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Let $f(x)= \begin{cases} \sin \left(\frac{\pi}{x} \right) &\text{ if } 0 < x < 1\\ 0 &\text{ if } x=0 \end{cases}$

Prove that $f$ is Riemann integrable on $[0, 1]$.

I know a function is considered Riemann integrable if it is continuous and bounded. In this case, f is continuous on the interval $[-1, 1]$ and is also bounded. I'm inclined to believe this answer would satisfy the question. Is this enough or is there more I could prove?

This is page $82$ question $3.27$ of Advanced Calculus: An Introduction to Linear Analysis by Leonard Richardson.

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  • $\begingroup$ notice $\sin(0) = 0,$ so $f$ is really not piecewise defined. I guess you want $f(1) = 0$ too. $\endgroup$ – Will M. Apr 21 '17 at 19:34
  • $\begingroup$ Are you sure you mean $\sin (x/\pi)$? Because if so, there doesn't seem much point to specifying $0$ at $x=0$ since $\sin(0/\pi) = 0$ anyway. $\endgroup$ – Zain Patel Apr 21 '17 at 19:34
  • $\begingroup$ @Jennifer You made a mistake in your edit. Its $\sin(\pi/x)$. $\endgroup$ – Arbuja Apr 21 '17 at 19:35
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    $\begingroup$ @ErinA If you already have the result that $f$ continuous on $(0,1)$ and bounded on $[0,1]$ implies $f$ integrable on $[0,1]$ then all you need to do is state $f$ is continuous on $(0,1)$. However, I suspect that you're meant to use $f$ continuous on $[0,1]$ implies integrable on $[0,1]$ -- which means proving $f$ is continous at $0$, else the question seems too easy. $\endgroup$ – Zain Patel Apr 21 '17 at 19:39
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    $\begingroup$ This question makes less sense now than originally stated. what are the values in $[-1,0)$? $\endgroup$ – Will M. Apr 21 '17 at 19:41

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