I have to show that the following function $f:[0,1]\rightarrow\mathbb{R}$ is Riemann Integrable:
$$f(x) =
\left\{
\begin{array}{ll}
1 & \mbox{if } x = \frac{1}{n} \\
0 & \mbox{otherwise}
\end{array}
\right.$$
For the upper and lower Riemann sum I am using the following definitions:
$$S_{l}(f,V)=\sum^{n}_{j=1}\inf_{I(j)}(f)(x_j-x_{j-1})$$
With $I(j)$ denoting the interval $[x_{j-1},x_j$] and $V$ is a partition $V=\{0,x_1,…,1\}$. The upper sum is defined with the supremum. I have shown that for any partition on $[0,1]$ the lower sum is $0$. But now I need to prove that for every $\epsilon>0$ there is a partition $V$ such that $S_{u}(f,V)<\epsilon$. Completing the proof is easy. I see that any partition on $[0,1]$ will only contain a limited number of points of the set $\{\frac{1}{n}:n\in\mathbb{N}\}$. But I can't make the proof concrete. Could anybody help me out?
Best Answer
Try the following:
The set $F=\{x\in [0,1]: f(x)>\epsilon \}$ is finite for every $\epsilon>0$. Then you can form a partition such that if an interval contains some $x\in F$ then it have no other. Finally you can choose the partition such that the sum of interval who contains some $x\in F$ is $<\epsilon$. Separate the interval wich cover $F$ and those which don't.
Can you continue from this?