(Context: I have been using this resource from UC Davis to help me out with Riemann integrability.)
So, I more or less get how Riemann integrability works when we specify $x$ over an interval, say $x>0$. But what if $x$ is a function of $n$? I came across this problem:
Prove that $f(x)$ is Riemann Integrable over $[0,1]$ and that $\int_{0}^{1}f(x) = 0$ given $$f(x)=\left\{\begin{matrix} 1 & x=\frac{1}{n} \\ 0 & \text{otherwise} \end{matrix}\right.$$
I'm not sure how to partition this (because of the $n$) to show this. Anyone have any insight?
The idea here is that all but finitely many of the points in $\{ 1/n : n \in \mathbb{N} \}$, where the discontinuities in $f$ are, are very close to $0$. So you can work this way. Let $\varepsilon >0$. Choose $N \in \mathbb{N}$ such that $1/(N+1)<\varepsilon/2$. Then make the first two points of the partition be $0$ and $1/(N+1)$.
Now you just have $N$ discontinuities to work with. Make your partition so that they are each surrounded by intervals of length at most $\varepsilon/2N$. Now you can show that the lower sum of this partition is $0$ while the upper sum is some positive number which is less than $\varepsilon$.