(Proof Verification) Convergent sequence of functions in $L^1 (\mu) \implies$ the sequence is uniformly integrable. This corresponds to Exercise $3.11$(b) in Folland's Real Analysis Book. I was hoping someone could have a look at my proof. Specifically, I wanted to make sure the final paragraph of my argument holds. 
The definition of uniform integrability provided is as follows: 
$\{f_\alpha\}_{\alpha \in A}$ is uniformly integrable if $\forall \epsilon>0$, $\exists \delta > 0$ such that $|\int_E f_\alpha d\mu| < \epsilon$ for all $\alpha \in A$ whenever $\mu(E) < \delta$ for any measurable $E$.  
The problem gives us a sequence $f_n \to f$ in $L^1$. We are to show the entire sequence is uniformly integrable. Here's my attempt: 
Set any $n \in \mathbb{N}. |f_n| = |f_n - f_m + f_m - f + f| \leq |f_n -f_m| + |f_m - f| + |f|$. 
$\implies \int_E |f_n| d\mu \leq \int_E |f_n -f_m| d\mu + \int_E |f_m - f| d\mu + \int_E |f| d\mu.$
$f \in {L}^1 \implies f$ is uniformly integrable. $(f_n)_{n \in \mathbb{N}}$ is a convergent sequence in $L^1$ suggests that $m \to \infty$, we have: 
$\int_E |f_n -f_m| d\mu \to 0$ and $\int_E |f_m - f| d\mu \to 0$. 
Given $\epsilon > 0$ chosen and large enough $m$, whenever $\mu(E) < \delta$, (the $\delta > 0$ based on uniform integrability of $f$) we have: 
$$ \int_E |f_n| d\mu \leq \int_E |f_n -f_m| d\mu + \int_E |f_m - f| d\mu + \int_E |f| d\mu < 3 \frac{\epsilon}{3} = \epsilon $$
This holds for all $n \in \mathbb{N}$ ($\because n$ was arbitrary) $\implies$ bounding holds for the entire sequence $(f_n)_{n \in \mathbb{N}}$, in other words, implying that $(f_n)_{n \in \mathbb{N}}$ is uniformly integrable. For greater detail, we could alternatively say that using the fact that is a Cauchy sequence, this holds for all $l \geq n$, which would leave us with the finite collection $(f_i)_{1 \leq i \leq n-1}$ to consider. To this, we could apply the procedure used to establish that a finite collection of $L^1$ functions is uniformly integrable and get the result.
UPDATE 
As very kindly pointed out by Friedrich Philipp, the above proof was founded on an incorrect premise (detailed in comments). I am leaving the proof I settled on eventually here for reference. 
Fix $\epsilon > 0$. There exists $N_{\epsilon} \in \mathbb{N}$ such that for all $k > N_{\epsilon}$, by the fact of $L^1$ convergence, we have: $|\int_E f_k - f d\mu| \leq \int_E |f_k - f| d\mu < \frac{\epsilon}{2}$. considering $f \in L^1$ as a single element collection, by part (a), we can say that $f$ is uniformly integrable. That is, $\exists \delta_1 > 0$ such that $|\int_E f d\mu| < \frac{\epsilon}{2}$ whenever $\mu(E) < \delta_1$
Thus $\forall k > N_{\epsilon}$, we have: 
$$|\int_E f_k d\mu| \leq  \int_E |f_k - f| + |\int_E f d\mu| < \epsilon$$ whenever $\mu(E) < \delta_1$. 
By part (a), we have that $(f_i)_{1 \leq i \leq N_{\epsilon}}$ is a finite collection in $L^1$ which is uniformly integrable, that is, $\exists \delta_0 > 0$ such that $|\int_E f_i d\mu | < \epsilon$ for all $i \in \{ 1, \dots, N_{\epsilon} \}$ whenever $\mu(E) < \delta_0$.  
Taking $\delta = \min \{\delta_0, \delta_1\}$, whenever $\mu(E) < \delta$, we have $|\int_E f_n d\mu| < \epsilon$ for all $n \in \mathbb{N}$, that is, $\{f_n \}_{n \in \mathbb{N}}$ is uniformly integrable.  
 A: Let $\{f_n\}\rightarrow f\in L^1(\mu)$ and $\epsilon > 0$. Choose $N\in\mathbb{N}$ such that $$\int |f_n - f|d\mu < \frac{\epsilon}{2}$$ for all $n\in\mathbb{N}$ with $n> N$.
First, note that, since $\{f_n\}_\{n=1\}^N$ is uniformly integrable by part a.) there exists a $\delta_1 > 0$ such that, for all $n \leq N$,
$$\left|\int_E f_n d\mu\right| < \epsilon$$ whenever $E\in M$ and $\mu(E) < \delta_1$.
Now, since $\{f\}$ is uniformly integrable by part a.) there exists a $\delta_2 > 0$ such that $$\left|\int_E f d\mu\right| < \frac{\epsilon}{2}$$ whenever $E\in M$ and $\mu(E) < \delta_2$. Then, for all $n>N$, we have
\begin{align*}
\left|\int_E f_n d\mu\right| &= \left|\int_E (f_n - f) + \int_E f d\mu\right|\\
&\leq \left|\int_E (f_n - f)d\mu\right| + \left|\int_E f d\mu\right|\\
&< \frac{\epsilon}{2} + \frac{\epsilon}{2}\\
&= \epsilon
\end{align*}
whenever $E\in M$ and $\mu(E) < \delta_2$.
So, taking $\delta=\min\{\delta_1, \delta_2\}$, we have that, for all $n$,
$$\left|\int_E f_n d\mu\right| <\epsilon$$
whenever $E\in M$ and $\mu(E) < \delta$.
Thus $\{f_n\}$ is uniformly integrable.
