Let $X$ be a nonnegative random variable and $p > 1$. I want to prove that $$ E ( X^p) < \infty \Longrightarrow E (f(X)^p) < \infty,$$ if $f(x) = O(x)$. This statement makes absolut sense and my proof goes as follows. Let $M, x_0$ be such that $$ f(x) \leq M \, x \quad \forall x \geq x_0.$$ Then, $$ E (f(X)^p) = E ( \mathbb{1}_{X \leq x_0} f(X)^p) + E (\mathbb{1}_{X > x_0} f(X)^p) \leq f(x_0)^p + E (\mathbb{1}_{X > x} M^p X^p) \leq f(x_0)^p + M^p E(X^p) < \infty.$$
Is this correct, and is there an even easier way to show this (or is the above statement even trivial)?
This is not even true. Let $f(x)=\frac 1 x$ for $x >0$ and $0$ for $x \leq 0$. Then the hypothesis holds with $x_0=1$ and $M=1$. Take $p=2$ for example. Can you think of an example where $X > 0, EX^{2} <\infty$ but $E\frac 1 {X^{2}}=\infty$?