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 $\lim_{x \rightarrow \infty} f(x) = +\infty$?