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For a countable set, it is easy to construct a countable cover of arbitrary small intervals to show that the set has measure zero. Can this reasoning be extended to sets with uncountably many isolated points? By definition of the Lebesgue measure, the cover of the points must consist of finitely many intervals.

The typical example of an uncountable set with Lebesgue measure zero is the Cantor set. However, for this set, the complement can easily be quantified and the set is constructed in countably many steps. How should uncountable point sets on which no order can be induced be dealt with?

John
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    Can you give an example of a set with uncountably many isolated points? – Umberto P. Sep 26 '14 at 16:12
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    @UmbertoP.: I was just thought about this! – Bumblebee Sep 26 '14 at 16:14
  • Let R be an equivalence relation with uncountably many equivalence classes (e.g. R(p,q) iff p-q is rational) and let S be a set with exactly one point in each class. (This set is in fact not Lebesgue measurable, but the aim is to show this by contradiction) – John Sep 26 '14 at 16:15

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As Umberto P. already hints at, there are no uncountable subsets $E$ of $\Bbb{R}$ consisting only of isolated points; indeed, the intersection of $E$ with each $[-n,n]$ is finite (why?), so that $E$ is countable (why?).

EDIT: As @TonyK points out, the argument above is incorrect. But it is still true that every set consisting solely of isolated points is countable. To see this, note that each $x \in E$ has an (open) neighborhood $U_x$ in $\Bbb{R}$ such that $E \cap U_x$ is finite (in fact, one can choose $U_x$ so that $E \cap U_x = \{x\}$. Now $\Bbb{R}$ is second countable and hence every subset is separable, so that each subset is Lindelöf (see e.g. Prove that a separable metric space is Lindelöf without proving it is second-countable), so that there is a countable subcover $(U_{x_n})_n$. But then $E \subset \bigcup_n [E \cap U_{x_n}]$ is countable, because each $E \cap U_{x_n}$ is finite.

Also note that the definition of Lebesgue measure requires countable covers by intervals, not finite covers (otherwise, $\Bbb{Q}$ would not have measure zero).

An interesting set of positive (but smaller than expected) measure is obtained by

$$ \bigcup B(x_n, 1/2^n), $$

where $(x_n)_n$ is an enumeration of $\Bbb{Q}$.

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PhoemueX
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  • Thank you (and Umberto P. and Nilan) very much for your response.

    I think I am starting to see your point. Any point $e\in E$ has a neighborhood that contains no other points in $E$. An intersection $\left[-n,n\right]\cap E$ is a finite union of such neighborhoods (for radii sufficiently large), hence can contain at most finitely many points. But then $E$ is a countable union of finite sets $\left[-n,n\right]\cap E$ and is therefore countable. Is this more or less correct?

    Also, you're right about the countable covers, my appologies.

     – John Sep 26 '14 at 18:21
  • "The intersection of $E$ with $[-n,n]$ is finite": I think this is wrong. For instance, $E$ can contain the (isolated) points ${1/r : r \in \mathbb N}$. – TonyK Sep 26 '14 at 18:27
  • @TonyK: Thank you very much, I was a bit too fast. I have now provided a correct proof. – PhoemueX Sep 26 '14 at 18:40
  • You're not really using the Lindelöfness of $\mathbb{R}$, but the hereditary Lindelöfness (every subspace is Lindelöf). While every second-countable space is hereditarily Lindelöf, not all Lindelöf spaces are. – user642796 Sep 26 '14 at 18:55
  • Thanks again. What implication does this have for the set I mentioned earlier? The equivalence relation $R$ clearly has uncountably many classes. But let $d_{i}=min_{x\in S\setminus e} d(e,x)$ for metric $d$. If you subsequently pick a $q\in(0,d_{i})\cap\mathbb{Q}$, then $e<e+q<x$ and $e+q\notin S$, so all points are isolated. – John Sep 26 '14 at 18:56
  • @ArthurFischer: Oh, you are right. It seems like I really can't get this answer right. John: I don't really understand your comment, do you mean $R$ when you write $S$? Also, how do you know that the minimum $\min_{x \in R\setminus{e}} d(e,x)$ actually exists (i.e. is attained)? – PhoemueX Sep 26 '14 at 19:07
  • By R I mean the equivalence relation R(p,q) iff p−q∈Q and by S a set with exactly one point in each equivalence class. EDIT: As there is a rational number between any two reals, while there is exactly one rational in $S$, the infimum must be strictly positive. – John Sep 26 '14 at 20:46
  • Your argument is not valid. For example, there is no rational in $\Bbb{R}\setminus \Bbb{Q}$, but still $\inf_{y \in (\Bbb{R}\setminus \Bbb{Q}) \setminus {x}} |y-x| = 0$. – PhoemueX Sep 26 '14 at 20:54