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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!

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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 $< \epsilon$.

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  • $\begingroup$ I need some help with writing that rigorously. This is not a homework problem, just that I want to see how to write it $\endgroup$ – Frank Booth Mar 9 '16 at 17:02
  • $\begingroup$ @IdiotfromPrinceton: I added some clarification. $\endgroup$ – copper.hat Mar 9 '16 at 17:15
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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 $m(S)\leq \epsilon \left ( \sum_{k=1}^{N}\frac{1}{2^{n-1}} +1\right )<2\epsilon$

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

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