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Let $\mu$ be a probability measure: $\mu(X) = \int_X \mu(dx) = 1$.

Consider a locally bounded function $f: \mathbb{R}^n \setminus \{0\} \times X \rightarrow \mathbb{R}_{> 0} \ $ such that:

  • $\exists \bar{z} \in \mathbb{R}^n \setminus \{0\} $ such that: $\int_X f(\bar{z},x) \mu(dx) < \infty $.

$$ $$

Prove that:

$$ \exists \delta > 0 \text{ s.t. } \ \ ||z-\bar{z}|| < \delta \ \Rightarrow \ \int_X f(z,x) \mu(dx) < \infty $$

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Why should this be true? You could (say) let $f(0,\cdot)=0$ and $f(x,\cdot)$ for $x \neq 0$ be any bounded function that is not measurable. Or even if f(x,⋅) is measurable, let it be a random variable with infinite expectation –  ShawnD Mar 9 '12 at 19:33
    
maybe I formulated it wrong: $f(x,\cdot)$ is measurable. I meant the expectation is finite. –  Adam Mar 9 '12 at 20:03
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Please edit your post to make clear what you mean. –  g.castro Mar 9 '12 at 20:11
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1 Answer

up vote 1 down vote accepted

Try $n=1$, $X = \mathbb R$ with $\mu(dx) = \frac{dx}{\pi (1+x^2)}$, $f(z,x) = (xz)^2$, $\overline{z} = 0$.

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Ok, I need $\bar{z} \neq 0$. –  Adam Mar 9 '12 at 21:38
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OK, $f(z,x) = (x(z-1))^2$ and $\overline{z} = 1$. It's just the same example translated. –  Robert Israel Mar 9 '12 at 22:32
    
You're right: that's why now I have domain $\mathbb{R}_{> 0}$. Sorry for changing the problem statement... but thanks for helping me in being precise! –  Adam Mar 9 '12 at 22:40
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So just add $1$ to $f(z,x)$. And before you change the problem statement again to avoid this particular example, please think about whether you could modify the example to fit the new rules. –  Robert Israel Mar 9 '12 at 22:59
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