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).

Improve this question
$\endgroup$
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
    Commented 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. $X_n = n$. $\endgroup$
    – Nate Eldredge
    Commented Jul 9, 2020 at 14:05
  • $\begingroup$ I believe that if $\Omega$ is atomless, then (ii) implies (i). $\endgroup$
    – Nate Eldredge
    Commented Jul 9, 2020 at 14:05
  • $\begingroup$ Thank you @NateEldredge for the good example! $\endgroup$
    – Sofia
    Commented Jul 10, 2020 at 19:10

1 Answer 1

Reset to default
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.

Improve this answer
$\endgroup$
0

Not the answer you're looking for? Browse other questions tagged .