f(x)=1 for $\displaystyle x=\frac{1}{n}$

f(x)=0 for otherwise

A)

prove that for 0<a<1 f is integrabile an $\displaystyle \int_{a}^{1}f(x)dx=0$

B)prove that for every $\displaystyle \epsilon>0 $there is a division P of the [0,1] section so

$\displaystyle S(P)<\epsilon$

C)prove that f is integrabile on [0,1] and $\displaystyle \int_{0}^{1}f(x)dx=0$

how i tried:

part A i have solved

its very easy i just couddnt see that 1/n is behind a and till 1 we have n-1 fractures

so its 0 exsept finite number of points so the integral is zero

i am stuck on part B:

i was told to look at $\displaystyle [\frac{\epsilon}{2},1]$section so from part A$\displaystyle \int_{\frac{\epsilon}{2}}^{1}f(x)dx=0$and because f is integrabile on the section then $\displaystyle \int_{\frac{\epsilon}{2}}^{1}f(x)dx=inf\{S(P)|P=di viding[\frac{\epsilon}{2},1]}$

“there exists deviding Q of $\displaystyle [\frac{\epsilon}{2},1]$ so $\displaystyle S(Q)<\frac{\epsilon}{2}$ (other wise$\displaystyle \frac{\epsilon}{2}$ would be lower bound of the upper sums and the integral will not be zero.”

why the integral would be zero???

i cant see how to get it

i know that if its integrabile then the integral equals the lower bound of the upper sums

but why it will be zero?