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