Let ${f_n}$ be a sequence of integrable functions on $\mathbb{R}$ such that $f_n\rightarrow f$ almost everywhere. Let ${f_n}$ be a sequence of integrable functions on $\mathbb{R}$ such that $f_n\rightarrow f$ almost everywhere. We also have $f\in L^1(\mathbb{R})$ and $\int_{\mathbb{R}}f_n\rightarrow \int_{\mathbb{R}}f$. 
I want to prove: For all $\epsilon>0$, there exist a measurable set $A\subset\mathbb{R}$ with finite measure, an integrable function $g\geq 0$, and an integer$N\in \mathbb{N}$ such that for all $n\geq N$, 
$|\int_{\mathbb{R}\backslash A}f_n|<\epsilon$ and $|f_n|\leq g$ on $A$. 
This question appears in one of the past papers of the real analysis course I am taking this semester. But I have no idea how to do it. 
 A: Because of $f\in L^{1}$, we get
$$
\int_{\mathbb{R}}\left|f\right|=\lim_{k\to\infty}\int_{\left[-k,k\right]}\left|f\right|.
$$
Thus, there is $k\in\mathbb{N}$ with $\int_{\mathbb{R}\setminus\left[-k,k\right]}\left|f\right|<\frac{\varepsilon}{3}$.
Furthermore, $f$ is uniformly integrable, i.e. there is $\delta>0$
with $\int_{B}\left|f\right|<\frac{\varepsilon}{3}$ as soon as $\lambda\left(B\right)<\delta$,
where $\lambda$ is the Lebesgue measure.
We know $f_{n}\to f$ pointwise
on $\left[-k,k\right]$. By Egoroff's theorem, there is thus a set
$A\subset\left[-k,k\right]$ with $\lambda\left(\left[-k,k\right]\setminus A\right)<\delta$
and with $f_{n}\to f$ uniformly on $A$.
In particular, we get
$$
\left|f_{n}\left(x\right)-f\left(x\right)\right|\leq C
$$
for all $n\in\mathbb{N}$ and $x\in A$ for some absolute constant
$C>0$. Hence, we can take $g=C\cdot\chi_{A}+\left|f\right|\in L^{1}$,
because of $\left|f_{n}\right|\leq\left|f_{n}-f\right|+\left|f\right|\leq C\chi_{A}+\left|f\right|$
on $A$.
Finally, we get
\begin{eqnarray*}
\left|\int_{\mathbb{R}\setminus A}f_{n}\right| & = & \left|\int_{\mathbb{R}}f_{n}-\int_{A}f_{n}\right|\\
 & \xrightarrow[\text{and }\int_{\mathbb{R}}f_{n}\to\int_{\mathbb{R}}f]{\text{dominated convergence, }\left|f_{n}\right|\leq g\text{ on A},\, f_{n}\to f\text{ pointwise}} & \left|\int_{\mathbb{R}}f-\int_{A}f\right|\\
 & = & \left|\int_{\mathbb{R}\setminus A}f\right|\\
 & \leq & \int_{\mathbb{R}\setminus\left[-n,n\right]}\left|f\right|+\int_{\left[-n,n\right]\setminus A}\left|f\right|\\
 & \leq & \frac{\varepsilon}{3}+\frac{\varepsilon}{3}.
\end{eqnarray*}
Hence, we get $\left|\int_{\mathbb{R}\setminus A}f_{n}\right|<\varepsilon$
for $n\in\mathbb{N}$ large enough, as desired.
A: Here's a quick idea that might or might not work. Since $f\in L^1$ you can take a set $X\subset\mathbb{R}$ of finite measure and such that $\left\lvert \int_{\mathbb{R}\setminus X} f\right\rvert<\epsilon$ and $\left\lvert \int_{\mathbb{R}\setminus X} f_n\right\rvert<\epsilon$ (for $n$ big enough), so all the action really takes place on $X$. 
Now since $X$ is finite measure, Severini-Egorov's theorem is available, and it should do the trick.
