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Let $\mu(\cdot)$ be a probability measure on $X \subseteq \mathbb{R}$.

Consider a measurable but not integrable function $F: X \rightarrow \mathbb{R}_{\geq 0}$.

(2) Prove that there exists a function $\phi: \mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0}$ such that:

(2a) $\phi(0)=0$, $\phi$ continuous and strictly increasing, $\lim_{x \rightarrow \infty} \phi(x) = +\infty$;

(2b) $$ \int_X \phi( F(x) ) \ \mu(dx) \ < \ +\infty $$

EDIT:

(1) Find an example in which there is no function $f: X \rightarrow \mathbb{R}_{\geq 0}$

($f(0)=0$, $f$ continuous and strictly increasing, $\lim_{x \rightarrow \infty} f(x) = +\infty$) such that:

$$ \int_X f(x) F(x) \ \mu(dx) \ < \ +\infty $$

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(1) is doubtful since $f$ cannot exist if the integral of $F$ on $X\cap[x,+\infty)$ is infinite for every $x$. –  Did Apr 16 '12 at 9:06
    
Please provide a counterexample then. For instance, if $\mu(dx) = dx$, function $F(x) =x^{-1}$ is not integrable over $[y,\infty)$ for every $y$. However $f(x) = x^{-1}$ makes the product $f(x) F(x)$ integrable. –  Adam Apr 16 '12 at 9:59
    
Neither the Lebesgue measure on the halfline as a probability measure nor the function $f:x\mapsto1/x$ as an increasing and going to infinity function (see (1a)) are the first examples I would suggest... Anyway, as I said, any $F$ and $\mu$ satisfying the condition I wrote would do, but you could try $F(x)=\mathrm e^x$ and $\mu(dx)=dx/F(x)$ on $x\gt0$. –  Did Apr 16 '12 at 10:08
    
Ok, my example was wrong. In your example, can you formalize that there are no $f$s making the product not integrable? –  Adam Apr 16 '12 at 10:30
    
The answer has been edited –  Adam Apr 16 '12 at 10:59

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