Sufficient condition for uniform (Lebesgue) integrability

by The Bosco   Last Updated October 18, 2019 03:20 AM - source

This exercise is from Bass's book on graduate real analysis. I am stuck in one question.

Let $\mu$ be a finite measure and $\sup_n \int |f_n|^{1+\gamma} d\mu<\infty$ for any $\gamma > 0$. Then, $\{f_n\}$ is uniformly integrable.

Uniformly integrable means that, for any $\epsilon$, there is a $M$ such that $\int_{\{x:|f_n(x)|>M\}} |f_n(x)|d\mu < \epsilon$.

By definition, in the sets we are integrating over, we get that $$|f_n(x)|>M\geq0 \to \frac{|f_n(x)|}{M} > 1\to\frac{|f_n(x)|^{1+\gamma}}{M^\gamma} > 1$$

So the integral is bounded as follows:

$$\int_{\{x:|f_n(x)|>M\}}|f_n(x)| d\mu < \frac{1}{M^\epsilon} \int_{\{x:|f_n(x)|>M\}} |f_n(x)|^{1+\epsilon}d\mu$$ $$\leq \sup_n \left[ \frac{1}{M^\epsilon} \int_{\{x:|f_n(x)|>M\}}|f_n(x)|^{1+\epsilon} d\mu \right] = \frac{1}{M^\epsilon} \sup_n \left[ \int_{\{x:|f_n(x)|>M\}}|f_n(x)|^{1+\epsilon} d\mu \right]$$ $$\leq \frac{1}{M^\epsilon} \sup_n \left[ \int |f_n(x)|^{1+\epsilon} d\mu\right] < \infty$$


$$\int_{\{x:|f_n(x)|>M\}}|f_n(x)| d\mu < \frac{1}{M^\epsilon} \sup_n \left[ \int_{\{x:|f_n(x)|>M\}}|f_n(x)|^{1+\epsilon} d\mu \right] < \infty$$

My problem is picking the $\epsilon$ that will give an $M$ such that the above is true.

Can I just let

$$M = M(\epsilon) = \left(\frac{1}{\epsilon} \sup_n \left[\int |f_n(x)|^{1+\epsilon} d \mu \right] \right)^{1/\epsilon}$$


Thank you!

Answers 1

Let $\epsilon> 0$. Choose $M > \frac 1\epsilon\sup_n\int|f_n|^2\,d\mu$. Then $$ \int_{\{|f_n|>M\}}|f_n|\,d\mu\,\le\frac 1M\int_{\{|f_n|>M\}}|f_n|^2\,d\mu\,\le\,\epsilon. $$

October 18, 2019 03:17 AM

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