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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?

Breton Thomas
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    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. –  Sep 17 '18 at 15:08

2 Answers2

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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.

mlainz
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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.

Yanko
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