Define a function on [0,1] by?

Question

f(x)= 0 if x=1/n for some n in the natural numbers,
and
f(x)=1 otherwise.

Show that f is integrable on [0,1] and the integral (with lower limit 0 and upper limit 1) of f = 1.
(hint: you have to create a partition of [0,1] as follows. Let epsilon>0 be given. One interval should have width epsilon/2 and contain the 'infinite tail end' of the sequence of points 1/n and the remaining intervals should contain the 'finite beginning' of the sequence 1/n and have total length epsilon/2. Now calculate U(f,P) and L(f,P), show that U(f,P) - L(f,P) = epsilon and argue that the given integral is true.)

I'm not very sure how to do this, even with the hint that follows.
Please help. thank you.

Answer 1

Which part of the hint do you not see? Put one small interval around the 'tail' of the sequence {1/n} and a bunch of small intervals around the remaining points in that sequence. Then there will be long intervals that avoid the sequence where the function has a constant vaue of 1. Now compute the upper and lower sums for that partition. The maxima on the intervals with points of {1/n} will be 1 and the minima will be 0. For the other intervals, both the max and min will be 1. Now add up the Riemann sums.

Answer 2

Here's the key point. You're trying to come up with a set of intervals that has the following properties:

1. The total width is epsilon. Hence the maximum difference between the upper and lower estimates of the aggregate area of the function over those intervals is epsilon*1 = epsilon.

2. Those intervals contain all the x = 1/n points. Consequently, in the complement of those intervals, f is identically = 1 and hence there's no difference between the upper and lower estimates.

From there it should be clear sailing.