TITLE: Is $f(x)=\sin(1/x),\;x\ne0$, $f(0)=0$, Riemann integrable on $[0,1]$?
QUESTION [9 upvotes]: Is the function $\sin \frac{1}{x}$ Riemann integrable on an interval containing $0$?

REPLY [9 votes]: This is late but will be useful for future readers.
PROOF
Let $\epsilon>0$ be given  but arbitrarily small. Choose $x_1\in(0,1]$ such that $1-x_1<\epsilon/2.$ Then, the set $P=\{0,x_1,1\} $ is a patition. The upper and lower Darboux sums are given by  
\begin{align}U(f,P)=\sum^{2}_{k=1}M_k (x_k-x_{k+1}),\;\;\text{where}\;\;M_{k}= \sup\{f(x):\;x_{k-1}\leq x<x_{k}\},\end{align}
\begin{align}L(f,P)=\sum^{2}_{k=1}m_k (x_k-x_{k+1}),\;\;\text{where}\;\;m_{k}= \inf\{f(x):\;x_{k-1}\leq x<x_{k}\}.\end{align}
By computation
\begin{align}U(f,P)&=\sum^{2}_{k=1}M_k (x_k-x_{k+1})\\&=M_1 (x_1-x_{0})+M_2 (x_2-x_{1})\\&= 1-x_{1},\;\;\text{where}\;\;M_{1}=0,\,M_{2}=1,\end{align}
\begin{align}L(f,P)&=\sum^{2}_{k=1}m_k (x_k-x_{k+1})\\&=m_1 (x_1-x_{0})+m_2 (x_2-x_{1})\\&=-(1-x_{1}),\;\;\text{where}\;\;m_{1}=0,\,m_{2}=-1.\end{align}
Thus,
\begin{align}U(f,P)-L(f,P)&=2(1-x_{1})<\epsilon.\end{align}
So, given any $\epsilon$, choose $P=\{0,x_1,1\} $ such that $1-x_1<\epsilon/2.$ Then,
\begin{align}U(f,P)-L(f,P)<\epsilon.\end{align}
Hence, $f$ is integrable.