Let $(X, \mathcal M, \mu)$ be a measure space with $\mu(X)$ finite, and $f,f_1,f_2,\dots$ be $L^1$ functions. Show that if $\{ f_n \}$ is uniformly integrable and $f_n \to f$ for a.e. $x \in X$, then $f_n \to f$ in $L^1$.
A sequence $\{ f_n \}$ is called uniformly integrable if for any $\epsilon > 0$ there exists $\delta > 0$ such that $A \in \mathcal M, \mu(A) < \delta \implies \int_A |f_n|d\mu < \epsilon$ for all $n$.
I know that by Egorov's theorem that for a given $\epsilon > 0$ there is $E \subseteq X$ such that $\mu(E) < \epsilon$ and $f_n \to f$ uniformly on $E^c$. Since $f_n$ is uniformly integrable, there is $\delta > 0$ such that $A \in \mathcal M, \mu(A) < \delta \implies \int |f_n|d\mu$ for all $n$. Thus we have that
$$\int |f_n - f| d\mu \leq \int_E |f_n - f| d\mu + \int_{E^c}\epsilon d\mu = \int_E |f_n - f| d\mu + \epsilon \mu(E^c)$$
The second integral of which is finite. The problem arises in trying to bound the first integral, i.e. $\int_E |f_n - f| d\mu$. Can anyone help?
You don't need Egorov's theorem. WLOG consider that $f=0$ and fix $\epsilon>0$
First see that, given $\epsilon>0$, you can find $\delta>0$ such that for all $n$, $\int_{X}|f_{n}|\mathbf{1}_{A}\,d\mu<\epsilon$ for all $A$ such that $\mu(A)<\delta$ and $\delta$ depends on $\epsilon$ only.
As almost everywhere convergece implies convergence in measure, you have that for $\delta$ as above, $\mu(|f_{n}|>\epsilon)<\delta$ for all $n\geq N (\epsilon,\delta)=N(\epsilon)$
Thus for all such $n\geq N(\epsilon)$, you have that
\begin{align}\int_{X}|f_{n}|\,d\mu &=\int_{X}|f_{n}|\mathbf{1}_{|f_{n}|>\epsilon}\,d\mu+\int_{X}|f_{n}|\mathbf{1}_{|f_{n}|\leq \epsilon}\,d\mu\\\\ &\leq \epsilon +\epsilon \cdot\mu(X)=(\mu(X)+1)\cdot \epsilon \end{align}
As $\epsilon>0$ was arbitrary, this shows that $f_{n}\xrightarrow{L^{1}} 0$
For the general case, notice that if $f_{n}$ is uniformly integrable then as $f\in L^{1}$, you have that $f_{n}-f$ is also uniformly integrable. (Sum of two uniformly integrable families is uniformly integrable) and $f_{n}-f\xrightarrow{a.e} 0$
The proof of the above is very easy. Consider $\epsilon>0$ and you find $\delta_{1}$ and $\delta_{2}$ such that $\int_{X}|f_{n}|\mathbf{1}_{A}<\frac{\epsilon}{2}$ for all n and $A$ such that $\mu(A)<\delta_{1}$ and $\int_{X}|f|\mathbf{1}_{B}<\epsilon$ for all $B$ such that $\mu(B)<\delta_{2}$. Then taking $\delta=\min(\delta_{1},\delta_{2})$ does the job for $|f_{n}-f|$.
If you still want to use Egorov's theorem then do the follwing:-
First choose $\delta$ for a given $\epsilon$ as in the definition of uniform integrability. Then you choose your set $E$ such that $\mu(E)<\delta$ and on $E^{c}$ you have uniform convergence. Then you have that $\sup_{x\in E^{c}}|f_{n}(x)-f(x)|<\epsilon$ for all $n\geq N(\epsilon)$ . Then you say that $\displaystyle\int_{X}|f_{n}-f|\,d\mu=\int_{E}|f_{n}-f|\,d\mu+\int_{E^{c}}|f_{n}-f|\,d\mu$.
The first term is now less than $\epsilon$ by uniform integrability and the 2nd term is less than $\mu(X)\cdot \epsilon$ due to uniform convergence. This holds for all $n\geq N(\epsilon)$.