Let $\displaystyle H(x)=k$ for $\displaystyle x =\frac{1}{k} \ (k \in \mathbb{N})$ and $\displaystyle H(x)=0$ elsewhere on $\displaystyle [0,1]$. Is $\displaystyle H$ riemann integrable?
Suppose that $\displaystyle T = \left\{ {\frac{1}{k}:k \in \mathbb{Z}^ + } \right\}$. That is the set of discontinuities.
$\displaystyle T$ is countable so the function is Riemann integrable.
Here is an outline of a proof. In any partition of $\displaystyle [0,1]$ the subinterval that contains 0 contains almost all points of $\displaystyle T$. The points outside of that can be covered by intervals the sum of their is as small as needed. Thus showing the integral is zero.