Is the limit of uniformly integrable functions integrable? If $\left\{f_n\right\}$ are uniformly integrable and $f_n\overset{a.e.}{\rightarrow}f$ ($f$ measurable), is $f$ integrable? Can "uniformly integrable" be weakened to "integrable"?

Notes on Additional Assumptions

*

*It should be noted that for this to be true, we need the hypothesis that $|f|<\infty$ a.e..

For example, if $(X,\mathfrak{m},\mu)$ is defined by $X=\{x\},\mathfrak{m}=\{\emptyset,X\},\mu(X)=1$, take $f_n:=n$.


*It also should be noted that we also need the hypothesis that $\mu(X)<\infty$.

For example, if the measure space $X$ is $\mathbb{R}$ with Lebesgue measure $m$, take $f_n:=\chi_{[0,n]}$.

A proof assuming the following hypothesis :


*

*$|f|<\infty$ (or only a.e. is also ok)


*$\mu(X)<\infty$

can be given as follow.
Proof
By uniform integratability, find $\delta>0$ with $\mu(E)<\delta$ implies $\int_E|f_n|<1$. By Fatou, $\chi_Ef\in L^1(\mu)$.
Since $\mu(X)<\infty$, by successively cutting away these (can be taken to be finitely many) $E$, we may just assume now that every non-empty measurable subset of $X$ has measure $\geq\delta$.
In this case, notice for $x\in X$, $\mu(f^{-1}(f(x)))\geq\delta$, which shows that $f(X)$ is a finite set since $\mu(X)<\delta$, then the condition $|f|<\infty$ applied to each fiber of $f$ shows that $f\in L^1(\mu)$. $\blacksquare$
