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

Let $F: X \rightarrow \mathbb{R}_{\geq 0}$ a measurable but non-integrable function, i.e.

$ \int_X F(x) p(x) dx = \infty $

where $X \subseteq \mathbb{R}^n$ is a closed unbounded set.

I'm wondering if the following is true:

$ \forall \text{ such } F(\cdot) \ \ \exists \text{ a strictly-increasing, concave } f: \mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0} \text{ with } f(0)=0 \text{ such that: } \\ \int_X f(F(x)) p(x) dx < \infty $

Is it true if, in addition, we require $\lim_{x \rightarrow \infty} f(x) = +\infty$?

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    C$a$n someone explain to me the answer given in http://mathoverflow.net/questions/93041/composed-function-made-le$b$esgue-integrable/9$3$046#93046 ??? Thanks!2012-04-04

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Consider the probability measure $m$ with density $p$ on the positive halfline, and any decreasing positive integrable function $c$ on the positive halfline for the Lebesgue measure, for example $c(u)=\mathrm e^{-u}$ for every $u\geqslant0$. You are interested in the finiteness of the integral $ I(f)=\int f(F(x))\mathrm dm(x). $ Writing $f(F(x))=\displaystyle\int\limits [u\leqslant f(F(x))]\mathrm du$ and using Fubini theorem, one gets $ I(f)=\int\mathrm du\int[u\leqslant f(F(x))]\mathrm dm(x)=\int m(A_u)\mathrm du,\qquad A_u=[f(F)\geqslant u]. $ Define the function $f$ by the identity $ c(f(t))=m(F\geqslant t), $ that is, in the example above, $ f(t)=-\log m(F\geqslant t). $ The functions $c$ and $t\mapsto m(F\geqslant t)$ are nonincreasing hence $f$ is nondecreasing. Furthermore, $f(0)=0$ and $m(F\geqslant t)\to0$ when $t\to+\infty$ hence $f(t)\to+\infty$ when $t\to+\infty$. Finally, for every positive $u$, $m(A_u)=m(F\geqslant f^{-1}(u))=c(u)$, hence $I(f)=\displaystyle\int c(u)\mathrm du$ is finite.

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    I'm still not sure if we can get a strictly-increasing $f$. Your $f$ is in fact only non-decreasing.2012-12-02