# 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.

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.

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 $< \epsilon$.

• I need some help with writing that rigorously. This is not a homework problem, just that I want to see how to write it – Frank Booth Mar 9 '16 at 17:02
• @IdiotfromPrinceton: I added some clarification. – copper.hat Mar 9 '16 at 17:15

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 $m(S)\leq \epsilon \left ( \sum_{k=1}^{N}\frac{1}{2^{n-1}} +1\right )<2\epsilon$

• This is not a finite cover. It works for Lebesgue, not Jordan. – copper.hat Mar 9 '16 at 17:30
• @copper.hat: Thanks, I fixed it. – Matematleta Mar 9 '16 at 17:41