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"?
 A: *

*Yes
$\int \left|f\right|\overset{\mathrm{a}}{\leq}\liminf\int |f_n|\overset{\mathrm{b}}{<}\infty$
a. Fatou's
lemma
b. Uniform integrability $\implies$ $\sup \int |f_n|<\infty$ (e.g.
Klenke, Theorem 6.24i)

*No, e.g. $f_n:=\mathbb{1}_{\left[-n,n\right]}$ (borrowed from Per Manne's comment below)
A: Put $f_n(x) = n(n+1)I_{(1/n+1, 1/n)}$, $x\in[0,1]$.  Then $f_n\to 0$ in $[0,1]$ and the sequence of the $f_n$ is $L^1$-bounded.  However, $f_n\not\rightarrow 0$ in $L^1$ as $n\to\infty$, since $\|f_n\|_1 = 1$ for all $n$.
A: Evan's accepted answer is almost correct and useful in most cases. However, it's missing a bit of nuance that is actually rather important when discussing other theorems like the Vitali Convergence Theorem. 
Regarding your first question:  The pointwise limit of uniformly integrable functions is only guarenteed to be integrable if their domain $E$ has finite measure. I am using $m(A)$ for any set $A$ to refer to Lebesgue measure. 
As a counterexample in the case where $m(E) = \infty$, consider Per Manne's example. Set $E = [0, \infty)$ and $f_n = \chi_{[0, n]}$, the characteristic (indicator) function for $[0, n]$. The family $\{f_n\}$ is uniformly integrable: given $\epsilon > 0$, set $\delta = \epsilon$. Then for a measureable set $A$, 
$$m(A) < \delta \implies \int_A |f_n| \le \int_A 1 < \delta = \epsilon$$ 
However, $f_n \to f \equiv 1$ pointwise on $E$, and $\int_E |f| = \infty$ so $f$ is not integrable. 
In general, another condition called tightness is used to justify the integrability of a pointwise limit of functions on any domain. Specifically, a family of measurable functions $F$ on a domain $E$ is tight over $E$ iff $\forall \, \epsilon > 0$, there exists $E_0 \subset E$ such that $$\int_{E \setminus E_0} |f| < \epsilon \, \, \forall \, f \in F $$
The proof of this is very similar to what Evan Aad described earlier, with an extra trick: I've sketched it below.


*

*Fix $\epsilon > 0$. Given $\{f_n\}$ measureable, uniformly integrable, and tight on $E$ where $\{f_n\} \to f$ pointwise on $E$, there exists some measurable set $E_0$ with finite measure s.t. 
$$ \int_{E \setminus E_0} |f_n| < \epsilon \text{ and by Fatou's Lemma } \int_{E \setminus E_0} |f| < \epsilon$$

*We have now reduced the problem to showing $f$ is integrable over $E_0$, as


$$\int_E |f| = \int_{E \setminus E_0} |f| + \int_{E_0} |f|$$
As Evan pointed out, uniform integrability implies $\sup \int_{E_0} |f_n| < \infty$ as $m(E_0) < \infty$, and then Fatou's Lemma implies $\int_{E_0} |f| < \int_{E_0} |f_n| < \infty$, completing the proof.  
As a remark, if $m(E) < \infty$, then any family of functions is obviously tight; this is why the pointwise limit of uniformly integrable functions is guarenteed to be integrable if their domain has finite measure. 
In response to your second question: As has been mentioned above, the pointwise limit of integrable functions is not necessarily integrable on any domain. An example of this with a domain of finite measure is $f_n = n$ on $E = [0,1]$. Then ${f_n} \to f \equiv \infty$ pointwise and clearly $f$ is not integrable even though $\int_E |f_n| = n$ for any $n$. 
Citation: I am drawing from Sections 4.6 and 5.1 in Royden and Fitzpatrick, Real Analysis. 
A: 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 $f\in L^1(\mu)$. $\blacksquare$
