# Composed function made Lebesgue integrable?

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:

$$\int_X f(F(x)) p(x) dx < \infty$$

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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.]

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Thanks for the answer. However, it would be great if you can be more precise and rigorous. For instance: 1. why the first equalities follow from Fubini's theorem? 2. why is it guaranteed that there exists a decreasing $\variphi$ such that $\mu(\{f(F(.))=\varphi(u)\})$? 3. what is the "image measure of $\mu$ by $F$"? 4. what do you mean by "this characterizes $f$"? 5. why such an $f$ would start at $0$ and go to $\infty$? 6. why "$\mu(\{f(F(.))>u\})$ always go to $0$ as $u \rightarrow \infty$?" I guess this is true if $f(F(.))$ is integrable, but this is the thesis, not the assumption – Frank Apr 3 '12 at 22:17
For 6., no, this is true as soon as $f\circ F$ is a.e. finite, you do not need integrability. The rest is fairly classical - is it homework? – Vincent Beffara Apr 4 '12 at 8:13
This is not homework. I'm still not clear with your answer. Please be more rigorous. Moreover, why $f(F(.))$ is supposed to be a.e. finite? – Frank Apr 4 '12 at 16:04
Well, $F$ is assumed to take values in $\mathbb R_+$, and $f$ goes from $\mathbb R_+$ to itself, so $f\circ F$ takes values in $\mathbb R_+$, in other words you do have the assumption that it is finite everywhere (not only a.e. - which would be enough for what you want to do). Some hints then (but if you really want to understand measure theory, you should go through the steps yourself): 1. because $g(x) = \int_0^\infty 1_{y<g(x)} dy$; 2. $\varphi$ is chosen arbitrarily, what you want is $f$; 4. means that there is a unique $f$ satisfying the equality; 5. because $F$ is not in $L^\infty$. – Vincent Beffara Apr 5 '12 at 11:25