# Uniform integrability for a single random variable

Let $X$ be a random variable. Are the following three equivalent?

• $X \in L^1$, i.e. $E |X| < \infty$.

• $X$ is uniformly integrable. That is, if given $\epsilon>0$, there exists $K\in[0,\infty)$ such that $E(|X|I_{|X|\geq K})\le\epsilon$, where $I_{|X|\geq K}$ is the indicator function$I_{|X|\geq K} = \begin{cases} 1 &\text{if } |X|\geq K, \\ 0 &\text{if } |X| < K. \end{cases}$

• For every $\epsilon > 0$ there exists $\delta > 0$ such that, for every measurable $A$ such that $\mathrm P(A)\leqslant \delta$, $\mathrm E(|X|:A)\leqslant\epsilon$.

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What are your thought at least about the first two? –  Ilya Jan 28 '13 at 7:33
@Ilya: the first implies the second, proven by contradiction? –  Ethan Jan 28 '13 at 7:38
You can prove it directly by using the Dominated Convergence theorem. –  Stefan Hansen Jan 28 '13 at 7:42
@Ilya: I was wrong in my first comment. I forgot that the measure is a probability measure here. Is uniform integrable only defined for probabilty space? –  Ethan Jan 28 '13 at 7:42
@StefanHansen: Can you elaborate? Thanks! –  Ethan Jan 28 '13 at 7:43

Here's a sketch showing the equivalence of the last two statements: Let $(\Omega,\mathcal{F},P)$ denote the probability space we are working on.
$2)\Rightarrow 3)$: For any $A\in\mathcal{F}$ and $K>0$ we have
$$E[|X|:A]\leq E[|X|: |X|\geq K]+KP(A).$$ Let $\varepsilon>0$ be given, and pick $K>0$ (given by the assumption) such that $$E[|X|:|X|\geq K]\leq \frac{\varepsilon}{2}$$ and pick $\delta=\frac{\varepsilon}{2K}$. Conclude.
$3)\Rightarrow 2)$: Use Markov's inequality and the assumption to conclude that $$P(|X|\geq K)\to 0\quad \text{for }K\to\infty.$$ Let $\varepsilon >0$ be given. Pick a $K>0$ such that $P(|X|\geq K)\leq \delta$. Conclude.