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!


1 Answer 1


$\newcommand{\ep}{\varepsilon}\newcommand{\R}{\mathbb R}\newcommand{\N}{\mathbb N}$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, \begin{equation} \lim_{M\to\infty}\sup_n\int_{|g_n|>M}|g_n|\,d\nu=0, \end{equation} which can be rewritten as \begin{equation} \lim_{M\to\infty}\sup_n\int_{|f_n|>Mh}|f_n|\,d\mu=0. \end{equation} 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 $\mu$. $\quad\Box$


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