I want to ask the complement to this question (Lebesgue-integrability of roots and powers of a function). Suppose $f$ is a Lebesgue measurable function and in particular that $f(x) \geq 0 \: \forall x \in \mathcal{X}$ and $\int_{\mathcal{X}} f(t) dt = 1$, i.e., $f$ is a probability density function (pdf). I want to know what are the conditions I need on $f$ such that $\int_{\mathcal{X}} f(t)^p dt < \infty$ for $ 0 < p < \infty$.
Edit
Following feedback from @Dog_69 (!) and @Clement C. and some digging on my own, here are a few more observations I hope will be useful. Let $\{g_n\}$ be "a sequence of nonnegative simple functions". According to this (http://mathworld.wolfram.com/LebesgueIntegrable.html) we would need the following two conditions on $f(x)^p$:
- $\sum_{n=1}^\infty\int_{\mathcal{X}} g_n(t) dt < \infty$, and;
- $f(x)^p = \sum_{n=1}^\infty g_n(x)$ a.e.
Now, if one could frame the counter-examples to my foolish claim (edited out) that the function is integrable for $ 0 <p \leq 1$ by showing that they do not conform to these conditions, it would be a start, I reckon.
More edits
According to this link Product of measurable and integrable functions, if $f$ is integrable and $g$ is bounded and measurable, then $fg$ will be integrable. Since here $f(x)$ is assumed to be integrable, a sufficient condition is that $g(x) = f(x)^{p-1}$ is bounded and measurable. I had erroneously claimed that for $ 0 < p < 1$ any pdf would integrable. Let us work on the counter-examples kindly provided by @Dog_69 and @Clement C. to find out what went south:
$f_0(x) = \frac{2}{x^2}$ with $\mathcal{X} = [2, +\infty)$. Take $p = 1/2$ and thus $f_0(x)^{1/2} = \frac{2}{x^2} \cdot \frac{x}{\sqrt{2}}$. Since $g_0(x) = \frac{x}{\sqrt{2}}$ is not bounded on $\mathcal{X}$ we have that it doesn't conform to the condition above.
$f_1(x) = \frac{1}{\pi(1+x^2)}$ with $\mathcal{X} = (-\infty, +\infty)$. For simplicity, again take $p = 1/2$ and hence we have $f_1(x)^{1/2} = \frac{1}{\pi(1+x^2)} \cdot \sqrt{\pi (1+x^2)}$. Again, $g_1(x) = \sqrt{\pi (1+x^2)}$ is not bounded on $\mathcal{X}$.
If one could show that the result in the link is an iff statement, we'd be done, because we'd have a necessary and sufficient condition on $f$. But unfortunately this is not the case. Consider $\int_{0}^{+\infty} t^2 h(t) dt$ where $h$ is, say, a Gamma pdf. Clearly this expectation exists, but $g(t) = t^2$ is measurable but unbounded on $\mathcal{X}$.