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?
$$(i) \int|f_n|dm=\int_{[0,\frac{1}{n}]}\sqrt{n}=\frac{\sqrt{n}}{n} \le 1$$ $$(ii) \text{For each } A, m(A) \lt \delta, \int_A f_n=\int_{A \cap [0,\frac{1}{n}]}\sqrt{n}=\sqrt{n}m\left(A\cap\left[0,\frac{1}{n}\right]\right)\le \frac{1}{\sqrt{n}} $$
From above two, it can be seen that $\{f_n\}$ is uniformly integrable