Can someone explain this Riemann Integral Example from Bartle's book, I dont quite understand what the author means by 'counting the tags twice' and I also dont understand his choice of delta.
Is $$ G(x)=\sum^\infty_{n=1}\frac1n\mathbb{1}_{\{\tfrac1n\}}(x)$$
Riemann integrable on $[0,1]$?
We take the interval $[a,b]$ and putting a partition onto it. The partition is sufficiently fine than no sub-interval has a width greater than $\delta_\epsilon.$ We look at the upper sum. That is the maximum point in each sub interval, multiplied by the width of the sub-interval, and sum across all of sub intervals.
In the specific example $G(x) = \frac {1}{n}$ when $x=\frac {1}{n} \forall n\in \mathbb N$ and $0$ otherwise. And we are evaluating over the interval [0,1]
We can ignore all of the points where $G(n) < \epsilon$ because, even if we multiply these by the entire length of the interval, the sum of these points will be less than $\epsilon.$
Now we need the sum over all partitions that include a point where $G(x) > \epsilon$
For any epsilon there is an $N$ such that $n>N \implies \frac {1}{n} < \epsilon$
Which implies that are $N$ points where $G(x) > \epsilon.$
It is possible for any of these points to be an endpoint of a sub-interval, which makes it also and endpoint of the neighboring sub-interval. Even if we count every point twice, there are less than $2N$ sub-intervals where $G(x)>\epsilon$
$G(x) < 1$ over ever one of these sub intervals.
The upper sum over these sub-intervals is less than $(2N)\delta_\epsilon$
Now lets choose $\delta_\epsilon = \frac {\epsilon}{2} = \frac {1}{2N}$
We broke our interval in two, and the upper sum of each piece is less than $\epsilon$
And we can make $\epsilon$ arbitrarily small.