If $X=[0,\infty)$, $m$ is the collection of Lebesgue measurable sets and $\mu$ is Lebesgue measure. Prove or disprove whether the sequence $$f_n(x)=\sqrt{n}\chi_{[0,1/n]}(x)$$ is uniformly integrable.

At a glance, my judgement is No, it is NOT uniformly integrable.

By definition,

Given a measure space $(X,m,\mu)$, A sequence $\{f_n\}_{n=1}$ is said to be uniformly integrable if $\forall \epsilon >0 $ $\exists \delta > 0 $ such that given $E \subset X$ ,measurable, if $\mu (E)< \delta $ then $\forall n$, $\int_E |f_n| d\mu < \epsilon$.

So I let $E \subset R$ such that $\mu (E)<\delta$. Hence I wish to show that $\int_E|f_n|<\epsilon$ $\forall n$

BUT $$\int_E|f_n|=\int_E \sqrt{n}\chi_{[0,1/n]} =\sqrt{n}\cdot \mu (E\cap[0,1/n])\leq\sqrt{n} \cdot min(\delta,1/n)$$

And Since $\sqrt{n} \delta$ is unbounded as $n \to \infty$, we conclude that the sequence of functions is not uniformly integrable

Does this make sense?

  • 1
    $\begingroup$ You give away too much in passing from $\int_E\sqrt{n}\chi_{[0,1/n]}$ to $\sqrt{n}\mu(E)$. A sharper upper bound would be $\sqrt{n}\mu(E\cap[0,1/n])\le\sqrt{n}\min(\delta,1/n)$. $\endgroup$ – John Dawkins Jul 11 '17 at 21:14
  • $\begingroup$ Any sequence $f_n$ in $L^1$ such that $\|f_n\|_1 \to 0$ is uniformly integrable. $\endgroup$ – zhw. Jul 11 '17 at 21:14
  • $\begingroup$ What @zhw. said, or: bounded in $L^2$. $\endgroup$ – Did Jul 11 '17 at 21:21
  • $\begingroup$ @zhw. very helpful. $\endgroup$ – J. Kyei Jul 11 '17 at 21:31
  • $\begingroup$ @JohnDawkins. You're right, I jumped a couple relevant steps. But if I get you well, even though $\sqrt{n}min(\delta,1/n)$ depends on $n$, we can still conclude that it is less than some $\epsilon$? I thought because it's dependent on $n$ it increases past any $\epsilon$ and hence not uniformly integrable $\endgroup$ – J. Kyei Jul 11 '17 at 21:37

$$(i) \int|f_n|dm=\int_{[0,\frac{1}{n}]}\sqrt{n}=\frac{\sqrt{n}}{n} \le 1$$ $$(ii) \text{For each } A, m(A) \lt \delta, \int_A f_n=\int_{A \cap [0,\frac{1}{n}]}\sqrt{n}=\sqrt{n}m\left(A\cap\left[0,\frac{1}{n}\right]\right)\le \frac{1}{\sqrt{n}} $$

From above two, it can be seen that $\{f_n\}$ is uniformly integrable

  • $\begingroup$ thank you. But Isn't our $\epsilon $ supposed to be a constant? $\endgroup$ – J. Kyei Jul 11 '17 at 21:39
  • $\begingroup$ I don't understand what you are asking. If you want to choose $\delta$, you can take $\delta=\epsilon$ always in this case. $\endgroup$ – tattwamasi amrutam Jul 11 '17 at 21:40
  • $\begingroup$ What I wanted to mean is that, Assume $\epsilon =1/2$. then we will wish to show that suppose $\exists \delta >0$ such that if $\mu (E)<\delta$ then $\int_E |f_n|<1/2$ $\forall n$. Now if we choose $n$ large enough so that $1/n=\mu ([,1/n]) <\delta$ then I should have my result. So let $E=[0,1/n]$ but $\int_E |f_n|\leq 1$ . To me me this is not the same as saying $\int_E |f_n| \leq 1/2$ $\endgroup$ – J. Kyei Jul 11 '17 at 22:10

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