Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider a probability measure $m$ over $W \subseteq{R^m}$, so that $m(W) = 1$.

Consider a function $f: X \times W \rightarrow \mathbb{R}_{\geq 0}$, with compact $X \subset \mathbb{R}^n$, such that the following proposition holds true.

For any $\epsilon > 0$ there exists $c > 0$ such that $m(\{w \in W \mid f(x,w) \geq c \}) < \epsilon \ $ for all $x \in X$.

In other words, the measure of $\{f\geq c\}$ can be made arbitrarily small, uniformly on $X$.

What are the (weakest) conditions to have the family $\{f(x,\cdot)\}_{x \in X}$ Uniformly Integrable?

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

[Old answer] It is convenient to consider $\epsilon=2^{-n}$. Let $c_n$ be the corresponding $c$. You need the sum $\sum_{n}2^{-n}c_n$ to converge.

[New answer] It is convenient to consider $\epsilon_n=\sup_{x\in X} m(\{w\in W\colon f(x,w)\ge 2^n\})$. We know that $\epsilon_n\to 0$ as $n\to \infty$. I claim that the condition $\sum_{n=0}^\infty 2^n \epsilon_n<\infty$ is sufficient for uniform integrability.

Fix $x$ and consider the sets $W_n=\{w\in W\colon 2^n\le f(x,w)< 2^{n+1}\}$, $n=0,1,2,\dots$. Note that $m(W_n)\le \epsilon_n$. Since $$\int_W f(x,w)\,dm(w) = \int_{\{f<1\}} f(x,w)\,dm(w) + \sum_{n=0}^\infty \int_{W_n} f(x,w)\,dm(w) $$ we can estimate the integral $$\int_W f(x,w)\,dm(w) \le 1 + \sum_{n=0}^\infty 2^{n+1}\epsilon_n<\infty$$ Moreover, if instead of $W$ we integrate over the set $\{f\ge 2^N\}$, the estimate becomes $$\int_{f\ge 2^N} f(x,w)\,dm(w) \le \sum_{n=N}^\infty 2^{n+1}\epsilon_n<\infty$$ which is uniformly small.

share|improve this answer
    
Thanks for the reply. Can you please explain it in a more detailed manner? –  Adam Jun 8 '12 at 19:30
1  
@Adam Consider the set where f is between $2^n$ and $2^{n+1}$. The integral of f over this set is at most $2^{n+1}c_n$. So the convergence of the series implies f is integrable. You also get a bound in terms of the tail of the series, which should help with uniformity. –  user31373 Jun 8 '12 at 20:40
    
Still, I'm not clear with your answer. I guess you are saying that for each $\epsilon \geq \epsilon_i := 2^{-i}$, let $c_i$ be the "$c$" of the Assumption. Then, a sufficient condition for the family $\{f(x,\cdot)\}_{x \in X}$ to be uniformly integrable is that $\sum_{i=0}^{\infty} 2^{-i} c_i < \infty$. Is that what you meant? Can you please give a rigorous proof of such a claim? Thanks, that would help. (I also guess that your $n$ is not the dimension of $X \subset \mathbb{R}^n$.) –  Adam Jun 8 '12 at 21:31
    
Moreover, why $\sum_{i=0}^\infty 2^{-i} c_i$??? We have $\int_W f(x,w) m(dw) = \sum_{i=0}^{\infty} \int_{\{ 2^i \leq f \leq 2^{i+1} \}} f(x,w) m(dw) \leq \sum_{i=0}^{\infty} 2^{i+1} c_i $. In this latter summation that has to be finite??? –  Adam Jun 11 '12 at 15:50
    
@Adam I was in a hurry when I posted my reply. Rewritten now. –  user31373 Jun 11 '12 at 16:09
show 3 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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