Let $(\Omega,\Sigma,\mu)$ be a finite measure space and let $\mathscr{F}=\{f_{\alpha}:\Omega\to \mathbb{R}:\alpha\in I\}$ be an arbitrary family of measurable real functions on $\Omega$.
$\mathscr{F}$ is uniformly integrable if the following two conditions hold:
\begin{equation}
(i) \sup_{\alpha}\int_{\Omega}|f_{\alpha}|d\mu=C<\infty \hspace{1cm}
(ii) \lim_{\mu_{(A)}\to 0}\int_{A}|f_{\alpha}|d\mu=0
\end{equation}
Remark: (i) follows form (ii).
I want to know the difference when $\mathscr{F}\subset L^1$ and when $\mathscr{F}$ is uniformly integrable since this two looks same to me.Please help.
what I understood is that here condition (ii) means that $\forall \epsilon>0 ,\hspace{0.5 cm} \exists \delta>0$ such that $\int_{A}|f_\alpha|d\mu<\epsilon$ whenever $f\in \mathscr{F}$ and for all $\mu(A)<\delta.$
It might help to consider a simple example:
Let $\Omega = (-2,2)$, $\Sigma$ be the Borel $\sigma$-algebra on $(-2,2)$, and $\mu$ be the Lebesgue measure restricted to $(-2,2)$.
Consider $\mathscr F = \{ k\chi_{(-1,1)} \text{ s.t. } k \in \mathbb Z^+ \}.$ Each function $f_k:= k\chi_{(-1,1)}$ is in $L^1(\Omega)$ since $$ \| f_k\|_{L^1(\Omega)} = 2k < +\infty,$$ so $\mathscr F \subset L^1(\Omega)$. However, $\mathscr F$ is not uniformly integrable since $\| f_k\|_{L^1(\Omega)}$ is not bounded by some constant independent of $k$.