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Let $X \subseteq \mathbb{R}^n$ be an unbounded set and let $p(x)$ be a probability density function on $X$, so that $\int_X p(x) dx = 1$.

Consider $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ and $\phi: \mathbb{R}^n \rightarrow \mathbb{R}_{\geq 0}$ such that $\phi(\cdot)$ is continuous, $\phi(0) = 0$ and $\lim_{|y| \rightarrow \infty} \phi(y) = \infty$.

$f(\cdot)$ and $\phi(\cdot)$ are such that

$$ \int_X \phi(f(x)) p(x) dx < \infty $$

I'm wondering if there exists $\epsilon \in \mathbb{R}^n \setminus \{0\}$ such that

$$ \int_X \phi(f(x) + \epsilon) p(x) dx < \infty $$

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1 Answer 1

So assuming $\epsilon\neq 0$, I think the following breaks this: Let $p(x)=\frac{1}{|x|^{|x|+1}\log^2x}$, up to a normalization constant. Let $\phi(x)=|x|^{|x|}$ and $f(x)=x$. Then everything is nice and integrable at $x=0$ but for $\epsilon\neq 0$, you will necessarly lose integrability either at 0 or infinity.

Edit: Sorry, we still have problems at 0 for $\epsilon=0$, in other words the function $\phi(f(x))p(x)$ isn't quite integrable. I would hope that this can be corrected by changing $\phi$ around 0.

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what do you mean by "losing integrability either at $0$ or at $\infty$"? –  Adam Apr 4 '12 at 21:10
1  
I'm using the fact that the integral of the function whose tail at infinity looks like $\frac{1}{x^a\log^b x}$ converges only for $a>1, b>0$ or when $a=1$ and $b>1$, and the opposite analysis holds for a function that looks like this near $x=0$. –  Alex R. Apr 4 '12 at 21:27
    
so you're saying that $|x+\epsilon|^{|x + \epsilon|}$ is integrable iff $\epsilon = 0$? –  Adam Apr 4 '12 at 21:34
    
$\phi(f(x)+\epsilon)p(x)$ is integrable iff $\epsilon=0$. –  Alex R. Apr 4 '12 at 22:32
    
Why is this integrable for $\epsilon = 0$? –  copper.hat Apr 4 '12 at 22:57

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