Choose your intervals so that the endpoints are 1/2, 1/3, 1/4, etc.
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?
A similar thing in slightly different words.Since , there exists a division Q such that . This follows from the definition of infimum. For the segment , choose any division. Since the value of f(x) does not exceed 1, the upper sum of the segment does not exceed . For a division for the whole segment [0, 1] take the union of any division on and Q. Then the corresponding sum will be .