Uniform integrability definitions I am confused with the way uniform integrability is defined in the context of random variables. Keeping with the analysis idea, I had expected this definition:
If $X$ is a random variable, given $\epsilon \ge 0$ and $A \subseteq \Omega$, there must be a $\delta$ such that $\int_{A} d\mu \le \delta$ and $\int_{A} X(\omega) d \omega \le \epsilon$. 
Instead this is what I find in most places including Wiki (when $K \subset L^1(\mu)$): 
$$\lim_{c \to \infty} \sup_{X \in K} \int_{|X|\ge c} |X| d \mu = 0$$
What is the reason for this different definition? Why does the bound on the value of $X$ even make an appearance?
 A: Three conditions are involved:

(C) For every $\varepsilon\gt0$, there exists a finite $c$ such that, for every $X$ in $\mathcal H$, $\mathrm E(|X|:|X|\geqslant c)\leqslant\varepsilon$.
(C1) There exists a finite $C$ such that, for every $X$ in $\mathcal H$, $\mathrm E(|X|)\leqslant C$.
(C2) For every $\varepsilon\gt0$ there exists $\delta\gt0$ such that, for every measurable $A$ such that $\mathrm P(A)\leqslant\delta$ and every $X$ in $\mathcal H$,  $\mathrm E(|X|:A)\leqslant\varepsilon$.

Then (C) is equivalent to (C1) and (C2). To wit:

*

*(C) implies (C1) since $\mathrm E(|X|)\leqslant c+\mathrm E(|X|:|X|\geqslant c)$.


*(C) implies (C2) since $\mathrm E(|X|:A)\leqslant c\mathrm P(A)+\mathrm E(|X|:|X|\geqslant c)$.


*(C1) and (C2) imply (C) since $\mathrm P(|X|\geqslant c)\leqslant\mathrm E(|X|)/c$.
The intuition might lie in the decomposition $A=(A\cap[|X|\lt c])\cup(A\cap[|X|\geqslant c])$ used to show that (C) implies (C2): for any given $c$, the expectation of $|X|$ on the first part is controlled uniformly over $X$ by something of the order of $\mathrm P(A)$ and the expectation of $|X|$ on the second part is controlled uniformly over $A$ thanks to condition (C).
A: There is an equivalent definition of uniform integrability which goes: Let $(\Omega,\mathcal{F},P)$ be a probability space. A set $\mathcal{H}$ of random variables is uniformly integrable if and only if
$$
\sup_{X\in \mathcal{H}}\int_\Omega |X| dP < \infty
$$
and 
$$
\forall \epsilon >0\, \exists \delta >0\;\forall A\in \mathcal{F}: P(A)\leq \delta \Rightarrow \sup_{X\in\mathcal{H}} \int_A |X| dP\leq \epsilon.
$$
Maybe this it what you are looking for.
