Computing Jordan Measure I have the following problem:
I need to compute the Jordan measure of the set $S=\frac{1}{n}, n=1,2,3,4,...$.
My thinking is this: Let $P$ be a partition of the interval $[0,1]$, and let $Q_i$ be the interval such that $Q_i=[P_i,P_{i-1}]$. I need to show that the sum of these rectangles are smaller than an epsilon. Let $m(S)$ define the Jordan measure. then $m(S)= \inf \sum Q_i\lt \epsilon$. 
I do not know if this is correct. I guess I do not know how to write the epsilon part rigorously. Help would be much appreciated.
Thanks in advance!
 A: The idea here is to guess and then prove.
Note that for any finite set, the Jordan measure is zero, since you can
pick arbitrarily small intervals around each point.
A good guess is that the Jordan measure of $S$ is zero.
The only catch is that there are an infinite number of points, but since
they cluster at $0$, we can 'capture' almost all of them with a small
interval around $0$. Then we can choose the partition so that the 
remaining (finite number) points are contained in suitably small intervals.
Addendum:
Choose $\epsilon>0$, then there is some $N$ such that ${1 \over n} < {1 \over 4}\epsilon$ for all $n \ge N$.
In particular, ${1 \over n} \in (-{1 \over2 } \epsilon, {1 \over 2}\epsilon )$
for all $n \ge N$.
Hence all but a finite number of the points are contained in the interval $[0,{1 \over 2}\epsilon]$.
For the remaining points ${1 \over N-1},...,1$,
choose an interval $I_k$ of length ${1 \over 2 (N-1)}\epsilon$ centred on each of
these points.
Then $S$ is contained in $[0,{1 \over 2}\epsilon] \cup I_{N-1} \cup \cdots \cup I_1 $, and the sum of the lengths is $< \epsilon$.
A: Let $\epsilon >0$.Choose $N\in \mathbb N$ so that $m(\left ( -\frac{1}{N}, \frac{1}{N}\right ))<\epsilon$
Now, take $\mathcal S=\left \{ \left (\frac{1}{n}- \frac{\epsilon}{2^{n}},\frac{1}{n}+ \frac{\epsilon}{2^{n}} \right ) \right \}_{0<n\leq N}\cup \left \{ \left ( -\frac{1}{N}, \frac{1}{N}\right ) \right \}$. 
$\mathcal S$ a finite cover of $S$ and so $m(S)\leq \epsilon \left ( \sum_{k=1}^{N}\frac{1}{2^{n-1}} +1\right )<2\epsilon$
