# A question relating to uniform integrability and tightness

Let $$(f_n)$$ be a sequence of integrable functions on $$\mathbb{R}$$. I want to show that $$(f_n)$$ is uniformly integrable and tight over $$\mathbb{R}$$ if and only if for each $$\varepsilon>0$$, there are positive numbers $$r$$ and $$\delta$$ such that for each open subset $$\mathcal{O}$$ of $$\mathbb{R}$$ and index $$n$$, if $$m(\mathcal{O}\cap (-r,r))<\delta$$, then $$\int\limits_{\mathcal{O}}|f_n|dm<\varepsilon$$.

If the given condition holds, then it is easy to show that $$f$$ is uniformly integrable and tight. Let $$\varepsilon >0$$. Then there exist $$\delta >0,r>0$$ such that for all open sets $$\mathcal{O}$$ with $$m(\mathcal{O}\cap (-r,r))<\delta\implies \int\limits_{\mathcal{O}}|f_n|dm<\varepsilon$$. Let $$A$$ be a measurable subset of $$\mathbb R$$ with $$m(A)<\frac{\delta}{2}$$. Thus there exists an open set $$\mathcal{O}$$ containing $$A$$ such that $$m(\mathcal{O}\setminus A)<\frac{\delta}{2}$$. Thus\begin{align} m(\mathcal{O}\cap (-r,r)) & = m((\mathcal{O}\setminus A) \cap (-r,r))\\ &\quad +m(A\cap (-r,r))\\ & <\frac{\delta}{2}+\frac{\delta}{2}\\ & =\delta.\end{align} Thus by hypothesis $$\int\limits_{\mathcal{O}}|f_n|dm<\varepsilon$$. Hence $$\int\limits_A |f_n|dm\leq \int\limits_{\mathcal{O}}|f_n|dm<\varepsilon$$. Also $$\mathbb R\setminus [-r,r]$$ is an open set such that $$(\mathbb R\setminus [-r,r])\cap (-r,r)=\emptyset$$. Thus $$\int\limits_{\mathbb R\setminus [-r,r]}|f_n|dm<\varepsilon$$ and so $$(f_n)$$ is tight. But how to prove the converse? Help please!

For the converse, fix a positive $\varepsilon$. From uniform integrability, we know that there exists a positive $\delta$ such that for all measurable set of measure smaller than $\delta$, $\sup_n\int_A\left\lvert f_n\right\rvert dm\lt\varepsilon$. From tightness, we know that there exists a set $E_0$ such that $\sup_n\int_{\mathbb R\setminus E_0}\left\lvert f_n\right\rvert dm\lt\varepsilon$. Choose $r$ such that $m\left(E_0\setminus [-r,r]\right)\lt \delta$.
Let $O$ be an open set such that $m\left(O\cap [-r,r]\right)\lt \delta$. With $A:=O\cap [-r,r]$, we get that $$\tag{*}\sup_n\int_{O\cap [-r,r]}\left\lvert f_n\right\rvert dm\lt\varepsilon.$$ Since $$\int_{O\cap [-r,r]^c}\left\lvert f_n\right\rvert dm\leqslant \int_{ [-r,r]^c}\left\lvert f_n\right\rvert dm =\int_{ E_0 \cap [-r,r]^c}\left\lvert f_n\right\rvert dm+\int_{ E_0^c \cap [-r,r]^c}\left\lvert f_n\right\rvert dm,$$ we by applying uniform integrability this time with $A=E_0 \cap [-r,r]^c$ and using tightness for the second term that $$\tag{**} \sup_n\int_{O\cap [-r,r]^c}\left\lvert f_n\right\rvert dm\leqslant 2\varepsilon$$ hence the combination of $(*)$ and $(**)$ yields $$\sup_n\int_O\left\lvert f_n\right\rvert dm\lt 3\varepsilon.$$