# Suppose $\mu$ is a finite measure and $\sup_n \int |f_n|^{1+\epsilon} \ d\mu<\infty$ for some $\epsilon$. Prove that $\{f_n\}$ is uniformly integrable

Problem

Suppose $\mu$ is a finite measure and $\sup_n \int |f_n|^{1+\epsilon} \ d\mu<\infty$ for some $\epsilon>0$. Prove that $\{f_n\}$ is uniformly integrable.

Background

A family $\{f_n\}$ is uniformly integrable if given $\epsilon>0$, there exists $M$ such that $$\int_{\{x: |f_n(x)|>M\}} |f_n(x) d\mu<\epsilon$$ for each $n$.

Along with the usual limit theorems (Monotone Convergence, LDCT, Fatou's Lemma), the following is also known:

Theorem (Vitali): Suppose $\mu$ is a finite measure. If $f_n \rightarrow f$ a.e., and $\{f_n\}$ is uniformly integrable, then $\int |f_n -f| \rightarrow 0.$ On the other hand, 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.

Attempt

Let $E=\{x: |f_n|\geq 1\}$ and $m\in \mathbb{N}$ such that $\frac{1}{m}<\epsilon$. Then $|f_n|^{1+1/n}\leq |f_n|^{1+\epsilon}$ for all $x \in E$ and $k\geq M$. Consequently, \begin{align*} \int_E |f_n|^{1+1/k} \ d\mu &\leq \int_E |f_n|^{1+\epsilon} \ d\mu \\ &\leq \sup_n \int_E |f_n|^{1+\epsilon}\\ &<\infty. \end{align*} Thus, since $|f_n|^{1+1/k} \rightarrow |f_n|$ on $E$ we see by the Lebesgue dominated convergence theorem that $\int_E |f_n|^{1+1/n}\ d\mu \rightarrow \int_E|f_n|\ d\mu.$ Furthermore, on $E^c,$ $$|f_n|^{1+1/n}\leq |f_n|< 1$$ so by the monotone convergence theorem we have $\int_{E^c} |f_n|^{1+1/n}\ d\mu \rightarrow \int_{E^c}|f_n| \ d\mu$. Combining these two facts we see that $$\int |f_n|^{1+1/n}\ d\mu \rightarrow \int_{E^c}|f_n| \ d\mu$$ and by the theorem above, $\{|f_n|^{1+1/k}\}$ is uniformly integrable. Since $\{|f_n|\}\subset \{|f_n|^{1+1/k}\}$, we have the result.

Remarks

This feels horribly wrong but I can't seem to put a dent in this problem any other way.

$$\int |f_n|1\{|f_n|>M\}d\mu \le M^{-\epsilon} \int |f_n|^{1+\epsilon}1\{|f_n|>M\}d\mu$$ $$\le M^{-\epsilon} \int |f_n|^{1+\epsilon}d\mu \le M^{-\epsilon} \sup_n\int|f_n|^{1+\epsilon}d\mu \rightarrow 0$$
as $M\rightarrow\infty$ uniformly in $n$.
• $|f_n|\times \frac{|f_n|^\epsilon}{M^\epsilon}\ge |f_n|$ on $\{|f_n|>M\}$... – d.k.o. Jun 21 '15 at 5:24