# 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:

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

Thus,

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

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!

Tags :

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