# Does the integrability of $f^3$ imply the integrability of $f^2$ and/or $f$?

We know that the integrability of $f$ implies the integrability of $f^2$, but the integrability of $f^2$ does not imply the integrability of $f$ (for example, the function $f(x) = 1$ when rational and $-1$ when irrational).

Question: However, does the integrability of $f^3$ imply anything about the integrability of $f$? And what about higher powers?

• I think it depends on the underline set. Do you consider $f$ as a function on a closed interval $[a,b]$ or on an infinite interval, say $[0,\infty)$ or $\mathbb{R}$? In the latter I can find a non-integrable function $f$ such that $f^3$ is integrable. – Yanko Sep 17 '18 at 15:04
• You can make implications but it depends on whether the domain Ω is bounded or not. – user540665 Sep 17 '18 at 15:08

If you consider $f$ on a closed interval click here.
The statement does not hold for functions on infinite intervals, for example the function $f:[1,\infty)\rightarrow \mathbb{R}$ with $f(x)=\frac{1}{x}$ is not integrable, but any power $f^n$ is integrable.
That is false. $$f = \begin{cases} \frac{1}{\sqrt{x}} &\quad 0 < x < 1\\ 0 \quad &\text{otherwise} \end{cases}$$
is (Lebesgue)-integrable, but $f^2$ isn't.
This can be generalized to arbitrary greater than $1$ powers. We can say something more if either the domain or the functions are bounded.