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!
 A: 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 
$$
\sup_n\int_O\left\lvert f_n\right\rvert dm\lt 3\varepsilon.
$$
