If $f$ is bounded and if $f^2$ is Riemann integrable, then is f Riemann integrable?

Question

This was on a past exam paper, and it was asking if $f$ is bounded and if $f^2$ is Riemann integrable, then is f Riemann integrable?

If I had to guess, I'd say no. I tried creating an $\epsilon$ argument, but I feel like I'm grasping at straws, so I'd appreciate a hint towards a slightly more concrete proof.

Answer 1

Hint: Consider $f \colon [0,1] \rightarrow \mathbb{R}$ given by

$$ f(x) = \begin{cases} 1 & x \in \mathbb{Q} \cap [0,1] \\ -1 & x \in [0,1] \setminus (\mathbb{Q} \cap [0,1]). \end{cases}. $$

Answer 2

No. Consider $$ f(x) = 2\chi_{\mathbb{Q}} - 1 $$ where $\chi_{\mathbb{Q}}$ is the indicator of the rational. $(f)^2 = 1$, which is Riemann integrable, but $f$ is not Riemann integrable by Lebesgue's criterion of Riemann integrability.