Show that $L^1$ convergence implies uniform integrability. Let $(X,\mathscr{M}, \mu)$ be a measure space and suppose $(f_n)$ is a sequence of $L^1$ functions ($L^1$ being the space of equivalence classes of absolutely integrable functions) converging to $f$ in $L^1$ ( that is $\displaystyle\lim\limits_{n \rightarrow \infty} \|f-f_{n} \|_{1}=\displaystyle\lim\limits_{n \rightarrow \infty} \int_{X} \left |f-f_{n} \right | d\mu=0$). Prove that the sequence $(f_n)$ is uniformly integrable, that is the sequence satisfies the following: 
i) $\displaystyle\sup\limits_{n} \|f_{n} \|_{1} < +\infty$
ii)  $\displaystyle\sup\limits_{n} \int_{|f_{n}| >M} \left | f_n \right | d \mu \rightarrow 0$ as $M \rightarrow \infty$
iii) $\displaystyle\sup\limits_{n} \int_{|f_{n}| < \delta} \left | f \right | d \mu \rightarrow 0$ as $\delta \rightarrow 0$.
I'm working on this problem and am having some difficulty. 
Part (i) is straight forward from $\left |f_n \right |< \left |f_n -f \right | +\left |f \right |$. Since the sequence is in $L^1$ it follows that $f$ is, so we have that $\int \left | f \right | =A <+\infty$. Let $\epsilon >0$, then we can find an $N$ so that for $n \geq N$ we have $\|f-f_{n} \|_{1} < \epsilon$. If $B = \max\{\|f-f_{1} \|_{1},\|f-f_{2} \|_{1},\ldots, \|f-f_{N-1} \|_{1},\epsilon\}$, then $\| f_n \| < A + B < +\infty$ for all $n$. 
Part (ii) I'm thinking that we should somehow use the fact that $\{{f_n} >M\} \subseteq \{ \left | f_n-f \right | > \frac{M}{2}\} \cup  \{\left | f \right | > \frac{M}{2}\}$ to break up the integral into pieces which we can control.
Part(iii) I'm thinking that we can again use $\left |f \right |< \left |f -f_n \right | +\left |f_n \right |$. For a fixed $n$ we can show that $\int_{|f_n| < \delta} |f_n| \rightarrow 0$ as $\delta \rightarrow 0$ so if we can somehow bound $\int_{|f_n| < \delta} |f-f_n|$ we will be done. 
Any help/advice/direction would be greatly appreciated. Thanks!
 A: I think to prove $1$ it suffice to use
$$|f_{n}-f|\ge ||f_{n}|-|f||$$
since you already know $f_{i},f\in L^{1}$, and $\lim_{n\rightarrow \infty}|f_{n}-f|\rightarrow 0$, if $\sup |f_{n}|=\infty$ then the above inequality will give a contradiction immediately. Here all  the norms are $L^{1}$ norms. 
For $2$ you can use $|f_{n}|\le |f_{n}-f|+|f|$. Then $$\sup \int_{|f_{n}|>M}|f_{n}|\le \sup \int_{|f_{n}|>M}|f_{n}-f|+\sup \int_{|f_{n}|>M}|f|$$
The second part goes to $0$ as $M\rightarrow \infty$, since for any $\epsilon$, for any $f_{n}$ there exist a $M_{n}$ such that $m(|f_{n}|\ge M_{n})<\epsilon$. The first part goes to $0$ for the same reason. You may pick $N$ to be the maximum of the two numbers to give a good enough bound. 
I don't have time to take a look at $3$, so sorry. 
A: For the second question, try to show that this strong convergent sequence$(f_n)$ is equi-integrable:
$\forall \epsilon>0, \exists \delta > 0$ such that for every measurable set $A \subset X$ with $\mu(A) < \delta$, we have
$$ \int_A |f_n| \mbox{ } d\mu < \epsilon, $$ for all $n \in \mathbb{N}$. This condition will imply statement two by Chebyshev's inequality
$$ \mu \{ |f_n| > t \} \leq \frac{1}{t} \int_X |f_n| \mbox{ } d\mu.$$
For the third part, we need a similar intermediate step. Try to show that 
$\forall \epsilon>0, \exists$ a measurable set $\Omega \subset X$ of finite measure such that 
$$ \int_{X \backslash \Omega} |f_n| \mbox{ } d\mu < \epsilon $$
for all $n \in \mathbb{N}$. Then, fix an $\epsilon > 0$. When $n > N$ is large enough, we get an estimate
$$ \int_{ \{ |f_n| < \delta \} } |f| \mbox{ } d\mu  
\leq  \int_{ \{ |f_n| < \delta \} }  |f - f_n| \mbox{ } d\mu 
+ \int_{ \{ |f_n| < \delta \} }  |f_n| \mbox{ } d\mu 
\leq ||f_n - f||_1 + \int_{X \backslash \Omega } |f_n| \mbox{ } d\mu
+  \mu(\Omega) \delta .$$
The third statement follows.
