f(x)=1 for

f(x)=0 for otherwise

A)

prove that for 0<a<1 f is integrabile an

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

C)prove that f is integrabile on [0,1] and

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 section so from part A and because f is integrabile on the section then

“there exists deviding Q of so (other wise 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?