Proving Riemann Integrability of a function with countably many discontinuities? (No measure theory) Let $E = \{1/n : n \in \mathbb{N}\}$, and suppose $f: [0,2] \rightarrow \mathbb{R}$ is continuous everywhere but $E$. Prove that $f$ is Riemann integrable.
I know the Lebesgue criterion for integrability (discontinuities have measure $0 \Rightarrow f \in \mathscr{R}$), but how would you do this without? It is my understanding that partitions need to have finitely many internal endpoints.
 A: Sketch: we can confine all but finitely many discontinuities of $f$ into a small interval $I$. Outside $I$, we can choose a partition fine enough to make the upper and lower sums arbitrarily close. Simultaneously, since $f$ is bounded, we can choose $I$ small enough that it cannot contribute too much to the difference between the lower and upper sums.
Details:
Since $f$ is bounded above and below, we can subtract off the lower bound in order to assume without loss of generality that $f \geq 0$. Having done so, choose $M$ such that $f \leq M$. 
Let $\varepsilon > 0$. Choose $N$ large enough that $M/N<\varepsilon/2$. Choose a partition $P$ of $[1/N,2]$ whose mesh is small enough that the lower and upper sums for $P$ differ by at most $\varepsilon/2$. (Here I assume we have already proven that a function with finitely many discontinuities is Riemann integrable.)
Now extend $P$ to a partition $P'$ of $[0,2]$ by adding in the points $0$ and $1/N$. The difference between the lower and upper sums on $[1/N,2]$ is still at most $\varepsilon/2$, while the difference between the lower and upper sums on $[0,1/N]$ is also at most $\varepsilon/2$, since $0 \leq f \leq M$ there. So the difference between the lower and upper sums for $P'$ is at most $\varepsilon$.
We can tweak this to look much like the proof of the Lebesgue criterion. Set up the same as above, but where $M/N<\varepsilon/3$ instead. Take $\delta>0$ small enough that $MN\delta<\varepsilon/3$. Take intervals of length at most $\delta$ around $1,1/2,\dots,1/N$ in your partition. Then refine the rest of the partition enough that the difference between the upper and lower sums is less than $\varepsilon/3$ there (which we can do because $f$ is continuous there). Then again the lower and upper sums for the overall partition is at most $\varepsilon$ ($\varepsilon/3$ from the small intervals near the "widely spaced" discontinuities, $\varepsilon/3$ from the small interval containing the "tightly packed" discontinuities, and $\varepsilon/3$ from the continuous part).
Here I add in the assumption that $f$ is bounded, because unbounded functions are never Riemann integrable (since there is always a subinterval of the partition where the upper or lower sum is infinite).
