Let $(\Omega,\mathcal{A},\mu$ be a measure space.
A set $\mathcal{F}$ of measurable, numerical functions is called equi-integrable if for any $\varepsilon > 0$ it exists a nonnegative, integrable function $h$ such that $$ \sup_{f\in\mathcal{F}}\int 1_{\left\{\lvert f\rvert\geq h\right\}}\lvert f\rvert\, d\mu < \varepsilon. $$
Now the task is to show the following:
$\mathcal{F}\subset\mathcal{L}_{\mu}^1$ finite $\implies \mathcal{F}$ equi-integrable.
My idea to prove that is:
Let $\mathcal{F}=\left\{f_1,\ldots,f_n\right\}$ with $f_i\in\mathcal{L}_{\mu}^1$ for $i=1,\ldots,n$. This means that $0\leq\lvert f_i\rvert=m_i<\infty$ a.s. for $i=1,\ldots.n$. Choose $$ h:=\max_{i\in\left\{1,\ldots,n\right\}}\left\{m_i\right\}. $$ Then for any $\varepsilon > 0$ and any $f_i\in\mathcal{F}$ it is $$ \int 1_{\left\{\lvert f_i\rvert\geq h\right\}}\lvert f_i\rvert\, d\mu=0<\varepsilon $$ and because this was not dependent on $i$, it is for any $\varepsilon >0$ $$ \sup_{f\in\mathcal{F}}\int 1_{\left\{\lvert f\rvert\geq h\right\}}\lvert f\rvert\, d\mu=0<\infty. $$
So $\mathcal{F}$ is equi-integrable.
Try $h=2\sum\limits_{f\in\mathcal F}|f|$. Let $A=\bigcap\limits_{f\in\mathcal F}\{f=0\}$. Then, for every $f$ in $\mathcal F$, $\{|f|\geqslant h\}=A$ and $\int\limits_A|f|\mathrm d\mu=0$.
Additionally, note that the argument based on the fact that for every $f$ in $\mathcal L^1_\mu$ there would exist some finite constant $c$ such that $|f|\leqslant c$ almost surely, is wrong (try $f(x)=1/\sqrt{x}$ on $\Omega=(0,1)$ with $\mathcal A$ the Borel sigma-algebra and $\mu$ the Lebesgue measure).