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.


$$ \int\_X F(x) p(x) dx = \infty $$


I'm wondering if the following proposition is true:


$ \forall \text{ such } F(\cdot) \ \ \exists $ a strictly-increasing, concave function 
$ f: R\_{\geq 0} \rightarrow R\_{\geq 0} $, with $f(0) = 0$, $\lim\_{y \rightarrow \infty} f(y) = + \infty$ such that:


$$ \int\_X f(F(x)) p(x) dx < \infty $$



Composed function made Lebesgue integrable?
Here is a way to start cooking such a function. From Fubini, we have for every non-negative function $g$ the equality $$\int g(x) p(x) dx = \int\_0^\infty du \int 1\_{g(x)>u} p(x) dx = \int\_0^\infty \mu(\{x:g(x)>u\}) du$$ (where $\mu$ is of density $p$).


So if you want $f\circ F$ to be integrable, pick any decreasing integrable function $\varphi$ on $\mathbb R\_+$ and try to ensure that $\mu(\{f\circ F>u\}) = \varphi(u)$. If everything is smooth, e.g. if the image measure of $\mu$ by $F$ has positive density, this characterizes $f$, which will be strictly increasing, start at $0$ and go to infinity.


If not you might have to fiddle a little bit, but notice that $\mu(\{f\circ F>u\})$ always goes to $0$ as $u\to\infty$ so you *will* find an $f$ satisfying integrability.


[I said nothing about the concavity of $f$. Once you have an increasing $f$, finding a concave and increasing one below $f+A$ for $A$ large enough should be doable, I did not check.]


