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!