Integrability of $\sin \frac1x$ on $[0,1]$ using Darboux sums 
Prove that $f:[0,1] \rightarrow\mathbb{R}:f(x)= \left \{\begin {array}{ll}
\sin \frac1x &, \textrm{if}~      x\in(0,1]\\
0 &, \textrm{if}~~x =0
\end{array}
\right.~~$ is Riemann integrable using Darboux sums.

Attempt. The proof that I am aware of uses the classic result that extends integrability from every $[a,1]$ to $[0,1]$ (if $f$ is bounded on $[0,1]$ and integrable on $[a,1]$ for all $a\in (0,1)$, then $f$ is integrable on $[0,1]$). I am interested in a proof that uses Darboux sums, especialy: $$U(f,P_n)-L(f,P_n) \to 0$$
for a specific sequence $(P_n)$ of partitions of $[0,1]$. So far I haven't found one that does the job. 
Note: the fact that such a partition exists neither means that we are looking for a good looking one, nor that it could be written in a closed form. 
Thanks in advance.
 A: Fix $n$ large. Then for $1/n\le x <y\le 1,$ the mean value theorem shows
$$\tag 1 |f(y)-f(x)|\le n^2|y-x|.$$
Now let $m\in \mathbb N$ be greater than $n^3.$ Set
$$P_n=\{0\} \cup\{1/n+k\frac{(1-1/n)}{m},k=0,1,\dots ,m\}.$$
Using $(1),$ we get
$$U(f,P_n)-L(f,P_n)\le \frac{2}{n} +\sum_{k=1}^{m} \left (n^2\cdot \frac{1-1/n}{m}\right)\cdot \frac{1-1/n}{m}$$ $$ \le \frac{2}{n} + m\frac{n^2}{m^2} < \frac{2}{n} + \frac{1}{n} = \frac{3}{n}.$$
A: Take $\epsilon > 0$ and $n \in \mathbb N$ such that $0 < \frac{1}{n \pi} < \frac{\epsilon}{4}$.
As $f$ is continuous on $[\frac{1}{n \pi},1]$ it is Riemann integrable on that interval and you can find a partition $P^\prime_n$ of $[\frac{1}{n \pi},1]$ such that
$$0 \le U(f,P_n^\prime)-L(f,P_n^\prime) < \frac{\epsilon}{2}$$
Now take $P_n = \{0\} \cup P^\prime_n$. You have 
$$\begin{aligned}
0 &\le U(f,P_n)-L(f,P_n)\\
 &=  \left(U(f,\{0, \frac{1}{n \pi}\}) - L(f,\{0, \frac{1}{n \pi}\})\right) + \left(U(f,P_n^\prime)-L(f,P_n^\prime)\right)\\
&< \frac{2}{n \pi} + \frac{\epsilon}{2} < \epsilon
\end{aligned}$$
as deisred.
