I know that continuous functions are Riemann integrable, piecewise functions are Riemann integrable and that monotonic functions are Riemann integrable.
I would like to know if there is a function $f$ that is Riemann integrable such that $f$ is not picewise continuous and such that $f$ is not monotonic.
Thank you.
Yes. Such a function exists. For example, let $f:[0,1]\rightarrow\mathbb{R}$ be defined by $f(x)=0$ if $x\in\{0,1\}\cup\left([0,1]\setminus\mathbb{Q}\right)$ and $f(x)=\frac{1}{q}$ if $x\in(0,1)\cap\mathbb{Q}$ and $x=\frac{p}{q}$, where $p,q\in\mathbb{N}$ and $p,q$ are relatively prime. $f$ is known as the Riemann function. It is well-known that the set of discontinuity of $f$ is $(0,1)\cap\mathbb{Q}$, which is of Lebesgue measure zero. By a characterization theorem (due to Lebesgue), $f$ is Riemann integrable. Clearly, $f$ is not piecewise continuous nor monotone on any non-empty open sub-interval of $[0,1]$.