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If $f$ is a bounded real function on $[a,b]$, and $f^2 \in \mathscr R$, does it follow that $f \in \mathscr R$?


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up vote 2 down vote accepted

No. Consider

$$f(x)=\begin{cases}1 & \mbox{if}\quad x\in\mathbb{Q}\\ -1 & \mbox{if}\quad x\notin\mathbb{Q}\end{cases}$$

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Thanks, makes perfect sense. –  Ernest Singleton Mar 10 '13 at 0:34
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