Vtyjkyuk

I have to show that the following function $f:[0,1]rightarrowmathbbR$ is Riemann Integrable:

$$f(x) =
left{
beginarrayll
1 & mboxif x = frac1n \
0 & mboxotherwise
endarray
right.$$

For the upper and lower Riemann sum I am using the following definitions:

$$S_l(f,V)=sum^n_j=1inf_I(j)(f)(x_j-x_j-1)$$

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 $frac1n:ninmathbbN$. But I can't make the proof concrete. Could anybody help me out?

Try the following:

The set $F=xin [0,1]: f(x)>epsilon $ is finite for every $epsilon>0$. Then you can form a partition such that if an interval contains some $xin F$ then it have no other. Finally you can choose the partition such that the sum of interval who contains some $xin F$ is $<epsilon$. Separate the interval wich cover $F$ and those which don't.

Can you continue from this?

$ou have the right idea. It's all about the technicalities. One possible partition set-up is as follows.

For every $epsilon>0$, there exists an integer $N_0$ such that $1/N_0<epsilon$. Let $N=maxN_0,5$.

Partition $[0,1]$ with $V=0,x_1, x_2,ldots,x_4N-5$ where $x_1=frac12N$, $x_4N-5=1$, $x_2k+1-x_2k=frac1N^3$, and $x_2k<frac12N-k<x_2k+1$.

Then, $$S_u(f,V)= 1cdot frac12N + frac1N^3 cdot (2N-1)<frac12N+frac2NN^3=frac12N+frac2N^2$$.

Since $Nge 5$, $frac2N^2<frac12N$. Therefore, $S_u(f,V)<frac1N<epsilon$.