A question regarding uniform integrability We know that if $(h_n)$ is a sequence of nonnegative integrable functions on a measurable subset $E$ of $\mathbb R$ with finite measure such that $h_n\to 0$ pointwise a.e., then $\lim\limits_{n\to \infty}\int\limits_E h_n dm=0$ iff $(h_n)$ is uniformly integrable.
I want to have an example of non-uniformly integrable $(h_n)$, where $h_n$'s are not nonnegative and $h_n\to 0$, $\lim\limits_{n\to \infty}\int\limits_E h_n dm=0$. This question is from Royden's Real Analysis. Any hint will be appreciated.
 A: Consider $E=(-3,3)$, and for $n=1,2, \ldots$ define the functions
\begin{align}h_n(x) :=
  \begin{cases}
-2n - n^2x, \; \text{ if } x \in \left[-\frac{2}{n}, \,-\frac{1}{n} \right] , \\
                                n^2x , \;  \text{ if } x \in \left[-\frac{1}{n} , \, \,\frac{1}{n}\right] ,  \\
                                   2n - n^2x , \;  \text{ if } x \in \left[\frac{1}{n} , \,\frac{2}{n}\right], \\
    0, \; \text{ if } x \in \left(-3, -\frac{2}{n}\right) \cup \left(\frac{2}{n}, 3 \right). \end{cases} 
\end{align}
Notice $\lim\limits_{n\to \infty} h_n(x)=0$ and $\lim\limits_{n\to \infty}\int\limits_{E} h_n dm=0$. Let $\delta>0$ be given. By the Archimedean Property, we may find a positive integer $N$ such that $N \delta > 4$. So we have that $m\left( \left[-\frac{2}{N}, \frac{2}{N} \right]\right)< \delta$ but notice $\int_{-\frac{2}{N}}^{\frac{2}{N}} \left| h_n \right| dm=2$ whenever $n \geq N$*. Hence the sequence of functions $\{h_n\}_{n=1}^\infty$ is not uniformly integrable.
*Note: Technically it suffices to observe that $\int_{-\frac{2}{N}}^{\frac{2}{N}} \left| h_N \right| dm=2$ since 
$\neg \left(\forall \varepsilon\left(\varepsilon>0 \implies \exists \delta 
\left( \delta>0 \land \forall n \forall A \left(m(A)<\delta \implies \int_A |\,h_n| dm < \varepsilon \right)\right) \right) \right)$
$\iff \exists \varepsilon \left( \varepsilon>0 \land \forall \delta \left( \delta>0 \implies \exists N \exists A \left(m(A)<\delta \land \int_A |\,h_N| dm \geq \varepsilon\right)\right) \right)$.
