Integrability of $\sin \frac1x$ on $[0,1]$ using Darboux sums

Question

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.

Answer 1

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

$$U(f,P_n)-L(f,P_n)\le \frac{2}{n} +\sum_{k=1}^{m} \left (n^2\cdot \frac{1-1/n}{m}\right)\cdot \frac{1-1/n}{m}$$ $$ \le \frac{2}{n} + m\frac{n^2}{m^2} < \frac{2}{n} + \frac{1}{n} = \frac{3}{n}.$$

Answer 2

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.