Proving the integral of a bounded function with finite discontinuities exists? (example) I have this problem:
Prove that if $f(x)>0$ for a finite number of points and f(x)=0 for all other points in $[0,1]$, $f$ is Riemann integrable and the integral equals $0$.
Intuitively, this seems clear to me. You can just make the intervals of the partition containing the non-zero points arbitrarily small and then since their width is small, the upper and lower sums are just $0$. 
However, I'm really struggling to prove this rigorously. I've looked at similar proofs (including on this site) for the general case where there are a finite number of discontinuities, but they're not clicking with me. Any help would be greatly appreciated!

Your idea is prefectly fine, and it is not difficult to implement it in full rigor.
If we have a finite number of points $x_1,x_2,\ldots,x_K$ in $[0,1]$, for any $N\in\mathbb{N}$ large enough the intervals with length $\frac{1}{N}$ centered at them do not overlap, hence for any uniform partition of $[0,1]$ in $2N$ subintervals, each subinterval contains at most one "troublesome" point. Let $M$ be the maximum of $f$ over $\{x_1,x_2,\ldots,x_K\}$. The lower Riemann sum associated with the previous partition is zero, and the value of the upper Riemann sum is bounded by $\frac{KM}{2N}$, which goes to zero as $N\to +\infty$. This proves that the Riemann integral of $f$ over $[0,1]$ is zero.
