Show that balls in $L^{1 + \delta}(\mu)$, with $\mu$ some finite measure, are uniformly integrable Can anyone give some suggestion/guideline to do this problem :
Suppose $\mu$ is a finite measure and for some $\delta > 0$ $$\sup_n \int |f_n|^{1 + \delta}d\mu < \infty.$$
Show that $\{f_n\}$ is uniformly integrable.
The information I have is 
1 Def : $\{f_n\}$ is uniformly integrable if for each $\epsilon > 0$, there exists $M$ such that $$\int_{\{x : |f(x)| > M\}}|f_n(x)| d\mu < \epsilon$$ for all $n \in \mathbb{N}.$ 


*Theorem : $\{f_n\}$ is uniformly integrable if and only if $\sup_n \int |f_n| d\mu < \infty$ and $\{f_n\}$ is uniformly absolutely continuous. (I think that this theorem might not be helpful, instead it might make the matter worse)

*Vitali : Let $\mu$ be a finite measure. If $f_n \rightarrow f$ a.e., each $f_n$ is integrable, $f$ is integrable, and $\int|f_n - f| \rightarrow 0$, then $\{f_n\}$ is uniformly integrable.
 A: Let $H=\sup_{n} \int |f_{n}|^{1+\delta} $ and $E=E_{n,M}=\{ x:|f_{n}(x)|>M \}$.Then we have
$$M^{1+\delta} \mu(E) \leq \int_{E} |f_{n}|^{1+\delta}  \leq H$$
and
$$\int_{E} |f_{n}| \leq \left( \int_{E} |f_{n}|^{1+\delta} \right)^{\frac{1}{1+\delta}} \left( \int_{E} 1 \right)^{\frac{\delta}{1+\delta}} \leq H^{\frac{1}{1+\delta}} \times\mu(E)^{\frac{\delta}{1+\delta}}$$
by H \"{o}lder's Inequality.
So we get
$$\int_{E_{n,M}} |f_{n}|\leq \frac{H}{M^{\delta}}$$ and hence $\{ f_{n} \}$ is uniformly integrable.
A: Suppose by contradiction that $\{f_n\}$ is not uniformly integrable; then there is $\epsilon > 0$ and a subsequence $\{f_{n_k}\}$ such that for every $k > 0$ $$\int_{\{|f_{n_k}| > k\}} |f_{n_k}|\,d\mu \ge \epsilon.$$
Let $p = \delta + 1$, let $q$ be such $p^{-1} + q^{-1} = 1$ and set $\sup_n\|f_n\|_{L^p} = C$, then by Holder's inequality
$$ \epsilon \le \int_{\{|f_{n_k}| > k\}} |f_{n_k}|\,d\mu \le \|f_{n_k}\|_{L^p}\mu(\{|f_{n_k}| > k\})^{\frac{1}{q}} \le C\mu(\{|f_{n_k}| > k\})^{\frac{1}{q}}.$$
We get a contradiction if we can show that $$\lim_{k \to \infty} \mu(\{|f_{n_k}| > k\}) = 0. \tag 1$$

To prove $(1)$ we argue by contradiction: if it does not converge to $0$ then we can find a positive $\gamma$ and a further subsequence such that $$\lim_{j \to \infty} \mu(\{|f_{n_{k_j}}| > j\}) \ge 2\gamma.$$
Let $j$ be so large that 


*

*$\mu(\{|f_{n_{k_j}}| > j\}) \ge \gamma$

*$\gamma j^p > C^p$. 


Then we have $$C^p \ge \int_{\{|f_{n_{k_j}}| > j\}} |f_{n_{k_j}}|^p \ge j^p\mu(\{|f_{n_{k_j}}| > j\}) \ge \gamma j^p > C^p.$$
