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 $\mathrm P(A)\leqslant \delta$, $\mathrm E(|X|:A)\leqslant\epsilon$.

  • $\begingroup$ What are your thought at least about the first two? $\endgroup$ – Ilya Jan 28 '13 at 7:33
  • $\begingroup$ @Ilya: the first implies the second, proven by contradiction? $\endgroup$ – Ethan Jan 28 '13 at 7:38
  • $\begingroup$ You can prove it directly by using the Dominated Convergence theorem. $\endgroup$ – Stefan Hansen Jan 28 '13 at 7:42
  • $\begingroup$ @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? $\endgroup$ – Ethan Jan 28 '13 at 7:42
  • $\begingroup$ @StefanHansen: Can you elaborate? Thanks! $\endgroup$ – 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.

  • $\begingroup$ Is the first one equivalent to the second or third? $\endgroup$ – Ethan Jan 28 '13 at 8:05
  • $\begingroup$ It's equivalent to both (this answer shows that the second and third are equivalent). But it is easiest to show that it's equivalent to the second, which you started doing in the comments. $\endgroup$ – Stefan Hansen Jan 28 '13 at 8:07

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