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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 $n \in \mathbb{N}.$

  1. 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)

  2. 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.

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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.

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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

  1. $\mu(\{|f_{n_{k_j}}| > j\}) \ge \gamma$
  2. $\gamma j^p > C^p$.

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

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  • $\begingroup$ What is $\sup_n ||f_n||_{L^p = C}$ ? I suppose that it is the Lebesgue space $L_p$ and its norm, isn't it ? Unfortunately, I suppose that I should do this without this fact. Actually, this is a problem for a chapter far before $L^2$ first mentioned. $\endgroup$ – user117375 Oct 6 '15 at 12:15
  • $\begingroup$ Which book are you following? As far as I know the proof of this fact relies heavily on Holder's inequality, which a tool you introduce once you have $L^p$ spaces. $C$ is just there for the notation, I decided to use it as a short hand for $\sup_n\big(\int|f_n|^{\delta + 1}\big)^{\frac{1}{\delta + 1}}$ which is bounded by assumption, so you should not worry about this. The proof boils down to an application of Holder's inequality and I don't see any easy argument to prove the result that does not involve it. $\endgroup$ – Giovanni Oct 6 '15 at 14:20
  • $\begingroup$ I use Real Analysis for Graduate Students by Richard F. Bass. It is an exercise in Chapter 7. We do not have $L^p$ space and Holder's inequality. Anyway, I will try to understand your proof. If you come up with more basic proof, please leaves the suggestion. Thank you very much. $\endgroup$ – user117375 Oct 6 '15 at 19:49
  • $\begingroup$ @user117375: yeah, I see that it is the chapter on limit theorems. I'll think about it :) $\endgroup$ – Giovanni Oct 6 '15 at 21:18

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