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.


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.


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.


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


1 Answer 1


$$\int |f_n|1\{|f_n|>M\}d\mu \le M^{-\epsilon} \int |f_n|^{1+\epsilon}1\{|f_n|>M\}d\mu$$ $$ \le M^{-\epsilon} \int |f_n|^{1+\epsilon}d\mu \le M^{-\epsilon} \sup_n\int|f_n|^{1+\epsilon}d\mu \rightarrow 0$$

as $M\rightarrow\infty$ uniformly in $n$.

  • $\begingroup$ I am sorry I am a bit slow.. Could you say a little bit more? Specifically, how did you get the first inequality? $\endgroup$
    – illysial
    Jun 21, 2015 at 5:22
  • $\begingroup$ $|f_n|\times \frac{|f_n|^\epsilon}{M^\epsilon}\ge |f_n|$ on $\{|f_n|>M\}$... $\endgroup$
    – d.k.o.
    Jun 21, 2015 at 5:24
  • $\begingroup$ ahhh thank-you! $\endgroup$
    – illysial
    Jun 21, 2015 at 5:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.