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Suppose that $E \subseteq [0, 1]$ has the property that for each $x \in E$, there is $r_x > 0$ such that $B(x, r_x) \cap E = (x-r_x, x+r_x) \cap E$ has Lebesgue measure zero. Is it necessarily the case that $E$ itself has Lebesgue measure zero?

A couple remarks:

1) If the points of $E$ were legitimately isolated from one another, $E$ would necessarily be countable, and hence have measure zero.

2) Because I am only interested in the question of "measure zero or not," measurability does not appear explicitly in the statement of the question above. And indeed, to answer the question in the negative, one does not need to produce a set of positive measure; producing a Lebesgue non-measurable set $E$ with the given property would just as well constitute a negative answer.

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    $\begingroup$ See also: Lebesgue density theorem. $\endgroup$ – GEdgar Jun 15 at 23:50
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$E$ is Lindelöf, as a separable (second countable) metric space.

So we have a cover of $E$ by these $(x-r_x,x+r_x)$ and so a countable subcover $\{(x-r_x,x+r_x)\cap E: x \in N \}$ exists for some countable subset $N$ of $E$.

But then (as we have a cover of $E$): $$\lambda(E) \le \sum_{x \in N} \lambda((x-r_x,x+r_x) \cap E) = 0$$

by countable subadditivity of the Lebesgue measure $\lambda$. So $\lambda(E)=0$ indeed.

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Yes, it is true. Let $U=\cup_{e\in E}B(e,r_e)$. As $U$ is an open subset of $[0,1]$, it is a countable disjoint union of intervals, say $U=\cup_{n\in\mathbb N}I_n$ with $I_n, I_m$ disjoint for $n\not=m$. Then one has $$ E=\bigcup_{n\in\mathbb N}E\cap I_n $$ where each set $E\cap I_n$ is null, and hence so is $E$.

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  • $\begingroup$ If you want to disjoint intervals you have to make sure that they are contained in one of the original intervals. If you drop the word disjoint your proof is correct. In any second countable space any union of open sets is the union of a countable subfamily. $\endgroup$ – Kavi Rama Murthy Jun 15 at 23:21
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    $\begingroup$ Thanks for your response! But I'm not sure I can see why each $E \cap I_n$ is necessarily null. $\endgroup$ – User190212 Jun 15 at 23:21
  • $\begingroup$ @KaviRamaMurthy in fact the intervals $I_n$ are the connected components of $U$ and there are countably many since each has positive measure. User190212 one can see this by observing that $I_n$ can be written as a countable union of sets $B(e,r_e)$ by construction. $\endgroup$ – pre-kidney Jun 15 at 23:24
  • $\begingroup$ I know this argument but I am just saying you are making things unnecessarily complicated. Why do you need disjoint intervals? $\endgroup$ – Kavi Rama Murthy Jun 15 at 23:27
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    $\begingroup$ One way to see it is by taking a countable compact exhaustion of $I_n$. Each compact set is covered by finitely many elements of the open cover $B(e,r_e)$ of $I_n$. $\endgroup$ – pre-kidney Jun 15 at 23:43

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