$\{f_n\}$ is uniformly integrable if and only if $\sup_n \int |f_n|\,d\mu < \infty$ and $\{f_n\}$ is uniformly absolutely continuous? Let $(X, \mathcal{A}, \mu)$ be a measure space. A family of measurable functions $\{f_n\}$ is uniformly integrable if given $\epsilon$ there exists $M$ such that$$\int_{\{x : |f_n(x)| > M\}} |f_n(x)|\,d\mu < \epsilon$$for each $n$. The sequence is uniformly absolutely continuous if given $\epsilon$ there exists $\delta$ such that$$\left|\int_A f_n\,d\mu\right| < \epsilon$$for each $n$ if $\mu(A) < \delta$.
Suppose $\mu$ is a finite measure. How do I see that $\{f_n\}$ is uniformly integrable if and only if $\sup_n \int |f_n|\,d\mu < \infty$ and $\{f_n\}$ is uniformly absolutely continuous?
 A: Assume that $\{f_n\}$ is uniformly absolutely integrable. Let $\varepsilon > 0$. Now
\begin{align}
\int_X |f_n| = \int_{X \cap \{f_n \geq M_\varepsilon\}} |f_n| + \int_{X \cap \{f_n < M_\varepsilon\}} |f_n| \leq \varepsilon + M_\varepsilon \mu(X)
\end{align}
for all $n$. Thus the supremum over $n$ is finite.
To get uniform absolute continuity, notice that
\begin{align}
\left| \int_A f_n \right| & \leq \int_{A \cap \{f_n \geq M_\varepsilon\}} |f_n| + \int_{A \cap \{f_n < M_\varepsilon\}} |f_n| \\
& \leq \varepsilon + M_\varepsilon \mu(A)
\end{align}
for all $n$. Now choose $\delta < \varepsilon/M_\varepsilon$.
Now assume $\sup_n \|f_n\|_1 < \infty$ and the uniform abs. continuity. Let $\varepsilon > 0$ and let $\delta > 0$ be such that $\mu(A) < \delta$ implies $ |\int_A f_n| < \varepsilon$ for all $n$. Since $\int|f_n| < \infty$, we have
\begin{align}
\lim_{M \to \infty} \mu \{ |f_n| > M \} = 0\,.
\end{align}
Thus we may choose $M_n$ so large that $\mu\{ |f_n| > M_n \} < \delta$. Now
\begin{align}
\int_{|f_n| > M_n } |f_n| &= \int_{f_n > M_n} f_n + \int_{f_n < -M_n} (-f_n) \\
&= \left| \int_{f_n > M_n} f_n \right| + \left| \int_{f_n < -M_n} f_n \right| \\
&< \varepsilon + \varepsilon
\end{align}
for all $n$ since the sets over which we integrate have measure less than $\delta$
A: Here I show the forward direction.  Once I see the reverse direction, I will edit.
Given $\varepsilon>0$, choose $M>0$ so that 
$$ \int_{\{x : |f_n(x)| > M\}} |f_n(x)|\,d\mu < \epsilon $$
for all $n$.  Then we have
$$ \int|f_n|\ d\mu=\int_{\{|f_n|\leq M\}}|f_n|\ d\mu+ \int_{\{|f_n|> M\}}|f_n|\ d\mu\leq M\mu(X)+\varepsilon<\infty.$$
Since $n$ was arbitrary, we have
$$\sup_n\int|f_n|\ d\mu<\infty.  $$
Now pick $\delta>0$ so that $\delta<\varepsilon/M$. Then for measurable $A$ with $\mu(A)<\delta$, we have
\begin{align*}
\left|\int_A f_n\ d\mu\right|&\leq \left|\int_{A\cap\{|f_n|\leq M\}} f_n\ d\mu\right| +\left|\int_{A\cap\{|f_n|> M\}} f_n\ d\mu\right| \\
&\leq \int_{A\cap\{|f_n|\leq M\}} |f_n|\ d\mu + \int_{A\cap\{|f_n|> M\}} |f_n|\ d\mu \\
&<M\delta +\varepsilon <2\varepsilon.
\end{align*}
By rescaling, we see that $\{f_n\}$ is uniformly absolutely continuous.
