeed true in the case of an arbitrary measure space.

Theorem: Let $(X,\Sigma,\mu)$ be a measure space and $\{f_n\}_{n\in\mathbb{N}}\subseteq \mathcal{L}^1_\mathbb{R}(\mu )$. If the limit $\lim_{n\to\infty}\int _Ef_nd\mu\in \mathbb{R} $ exists for all $E\in \Sigma$, then $\{f_n\}_{n\in\mathbb{N}}$ uniformly integrable.

The definition of uniformly integrable I'm using is: $\{f_n\}_{n\in\mathbb{N}}\subseteq \mathcal{L}^1_\mathbb{R}(\mu )$ is uniformly integrable if

$$(\forall \varepsilon >0)(\exists w\in \mathcal{L}_\mathbb{R}^1(\mu ))\Big(\sup _{n\in\mathbb{N}}\int _{\{|f_n|>|w|\}}|f_n|d\mu <\varepsilon \Big).$$

We can show that if $\mu$ is finite, then $\{f_n\}_{n\in\mathbb{N}}$ is uniformly integrable if and only if

$$\lim_{M\to \infty }\sup _{n\in\mathbb{N}}\int _{\{|f_n|>M\}}|f_n|d\mu =0$$

According to the 4.5.6. Theorem of the book "Measure Theory" written by V.I. Bogachev (see this link), the previous theorem is true when $\mu$ is finite. The author also shows in the proof of this theorem a version of the following lemma:

Lemma: Let $(X,\Sigma,\mu)$ be a measure space and $\{f_n\}_{n\in\mathbb{N}}\subseteq \mathcal{L}^1_\mathbb{R}(\mu )$. Then there're a subset $\{X_k\}_{k\in\mathbb{N}}\subseteq \Sigma$ of pairwise disjoint sets with $\mu(X_k)<\infty$ for all $k\in\mathbb{N}$, a sequence $(a_k)_{k\in\mathbb{N}}$ of $\mathbb{R}$ and a finite measure $\nu:\Sigma\to \mathbb{R}$ such that $\forall n\in\mathbb{N}$ the function $g_n:X\to \mathbb{R}$ given by $g_n:=\Sigma _{k=0}^\infty a_k^{-1}\mathbf{1}_{X_k}f_n$ is $\nu$-integrable and satisfies $\int _E|g_n|d\nu =\int _E|f_n|d\mu $ and $\int _Eg_nd\nu =\int _Ef_nd\mu $ for all $E\in \Sigma$.

In my opinion, the previous lemma allow us to reduce that theorem to the case in which $\mu$ is a finite measure and, therefore, obtain the desired result. However, maybe I'm doing some mistakes (since I didn't see that theorem in any book) and that theorem is in fact false.

Thank you for your attention!