The following is equivalent.
$(1) \mathscr{F}=\{f_{\alpha},\alpha\in I\}$ is uniformly integrable.
$(2) \lim_{\lambda\to \infty}\int_{[|f_{\alpha}|>\lambda|]}{|f_{\alpha}|d\mu=0},\text{ uniformly in }\alpha\in I.$
(3) There exists a convex function $\Phi:\mathbb{R}\to \mathbb{R^+}$ such that $\Phi(0)=0,\Phi(-x)=\Phi(x)$ and $\frac{\Phi(x)}{x}\uparrow \infty$ as $x\uparrow \infty$ in terms of which $$C_{1}=\sup_{\alpha}\int_{\Omega}\Phi(f_{\alpha})d\mu<\infty$$
Can someone explain what $(3)$ is saying? I mean I don't know why but I am not getting it. The existence of such convex function.
Basically, (3) says that $\mathcal F$ is a bit more than bounded in $\mathbb L^1$, because $\int_{\Omega}\Phi(f_{\alpha})d\mu$ has to be bounded independently of $\alpha$ for a function $\Phi$ that growth a bit quicker than $\lvert x\rvert$ (keep in mind the examples $\varphi(x)=\lvert x\rvert^p$ for $p>1$).
The existence of such function $\Phi$ is not trivial to show, see for example here.