If Riemann integrable, then it has finite number of discontinuities I know that any bounded function with a finite number of discontinuities is Riemann integrable over some interval. Is vice versa i.e., 
If a bounded function is Riemann integrable, then it has a finite number of discontinuities?
Thanks.

No, for instance, the function
$$f(x) = \left\{ \begin{array}{ll}
e^{-x^2}, & x \notin \mathbb{Z}, \\
2, & x \in \mathbb{Z}
\end{array}\right.$$
Has countably many discontinuities, but is Riemann integrable over $\mathbb{R}$.
What is true is that a function is Riemann integrable iff the set of discontinuties is a set of Lebesgue measure zero. Very loosely speaking, this implies that you can have at-most countably many discontinuities on sort of "conventional" intervals. If you look for more pathological examples, you can generate uncountably many discontinuities and still be OK; examples are in the comments.
