Prove that a function is Riemann Integrable

Question

Let $ f: [0,1] \rightarrow \Bbb R$ defined by:

$f(x)= \sin(\frac{1}{x}) $ if $ x \notin \Bbb Q$ , $ 0 $ if $x \in \Bbb Q$

I have to prove that this function is Riemann integrable in that interval.

It is bounded in $ [-1,1]$ , and it's not continuous.

I have tried seeing if I could make $ U(f,P) - L(f,P) < \epsilon$ for some partition $P$ , but I wasn't able to evaluate $ M_k $ and $ m_k $ , as for some intervals they could be $ 0 , 1$ or $-1$.

I also know that, if it was just $\sin(\frac{1}{x})$ , if would be integrable in that interval , as it would be continuous in $(0,1]$ , and we could ignore the discontinuity in $\{0\}$ as it is just a point.

Thank you in advance.

Answer 1

You cannot prove it, since it is false. The set of points at which that function is discontinuous has Lebesgue measure greater than $0$.

Answer 2

Use the Darboux integral approach.
Define $g(x)=sin(\frac{1}{x})$. Let $f_+(x)=f(x)$ for $x\ge 0$ and $=0$ for $x\lt 0$. Let $f_-(x)=f(x)-f_+(x)$. Similarly define $g_+(x)$ and $g_-(x)$. Then The upper Darboux sum of $f_+(x)$ will converge to the integral of $g_+(x)$, while the lower sum converges to $0$. Similarly the upper sum of $f_-(x)$ converges to $0$, while the lower sum converges to the integral of $g_-(x)$. Combing these results, the upper sum of $f(x)$ converges to the integral of $g_+(x)$, whille the lower sum converges to the integral of $g_-(x)$.

Therefore the upper and lower Darboux sums for $f(x)$ don't converge to the same value, thus not being Darboux integrable (or Riemann integrable).