# Is this theorem true in the case of a general measure space?

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.

$$\newcommand{\ep}{\varepsilon}\newcommand{\R}{\mathbb R}\newcommand{\N}{\mathbb N}$$The answer is yes, the $$f_n$$'s are uniformly integrable wrt to $$\mu$$.
Indeed, let us follow the proof of Theorem 4.5.6 in Bogachev's book. Note that the measure $$\mu$$ is $$\sigma$$-finite on the set $$X_0:=\bigcup_{n\in\N}f_n^{-1}(\R\setminus\{0\})$$ and $$\int_E f_n\,d\mu=\int_{X_0\cap E} f_n\,d\mu$$ for all $$n$$ and all $$E\in\Sigma$$. So, without loss of generality $$\mu$$ is $$\sigma$$-finite. So, there is a $$\Sigma$$-measurable partition $$(X_k)_{k\in\N}$$ of $$X$$ such that $$\mu(X_k)<\infty$$ for all $$k$$.
Let now $$d\nu:=h\,d\mu$$, where $$h=a_k:=2^{-k}/(1+\mu(X_k))\in(0,\infty)$$ on $$X_k$$, and let $$g_n:=f_n/h$$. Then $$\nu$$ is a finite measure and $$\int_E f_n\,d\mu=\int_E g_n\,d\nu$$ for all $$E\in\Sigma$$. By Theorem 4.5.6 in Bogachev's book, the $$g_n$$'s are uniformly integrable wrt to $$\nu$$; that is, $$$$\lim_{M\to\infty}\sup_n\int_{|g_n|>M}|g_n|\,d\nu=0,$$$$ which can be rewritten as $$$$\lim_{M\to\infty}\sup_n\int_{|f_n|>Mh}|f_n|\,d\mu=0.$$$$ Therefore and because $$h>0$$ and $$\int h\,d\mu=\int d\nu<\infty$$, we conclude that the $$f_n$$'s are uniformly integrable wrt to $$\mu$$. $$\quad\Box$$