0
$\begingroup$

Working with the epsilon-delta-criterium, a family $(X_i)_{i \in I}$ on $(\Omega,A,P)$ is uniformly integrable if

i) $sup_{i \in I} E(X_i) <\infty$

ii) $\forall \epsilon>0$ ex. $\delta>0$ s.t. $P(A)<\delta \Rightarrow \int_{A} |X_i|dP < \epsilon $

I easily came up with an example where i) is fulfilled, but ii) isn't ($X_n=1_{[0,\frac{1}{n}]} \cdot n$).

But I am looking for an example where ii) is fulfilled but i) isn't.

I would be really happy if someone had an answer because it would help me understand why we need (i).

$\endgroup$
4
  • $\begingroup$ What are you taking the supremum of in (i)? Anyway, any example that satisfies (i) can easily be modified not to do so without breaking (ii). This is not research level. $\endgroup$
    – LSpice
    Jul 9, 2020 at 12:21
  • $\begingroup$ Let $\Omega$ be a single point having probability 1. Then (ii) is trivially satisfied because for any $\delta < 1$, the only $A$ with $P(A) < \delta$ is $A=\emptyset$. So now you can choose any family which violates (i), e.g. $X_n = n$. $\endgroup$ Jul 9, 2020 at 14:05
  • $\begingroup$ I believe that if $\Omega$ is atomless, then (ii) implies (i). $\endgroup$ Jul 9, 2020 at 14:05
  • $\begingroup$ Thank you @NateEldredge for the good example! $\endgroup$
    – Sofia
    Jul 10, 2020 at 19:10

1 Answer 1

1
$\begingroup$

Thanks @NateEldredge: Let Ω be a single point having probability 1. Then (ii) is trivially satisfied because for any δ<1, the only A with P(A)<δ is A=∅. So now you can choose any family which violates (i), e.g. Xn=n.

$\endgroup$
0

Not the answer you're looking for? Browse other questions tagged or ask your own question.