TITLE: Sufficient condition for Uniform integrability
QUESTION [5 upvotes]: The following is a problem that I've been struggling hard to prove. Any help would be appreciated.
Suppose $\mu$ is a finite measure and for some $\gamma>0$
$$
\sup_n \int |f_n|^{1+\gamma} \ d\mu < \infty
$$
Prove that $\{f_n\}$ is uniformly integrable, i.e given an $\epsilon>0$ there exists $M$ such that $\displaystyle \sup_{n} \int_{\{x:|f_n(x)|> M \}} |f_n(x)|\  d\mu < \epsilon$
My attempt to solve: I wrote $\int |f_n|^{1+\gamma} \ d\mu = \int_{\{x:|f_n(x)|> M \}} |f_n|^{1+\gamma} \ d\mu + \int_{\{x:|f_n(x)|\leq M \}} |f_n|^{1+\gamma} \ d\mu$. How do I get rid of $1+\gamma\ ?$

REPLY [5 votes]: Hint: For $x$ in the set $\{|f_n|  > M\}$, we have $$|f_n(x)| \le |f_n(x)| \frac{|f_n(x)|^\gamma}{M^\gamma} = \frac{|f_n(x)|^{1+\gamma}}{M^\gamma}.$$