I am trying to solve the following problem: A sequence $\{f_n\}_{n \in \mathbb{N}}$ is said to be uniformly integrable in $L(X,d\mu)$ if $$\lim_{t \to \infty}\sup_{n\ge1}\int_{\{|f|>t\}}|f_n| \, d\mu = 0.$$ Now, assume that $\mu(X) < \infty$ and $f_n \to f$ almost everywhere. If $\{f_n\}_{n \in \mathbb{N}}$ is uniformly integrable show that $f \in L(X,d\mu)$ and $$\lim_{n \to \infty}\int_{X}f_n \, d\mu = \int_{X}f \, d\mu.$$
My Intuitive Idea: Since, we have a finite measure space and pointwise convergent functions, we can use Egorov's Theorem to get uniform convergence. This allows us to get a convergence in the integral except for a set of small measure. Then, use uniform integrability to show that the integral over this small measure set vanishes.
My Attempt: Fix $\epsilon > 0$, then there exists a set $E \subset X$ such that $m(X/E) < \epsilon$ and $f_n$ converges $f$ uniformly on $E$. Then, \begin{equation} \begin{split} \lim_{n \to \infty}\int_{X}f_n \, d\mu &= \lim_{n \to \infty}\int_{E}f_n \, d\mu + \lim_{n \to \infty}\int_{X/E}f_n \, d\mu \\ &= \int_{E}\lim_{n \to \infty}f_n \, d\mu + \lim_{n \to \infty}\int_{X/E}f_n \, d\mu \\ &= \int_{E}f \, d\mu + \lim_{n \to \infty}\int_{X/E}f_n \, d\mu \end{split} \end{equation}
I do not know how to handle $$\lim_{n \to \infty}\int_{X/E}f_n \, d\mu.$$
My Questions:
(1) Am I on the right track?
(2) What is the intuitive meaning of uniform integrability?
(3) Is the result still true for $\sigma-$finite measure spaces?
Thanks, in advance.