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In his nice answer about uniform integrability, Did did show that condition (C) below is equivalent to (C1) and (C2) together.

(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 $\mathcal H$, $\mathrm E(|X|:A)\leqslant\varepsilon$.

I am wondering why (C2) does not imply (C), alone. Would anyone be so nice as to give me a counterexample? I mean, is there a family of random variables $\mathcal H$ such that (C2) holds but (C) (or (C1)) does not hold?

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André Caldas
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  • @RideTheWavelet: Why does your example satisfy (C2)? – André Caldas Jun 27 '17 at 14:38
  • Hint: Consider some probability space such that $P(A)\leqslant\delta$ implies $P(A)=0$. (But why ask a new question?) – Did Jun 27 '17 at 17:20
  • @AndréCaldas I made an error. I was thinking that $\mu(A)=\int_{A}f,\mathrm{d}\lambda$ is absolutely continuous with respect to Lebesgue measure ($\lambda$) (it is, and this implies a (C2)-like property for $\mu$), but the relevant measure would be $\mu'(A)=\int_{A}|x|f(x)\lambda(\mathrm{d}x),$ which I believe is not AC. – RideTheWavelet Jun 27 '17 at 23:16
  • @Did: Thank you. I made your comment into an answer. :-) – André Caldas Jun 28 '17 at 12:10
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    @Did: 1. Because I didn't think the answer was so easy. 2. Because I thought it was a different question. 3. Because I like things well documented in a way that questions and answers can be easly found. 4. Because I didn't want to be called attention for discussing a different question in the comments. :-) – André Caldas Jun 28 '17 at 12:15

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Following Did's comment to my question...

Just take the probability space $(\Omega, \sigma, P)$, where $\Omega = \{a\}$ is a singleton. Or just take a Dirac delta $P = \delta_a$ over any nonempty $\Omega$. In this case, any set of random variables will satisfy (C2).

In particular, $X_n = n$ will satisfy (C2), but not (C1).

Of course, this can easily be adapted to any $P$ with an "atom".

Intuition failed to me because I was under the impression that I could always break up the spaces into "small bits". Suppose that (C2) holds and also that for any $\delta > 0$ there are $A_1, \dotsc, A_n \in \sigma$ such that $P(A_j) < \delta$ and $$ \Omega = A_1 \cup \dotsb \cup A_n. $$ Take $\delta$ as in (C2) for $\varepsilon = 1$. Then, for any $X \in \mathcal{H}$, $$ E(|X|) \leq E(|X|; A_1) + \dotsb + E(|X|; A_n) \leq n. $$

Thank you, Did! :-)

André Caldas
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