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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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Perhaps not stated properly? You do not assume $f(x,\cdot)$ is integrable for each $x$? – GEdgar May 17 '12 at 0:32
I was thinking on $X = [0,1]$, $W = \mathbb{R}_{\geq 0}$, $f(x,w) = 1/x$ if $w \in [1/x,1/x+1]$, $0$ otherwise. But such $f(\cdot)$ is not locally bounded. – Adam May 17 '12 at 0:32
Yes. According to GEdgar, I meant $f(x,\cdot)$ integrable. – Adam May 17 '12 at 0:37
up vote 1 down vote accepted

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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I guest that even if we have assumed that $\forall x \in X \ $ $\int_W f(x,w) \mu(dw) < \infty$, we can not find $\delta > 0$ such that $\sup_{x \in X} \int_W \sup_{\xi \in X \cap (\{x\}+\delta \mathbb{B})} f(\xi,w) \mu(dw) < \infty$... Right? – Adam May 17 '12 at 0:57
Assuming your $\mathbb B$ is the open unit ball of ${\mathbb R}^n$, that's correct: in my example take $x = 0$ and you get $\sup_{\xi \in [0, \delta)} f(\xi, w) = w^2 e^{-1}$ for $w > 1/\delta$. – Robert Israel May 17 '12 at 1:34
To be precise, I meant $\mathbb{B}$ as the closed ball, but it should be the same thing: a $\delta$ does not exists. – Adam May 17 '12 at 2:44
Sorry for asking again. It would be interesting to find a strictly positive $f:X \times W \rightarrow \mathbb{R}_{>0}$ such that, assuming that $\max_{x \in X} \int_W f(x,w) \mu(dw) < \infty $, does not exist $\delta>0$ such that $\max_{x \in X} \int_W \max_{\xi \in X \cap \{x\}+\delta \overline{\mathbb{B}}} f(\xi,w) \mu(dw) < \infty $. – Adam May 17 '12 at 5:39
Can it be like $(x-a)(\text{something not integrale}) + \text{something integrable}$? So that for $x=a$ we get $\int_W (\cdot) < \infty$, but for $x=a+\delta$ we get $\infty$. Is this the idea behind an example like that? Thanks. – Adam May 17 '12 at 5:43

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