The proof is incorrect. In the fist part of the proof, the $\delta$ you picked does not guarantee that for all measurable sets $A$, $\mu(A) < \delta$ implies $|\int_A f_n\, d\mu| < \epsilon$. The second part of the proof is almost correct, but the condition $\lim\limits_{M\to \infty} \mu(E_M) = 0$ follows from the assumption $\sup_n \int |f_n|\, d\mu < \infty$, not that $\mu$ is finite. Indeed, if $\alpha := \sup_n \int |f_n|\, d\mu < \infty$, then $\mu(E_M) \le \frac{\alpha}{M}$ for all $M > 0$; as a consequence, $\lim\limits_{M\to \infty} \mu(E_M) = 0$. If you make this fix, then the second part of the proof will be correct.
To prove the forward direction, fix $\epsilon > 0$ and choose $M > 0$ such that
$$\sup_n \int_{E_M} |f_n| \, d\mu < \frac{\epsilon}{2}.$$
Then
$$\int |f_n|\, d\mu = \int_{E_M} |f_n|\, d\mu + \int_{E_M^c} |f_n|\, d\mu < \frac{\epsilon}{2} + M\mu(X) < \infty$$
for all $n\in N$.
Set $\delta = \frac{\epsilon}{2M}$. For all measurable sets $A$, $\mu(A) < \delta$ implies
$$\left|\int_A f_n\, d\mu\right| \le \int_A |f_n|\, d\mu = \int_{A\cap E_M} |f_n|\, d\mu + \int_{A\cap E_M^c} |f_n|\, d\mu < \frac{\epsilon}{2} + M\mu(A) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$
Since $\epsilon$ was arbitrary, $\{f_n\}$ is uniformly absolutely continuous.
Let $p=1+\gamma$. Then $\frac1q=\frac{p-1}{p}=\frac{\gamma}{1+\gamma}$ so $q=\frac{1+\gamma}{\gamma}$. By Holder's inequality,
$\int_A |f_n|\,d\mu \leq \|f_n\|_p\|1\|_q=\left[\int_A |f_n|^{1+\gamma}\,d\mu\right]^{\frac{1}{1+\gamma}}\left[\mu(A)\right]^{\frac{\gamma}{1+\gamma}}\leq \left[\sup_n\int_X |f_n|^{1+\gamma}\,d\mu\right]^{\frac{1}{1+\gamma}}\left[\mu(A)\right]^{\frac{\gamma}{1+\gamma}} .$
It should be clear how to pick $\delta(\epsilon)$ so that the above is bounded above by $\epsilon$.
Best Answer
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 $\delta$