Suppose $\mu$ is a finite measure and for some $\gamma>0$, we have $$\sup_{n}\int|f_n|^{1+\gamma}d\mu<\infty.$$ I'm trying to prove that $\{f_n\}$ is uniformly integrable.
$\{f_n\}$ is uniformly integrable if for a given $\epsilon$ there exists $M$ such that $$\int_{\{x:|f_n(x)| > M\}}|f_n(x)|d\mu<\epsilon$$ Is the proof below correct?
Let $\epsilon>0$, and assume the conditions above. Let $C = \sup_{n}\int|f_n|^{1+\gamma}d\mu$ and pick $M>\frac{C^{1/\gamma}}{\epsilon^{1/\gamma}}$.
We have that $$\int_{\{x:|f_n|>M\}}|f_n|d\mu\\
\leq \int_{\{x:|f_n|>M\}}|f_n|\frac{|f_n|^\gamma}{M^\gamma}d\mu \\
=M^{-\gamma}\int_{\{x:|f_n|>M\}}|f_n|^{\gamma + 1}d\mu \\
\leq M^{-\gamma}\int|f_n|^{\gamma + 1}d\mu \\
\leq M^{-\gamma}C < \frac{\epsilon}{C}C = \epsilon
$$
Therefore, $\{f_n\}$ is uniformly integrable.