The question is following:
Let $f:[0,1]\rightarrow \mathbb{R}.$
$f(x)=x,$ if $x=1/n, n\in\mathbb{N}$
$f(x)=0,$ otherwise.
Is $f$ Riemann-integrable? If it is, what is its value?
I know that the basic idea of Riemann-integral is to find two step-functions $h$ and $g$ which both integrate to the same value and $h(x) \leq f(x) \leq g(x)$. This time I can't find the upper step-function. However I guess that this function is Riemann-integrable as it's discontinuous only on $1/n$. Also I'm guessing that it's value is $0$. Am I correct and how could I proof it?
To do this from scratch, let $\epsilon>0$ and consider the partition $Q$, such that:
$1).\ \epsilon$ is a partition point, so that if $n>1/\epsilon,\ 1/n\in [0,\epsilon].$ Let $N\ge 2$ be the smallest integer for which this happens
and
$2).\ $ there are $x_{k}\in Q,$ such that $x_k<1/k<x_{k+1}$ and $x_{k+1}-x_k<\epsilon/2^k$ for $k<N.$
Then, $U(f,Q)=\epsilon\sum^N_{k=1}\frac{1}{k2^k}+\frac{\epsilon }{N}<\epsilon.$ Now clearly $L(f,Q)=0$ so $U(f,Q)-L(f,Q)<\epsilon.$
Since $L(f,Q)\le\underline \int f\le \overline \int f\le U(f,Q)$, we conclude that $\overline \int f- \underline \int f<\epsilon$, and since $\epsilon$ is arbitrary, $\int f$ exists and is equal to $0.$