Give an example of a finction $f(x)$ such that $f(x)>0$ $\forall x\in[a,b]$, where $\sqrt{f}$ is Riemann integrable but $f$ is not.
I don't know the example. Please help.
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2If $g : [a, b] \to [c, d]$ is Riemann integrable on $[a, b]$ and $\varphi : [c, d] \to \mathbb{R}$ is continuous, then $\varphi \circ g$ is Riemann integrable as well. What happens if you apply this statement to $g = \sqrt{f}$ and $\varphi(x) =x^2$? (Also recall that Riemann integrable functions are bounded.) – Sangchul Lee Jan 23 '18 at 15:15
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@SangchulLee Would you please give me a specific example. – rama_ran Jan 23 '18 at 15:36
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What I meant was that if $\sqrt{f}$ is Riemann integrable then $f$ is also Riemann integrable. But of course, there are examples where $\sqrt{f}$ is improperly Riemann integrable but $f$ is not. (One such example is $f(x) = 1/x$ as in the answer below.) – Sangchul Lee Jan 23 '18 at 15:39