Integration of upper functions (from Apostol)

[travis0868]

Homework Statement

(10.4 in Mathematical Analysis by Apostol)
This exercise gives an example of an upper function f on the interval I = [0,1] such that -f (not a member of) U(I). Let {r1, r2, ...} denote the set of rational numbers in [0,1] and let I_n = [r_n - 4^-n, r_n + 4^-n] (intersect) I. Let f(x) = 1 if x (is a member of) I_n for some n, and let f(x) = 0 otherwise.

a. Let f_n(x) = 1 if x (is a member of) I_n, f_n(x) = 0 if x (is not a member of) I_n and let s_n = max(f_1,...,f_n). show that {s_n} is an increasing sequence of step functions which generates f. This shows that f (is a member of U(I).

b. Prove that Integral f <= 2/3

The Attempt at a Solution

I don't understand part b. Suppose that your set of rational numbers has r_1 = 0. Then I_1 = [0 - 1, 0 + 1] intersect [0,1] = [0,1]. Thus s_n = 1 over [0,1]. The step integral of s_n over [0,1] equals 1. As n->infinity, the integral remains 1. Thus Integral f = 1.

What am I missing here?

Travis

 

[Dick]

I think that's the reason for the 4 in 4^(-n) and starting the indexing with n=1. If n=1 then the interval is [r1-1/4,r1+1/4]. But this kind of stuff is just technical detail. You do get the actual picture, right?

 

[travis0868]

Thanks a lot, Dick. I get it now.