$\lim_{n\to \infty}\int_{A}f_n\;dm=\int_A \lim_{n\to \infty}f_n\;dm$ We say that functions $f_n$, for $n=1,2,\dots$ integrable on a measurable set $A$ are equivalent-integrable if for every $\epsilon>0$ there is $\delta>0$ such that for every measurable subset $B$ of $A$, we have $\int_{B}|f_n|<\epsilon$ for $n=1,2,\dots$ whenever $m(B)<\delta$. Show that if ${f_n}$ is a pointwise convergent sequence of equipment-integrable functions on a set $A$ of finite measure, then
$$\lim_{n\to \infty}\int_{A}f_n\;dm=\int_A\lim_{n\to \infty}f_n\;dm$$
I feel we need to use Egorov's theorem.
Egorov's theorem said that, for any finite measure space, Given $f_1,f_2,f_3,\dots f_n\dots$ converges pointwise to $f$ on $A$, then for all $\epsilon>0$ there exists $E\subset A$ such that $m(A\cap E^c)<\epsilon$ and $f_n$ converges to $f$ uniformly on $E$.
How do I connect the dots? Hints would be great.
 A: Let $\epsilon>0$ be given and choose a $\delta>0$ such that the definition of uniformly integrability is satisfied. For this $\delta>0$, apply the Egorov theorem, then
\begin{align*}
\int_{A}|f_{n}(x)-f(x)|&=\int_{A-E}|f_{n}(x)-f(x)|+\int_{E}|f_{n}(x)-f(x)|\\
&=I_{n}+J_{n}.
\end{align*}
It is easy to conclude that $J_{n}\rightarrow 0$ by using LDCT.
For that $I_{n}$, we have
\begin{align*}
I_{n}\leq\int_{A-E}|f_{n}(x)|+\int_{A-E}|f(x)|\leq\epsilon+\int_{A-E}|f(x)|.
\end{align*}
Now note that $f$ is integrable on $A$ (explain later). So we can choose $\delta>0$ in such a way that $f$ is absolutely continuous on there:
\begin{align*}
\int_{B}|f(x)|<\epsilon,~~~~B\subseteq A,~~~~m(B)<\delta.
\end{align*}
For then $J_{n}\leq 2\epsilon$.
That $f$ being integrable actually needs some further observation. Let $B_{l}=\{x\in A:|f(x)|\geq l\}$. Note that $B_{l}\downarrow\emptyset$ and $A$ is of finite measure, then $m(B_{l})\rightarrow 0$. Hence, there is some $l$ such that $m(B_{l})<\delta$, then
\begin{align*}
\int_{A}|f(x)|&=\int_{A-B_{l}}|f(x)|+\int_{B_{l}}|f(x)|\\
&\leq m(A)\cdot l+\liminf_{n}\int_{B_{l}}|f_{n}(x)|\leq m(A)\cdot l+\epsilon<\infty.
\end{align*}
