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Let $\mu(\cdot)$ be a probability measure on $W \subseteq \mathbb{R}^m$, so that $\int_W \mu(dw) = 1$.

Consider a locally bounded function $f: X \times W \rightarrow \mathbb{R}_{\geq 0}$, with compact $X \subset \mathbb{R}^n$, such that $\forall w$ $f(\cdot,w)$ is continuous, $\forall x$ $f(x,\cdot)$ is integrable.

Find $f(\cdot)$ such that $ \int_W \sup_{x \in X} f(x,w) \mu(dw) = \infty $

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    Yes. According to GEdgar, I meant $f(x,\cdot)$ integrable.2012-05-17

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Take $\mu(dw) = 1/w^2\ dw$ for $w \in (1,\infty)$, $X = [0,1]$, and $f(x,w) = x w^3 e^{-wx}$. Note that $\sup_{x \in [0,1]} f(x,w) = f(1/w,w) = w^2 e^{-1}$.

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    Yeah, I see. You may be interested to [this question](http://math.stackexchange.com/questions/144824/induction-on-uniform-boundedness) about iterating uniform boundedness. :-)2012-05-17