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To demonstrate that ANY countable set has measure zero, is it sufficient to show that the natural numbers have a measure zero? If so, why; and, if not, why not?

Thank you :)

Rafael Vergnaud
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    Are you referring specifically to Lebesgue measure? Because this isn't true for an arbitrary measure. – carmichael561 Dec 10 '18 at 02:04
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    No, it is not sufficient. But, if you have a countable set, you can write it as something similar to the natural numbers. Specifically, you can sequence the set, so that each element has a unique corresponding natural number, i.e $S = {a_1, a_2 ,a_3, \cdots}$ – rubikscube09 Dec 10 '18 at 02:05
  • Hey Carmichael. I'm not sure. My book has only mentioned the term "measure zero," without additional qualifying adjectives. To quote, "A set $A \subseteq \mathbb{R}$ has measure zero if, for all $\epsilon > 0,$ there exists a countable collection of open intervals $O_n$ with the property that $A$ is contained in the union of all of the intervals $O_n$ and the sum of the lengths of all of the intervals is less than or equal to $\epsilon.$" – Rafael Vergnaud Dec 10 '18 at 02:06
  • Hey, Rubik. So, the proof would essentially be identical to the proof that would show the natural numbers to have a measure zero, correct? Essentially take the union of the intervals $(a_i - \frac{\epsilon}{2^{i+1}}, a_i + \frac{\epsilon}{2^{i+1}}).$ The sum would be $\sum_{i = 1}^{\infty} \frac{\epsilon}{2} (\frac{1}{2})^{i} = \epsilon.$ – Rafael Vergnaud Dec 10 '18 at 02:09
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    Yes it is identical. To answer the other question, your book is referring to the Lebesgue measure on $\mathbb{R}$. There are other measures of course. – rubikscube09 Dec 10 '18 at 02:14
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    Thanks, Rubik! :) – Rafael Vergnaud Dec 10 '18 at 02:14
  • Does this answer your question? A countable set has outer measure zero. Explanation of Royden example. – Mittens Sep 17 '20 at 19:40

1 Answers1

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(Asuuming we are talking about Lebesgue measur)

Let $a_i$ be a sequence of a countable set $A$, then $A_i=(a_i-2^{-i-1}\varepsilon,a_i+2^{-i-1}\varepsilon)$.

Now $A\subseteq \bigcup A_i$, and $\mu\left(\bigcup A_i\right)\le\sum2^{-i}\varepsilon=\varepsilon$

Showing that $\Bbb N$ is not enough because measure does not preserve by bijection, a simple example is the cantor set and the interval [0,1], where they have the same cardinality(hence there exists bijction between them) but one is measure 0 and the other 1.

ℋolo
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  • Hey, Holo. Thanks for the response. My proof was the following: take the union of the intervals $(a_i - \frac{\epsilon}{2^{i+1}}, a_i + \frac{\epsilon}{2^{i+1}}).$ The sum would be $\sum_{i = 1}^{\infty} \frac{\epsilon}{2} (\frac{1}{2})^{i} = \epsilon.$ Is this correct? My summation looks different than your's! – Rafael Vergnaud Dec 10 '18 at 02:11
  • @RafaelVergnaud you did the exact same things as I did, but you have a little(algebra) mistake, the summation of yours should be with $[...])^{i-1}$, then the $\epsilon/2$ will make it $[...])^i$ – ℋolo Dec 10 '18 at 02:14
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    Right. Got it! Thanks a lot Holo! – Rafael Vergnaud Dec 10 '18 at 02:15
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    @RafaelVergnaud I added an example where 2 sets has the same cardinality but not the same measure to explain why to show that $\mu(\Bbb N)=0$ is not enough – ℋolo Dec 10 '18 at 02:29