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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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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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    Why is this integrable for $\epsilon = 0$?2012-04-04