# Making bounded an unbounded integral

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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Any bounded increasing concave function $f$ such that $f(0)=0$ solves the problem, for example $f:t\mapsto t/(1+t)$. – Did Apr 2 '12 at 16:00
Alternatively, it might be possible to pick any strictly-increasing, concave function $g(x)$ that is integrable w.r.t. $p(x)$, and then compose it with $F^{-1}(x)$ to get $f(x) = g(F^{-1}(F(x))) = g(x)$. I'm not sure under what conditions $g(F^{-1}(y))$ will also be strictly-increasing and concave though. – Mr. F Apr 2 '12 at 16:06
Is the result true if $\lim_{y \rightarrow \infty} f(y) = \infty$? – Adam Apr 2 '12 at 16:09
I'm not clear about your comment. Of course the result is not true for any such $f(\cdot)$. I'm wondering if there exists at least one. For instance, if $p(x) \sim 1/x^2$ and $\Phi(x) \sim x$ (not integrable), then we can pick $f(y) \sim y^{1/2}$ so that $f(\Phi(x)) \sim x^{1/2}$ and $f(\Phi(x))p(x) \sim 1/x^{3/2}$ (integrable)... moreover, if $\Phi(\cdot)$ is exponential, then we can take $f(\cdot)$ logarithmic... is all that "general"? There exists a proof of that? – Adam Apr 3 '12 at 18:13
I meant $F(\cdot)$ in place of $\Phi(\cdot)$... – Adam Apr 3 '12 at 18:19

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.
is it possible to recover a strictly-increasing $f$? – Adam Apr 23 '12 at 9:49
Adam: try!   – Did Apr 23 '12 at 9:53
Are you saying that your answer already gives $f$ strictly increasing? – Adam Apr 23 '12 at 10:07