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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?

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[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.

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Thanks for the reply. Can you please explain it in a more detailed manner? – Adam Jun 8 '12 at 19:30
@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

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