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Let $R \subset \mathbb R^n$ a "brick" and $f ∶ R → \mathbb R$ a bounded function. We denote by $ℛ(R)$ the space of Riemann-integrable functions on R. Is the following statement true ? $$f^3 \in ℛ(R) \implies f \in ℛ(R)$$

I know that $f^2 \in ℛ(R) \implies f \in ℛ(R)$ is false by taking $n=1$, $R=[0,1]$ and $f(x) = 1$ if $x \in \mathbb Q$ and $f(x)=-1$ if $x \in \mathbb R \backslash \mathbb Q$. We have that $f^2=1 \forall x$ but $f$ is not integrable.

But I do not have any idea for the $f^3$ statement. I thought of taking something maybe with $n=3$ and also thought of $f(x)=\lfloor{1/x}\rfloor \mod 3 $ that can be interesting but it does not work. Maybe the statement is true. Any ideas ?

Kilkik
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    The cube root is a continuous function on $\Bbb R$ – Martin R Jun 30 '21 at 12:36
  • @MartinR But we could have said the same thing for $f^2$ since the square root is a continuous function on $\mathbb R^+$ but I gave a counter example. – Kilkik Jun 30 '21 at 12:43
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    Keep in mind the subtlety of roots vs principal roots here. The square root of $x^2$ (for $x\in\mathbb{R}$) is both $\pm x$. However, the principal square root of $x^2$ is $|x|$. The principal square root is well-defined as a function. The square root itself is not well-defined as a function since it is multi-valued. You don't work with general relations in analysis (or most of math), you work with well-defined functions. To that end, you are forced to adopt the principal square root (or the other one, but not both simultaneously like you're thinking). – Cameron Williams Jun 30 '21 at 12:50
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    @John: The difference is that the cube root is defined and continuous on all of $\Bbb R$. – Martin R Jun 30 '21 at 12:52
  • So ultimately what you can conclude is that if $f^2$ is integrable, $|f|$ is integrable (and so is $-|f|$). You need extra information about $f$ to conclude that it is actually integrable. – Cameron Williams Jun 30 '21 at 12:59

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Let $g(x)=x^{1/3}$, if $f^3$ is integrable then $g\circ f^3=f\Rightarrow \text{$f$ is integrable.}$

logarithm
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