Proof of Criterion for Uniform Integrability Defintion (Uniform Integrability): A family $\mathcal{F}$ of integrable functions is uniformly integrable if $\forall \varepsilon > 0 $ there is a $M_\varepsilon>0$, such that
$\int_{\{|f|>M_\varepsilon\}}|f|\,\mathrm{d}\mu < \varepsilon,\ \forall f\in\mathcal{F}$ 
Let $(X,\mathcal{A},\mu)$ be a finite measure space. A family of integrable functions $\mathcal{F}$ is uniformly integrable if and only if for all $\varepsilon > 0$ there exists a $\delta>0$ sucht that for all $A\in\mathcal{A}$ we have that
$\mu(A)<\delta\ \Rightarrow\ \int_A|f|\,\mathrm{d}\mu<\varepsilon ,\ \forall f\in \mathcal{F}$
Here is my proof and I would like to know if there are errors and would be thankful for any improvements.
"$\Rightarrow$":
Because of uniform integrability we can choose $\varepsilon/2 >0$ and get a $M_{\varepsilon/2}$ such that for arbitrary $A\in\mathcal{A}$
$\int_A |f|\,\mathrm{d}\mu =\int_{\{|f|>M_{\varepsilon/2}\}\cap A}|f|\,\mathrm{d}\mu + \int_{\{|f|\le M_{\varepsilon/2}\}\cap A}|f|\,\mathrm{d}\mu\le\mu(A) M_{\varepsilon/2} + \int_{\{|f|>M_{\varepsilon/2}\}\cap A}|f|\,\mathrm{d}\mu$
Note that we have $\mu(A)<\infty$ since the measure space is finite. Now we can pick a $\delta:=\frac{\varepsilon}{2}\frac{1}{M_{\varepsilon/2}}$ from which it follows by the previous equation that for all $A$ satisfying $\mu(A)<\delta$ 
$\int_A |f|\,\mathrm{d}\mu\le \varepsilon/2 + \varepsilon/2\le \varepsilon $
which is what we wanted to show.
"$\Leftarrow$":
We have, since each member in $\mathcal{F}$ is integrable, that
$\mu\left(\{|f|>m\}\right)\rightarrow 0$ for $m\rightarrow\infty$ and arbitrary $f\in\mathcal{F}$. Any suggestions on how to prove this statement rigorously?
This is equivalent to (just use the definition of a limit and exchange $\delta$ for $\epsilon$)
$\forall \delta >0\ \exists M_\varepsilon\in \mathbb{N}\ \forall n\ge M_\varepsilon\colon \mu\left(\{|f|>n\}\right)\le |\mu\left(\{|f|>n\}\right)|<\delta$
which proves this direction.
 A: The Falrach link identifies an additional requirement $\sup_{f \in \mathcal{F}} \int |f|d\mu<\infty$ for the posted condition to imply UI.  This additional requirement immediately enables the Markov inequality approach from my hint in comments above.  Here is a simple counter-example that shows what can go wrong without that additional requirement:  
Counter-example:
Define $X=0$ (a 1-element set) with $\mu(X)=1$. For $n \in \{1, 2, 3, ...\}$ define $f_n:X\rightarrow\mathbb{R}$ by $f_n(x)=n$.  Clearly $\int_X |f_n|d\mu = n$ for all $n \in \{1, 2, 3, ...\}$ and so $\{f_n\}_{n=1}^{\infty}$ is not UI.  But the functions $\{f_n\}_{n=1}^{\infty}$ satisfy the condition of the above post trivially: For all $\epsilon>0$ we can choose $\delta=1/2$ and indeed for any set $A \subseteq X$ that satisfies $\mu(A)<\delta$ we immediately have $\int_A |f_n|d\mu<\epsilon$. This is because the only subset of $X$ with measure less than $1/2$ is the empty set!
So the additional requirement $\sup_{f \in \mathcal{F}} \int |f|d\mu$ is quite needed in general.
A "Covering Property" that implies the additional requirement:
Suppose $\mathcal{F}$ is a family of integrable functions $f:X\rightarrow\mathbb{R}$ such that for all $\epsilon>0$ there is a $\delta>0$ such that $A \subseteq X$ with $\mu(A)<\delta$ implies $\int_A |f|d\mu < \epsilon$ for all $f \in \mathcal{F}$.  Now fix $\epsilon=1$ and choose the corresponding $\delta$ so that $\mu(A)<\delta$ implies $\int_A |f|d\mu < 1$ for all $f \in \mathcal{F}$. 
If there exists a finite sequence of sets $\{A_1, ...,A_m\}$ (for some positive integer $m$) such that $\cup_{i=1}^m A_i = X$, $A_i\subseteq X$ for all $i \in \{1, ...,m\}$, and $\mu(A_i)<\delta$ for all $i \in \{1, ..., m\}$ then for all $f \in \mathcal{F}$: 
$$ \int_X |f|d\mu \leq \sum_{i=1}^m \int_{A_i}|f|d\mu \leq m$$
So the additional requirement always holds in this case. 
Such a covering by finite sets is always possible when $X$ is a compact subset of $\mathbb{R}^k$ for some positive integer $k$ (and when we use the standard measure for $\mathbb{R}^k$).
