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Assume we are working in the reals under the standard Lebesgue measure.

Must a Borel set B of nonzero measure contain an interval as a subset?

I conjecture yes. Is the following line of reasoning valid? Borel sets are generated by countable intersections, countable unions, and complements of open sets, but all sets of nonzero measure must be a union of uncountably many points. The union may contain an interval or be composed entirely of discrete points (each with some neighborhood containing no other point from the union). Since Borel sets only allow for the mentioned countable operations and B has nonzero measure, the nonzero measure cannot be due to an uncountable union of discrete points, so the nonzero measure must be due to some interval contained in B.

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3 Answers 3

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The set $\mathbb{R} \setminus \mathbb{Q}$ or all irrational numbers is a G$_\delta$-set of full/infinite Lebesgue measure (or is a co-null set if, you prefer) which includes no intervals.

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  • $\begingroup$ can you explain how the irrational $R\setminus Q$ can have a positive finite Lebesgue measure ut do not contain any interval? $\endgroup$
    – user652838
    Commented Nov 10, 2020 at 17:18
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This fat Cantor set is a counterexample: closed, empty interior, yet positive Lebesgue measure.

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  • $\begingroup$ Wow, I think you are right. That's pretty cool. $\endgroup$ Commented Mar 26, 2013 at 1:52
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    $\begingroup$ I wonder what was wrong with my reasoning. Countable unions (and intersections) of countable sets are countable. However, countably many countable unions of countable sets may be uncountable. I am not sure, but I think this may be what was false with my reasoning. $\endgroup$ Commented Mar 26, 2013 at 1:56
  • $\begingroup$ Wait, a countable union of countable set is countable. So a countable union of countable unions of countable sets is countable. Since $\mathbb{N}\times \mathbb{N}\times\mathbb{N}\simeq \mathbb{N}\times\mathbb{N}\simeq\mathbb{N}$. $\endgroup$
    – Julien
    Commented Mar 26, 2013 at 2:01
  • $\begingroup$ To begin with: I don't understand your alternative: "the union may contain an interval or be composed of discrete points". $\endgroup$
    – Julien
    Commented Mar 26, 2013 at 2:12
  • $\begingroup$ You are certainly correct about countable iterations of countable unions being countable. As for my alternative, an uncountable union of points can either contain an interval as a subset, or not. My mistake, which you correctly pointed out, was that the latter case is not equivalent to each point containing a neighborhood such that no other point lies in that neighborhood. In retrospect, my interpretation was quite ridiculous since it ignores the whole concept of a limit point. For example, any neighborhood about a rational must contain another rational. And the rationals are 'just' countable! $\endgroup$ Commented Mar 26, 2013 at 2:29
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Even though this is an old question, and the right answer is already here, I thought I write down something that I would have found very helpful when I first saw this question.
Even though there might not be any intervals actually contained in $B$, there are intervals almost contained in the following sense.

Claim. If the Lebesgue measure of a set $E \subset \mathbb{R}$ is $m(E)>0$, then for every $\epsilon > 0$ there is an interval $I$ such that $m(E \cap (a,b)) > (1-\epsilon) m(a,b)$

Proof. We use the outer characterisation of the measures: $$m(E) = \inf \left\{\sum_{n\geq 1} m(I_n), E \subset \bigcup_{n \geq 1} I_n \right\}$$ where the $I_n$ are intervals, and $m(I_n)$ is the length of the interval.
Given $\epsilon > 0$, there is $(I_n)_{n\geq1}$ disjoint such that $\sum_{n}m(I_n) < (1+\epsilon)m(E) = (1+\epsilon) \sum_n m(E \cap I_n) $, so in particular there must be $N \geq 1$ such that $m(I_N) \leq (1 + \epsilon) m(E \cap I_N)$, from which after rearranging one gets $$(1 - \epsilon)m(I_N) \leq m(I_n) - \epsilon \cdot m(I_n \cap E) < m(E \cap I_n)$$

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