Measure theory: explanation for a theorem similar to Egorov 

Let $ E$ be a measurable set with $m(E)\lt \infty$,and {$f_n$} a
  sequence of measurable functions defined on $E$.Let $f$ be a
  measurable (real-valued) functions such that $f_n(x)\rightarrow f(x)$
  for each $x\in E$.Then given $\epsilon \gt0$ and $\delta \gt0$,there
  is a measurable set $A\subset E$  with  $m(A)\lt \delta$ and an
  interger $N$ such that $\vert f_n(x)-f(x)\vert \lt \epsilon $ $\forall
> x\in E-A$ and all $n\ge N$.


If $f_n(x)\rightarrow f(x)$ $\forall x\in E$ then since $E-A\subset E$ then it follows trivially that $\vert f_n(x)-f(x)\vert \lt \epsilon $ $\forall x\in E-A$.
Would this line of thinking be correct?
This thing has created a mess in my mind.I want to know  what this theorem  means?What are its applications?If possible please
 explain it with the help of an example.
I'm also looking for  some suggestions for how to go through the proof of this theorem.
THANKS!!
 A: In plain English, this theorem roughly says: "If $f_n$ converges pointwise to $f$ on a set $E$ of finite measure, then for sufficiently large $n$, $f_n(x)$ is close to $f(x)$ everywhere except on a small subset of $E$."
For an example, take $E = (0,1)$ and take $f_n(x) = 1/(nx)$, which converges pointwise to $f(x) = 0.$ Given $\epsilon > 0 $ and $\delta > 0$, we could pick $A = (0,\delta/2)$ and we could pick $N$ to be any integer greater than $2/(\delta \epsilon)$. Then for any $x \in E - A = [\delta / 2, 1)$ and for any $n \geq N$, we have $|f_n(x) - f(x)| = 1/(nx) < 1/(\tfrac{2}{\delta \epsilon}\times \tfrac{\delta}{2}) = \epsilon$.
I can't think of a great application off the top of my head, but I can at least come up with a proof:
Fix an $\epsilon > 0$. For each $n \in \mathbb N$, define the measurable set,
$$ A_n = \{ x \in E \ : \ | f_m(x) - f(x) | \geq \epsilon {\rm \ \ for \ some \ \ } m \geq n \}.$$
Thus
$$ A_1 \supset A_2 \supset A_3 \supset A_4 \supset \dots $$
Since $f_n$ converges to $f$ pointwise, we know that for every $x \in E$, it is possible to find a sufficiently large $n \in \mathbb N$ such that $|f_m(x) - f(x) | < \epsilon$ for all $m \geq n$. Said another way: for every $x \in E$, there exists an $n \in \mathbb N$ such that $x \notin A_n$. Thus,
$$ \cap_{n \geq 1} A_n = \emptyset.$$
Since $m(E) < \infty$, a simple argument using countable additivity of the measure shows that
$$ \lim_{n \to \infty} (A_n) = 0,$$
i.e. for every $\delta > 0$, there exists an $n' \in \mathbb N$ such that 
$$ m(A_{n'}) < \delta.$$
Take $N = n'$ and $A = A_{n'}$. Then $m(A) < \delta$, and
$$ x \in E - A \ \implies \ |f_m(x) - f(x) | < \epsilon {\rm \ \ for \ all \ } m \geq N.$$
