Let $ \left\{ {r_i } \right\}_{i = 1}^\infty = \mathbb{Q}$ an enumeration of $\mathbb{Q}$. Let $ J_{n,i} = \left( {r_i - \frac{1} {{2^{n + i} }},r_i + \frac{1} {{2^{n + i} }}} \right) $ $ \forall \left( {n,i} \right) \in {\Bbb N}^2 $ Then define $$ A_n = \bigcup\limits_{i \in {\Bbb N}} {J_{n,i} } $$ and $$ A = \bigcap\limits_{n \in {\Bbb N}} {A_n } $$

i) Prove that A has measure zero ii) prove that A cannot be writted as a countable union of sets with null jordan content, where the null jordan content, is the same definition as null measure, only instead of covering with countable intervals, only done with finite, i.e given any $ \varepsilon > 0 $ there exist finite intervals, such that cover the set, and the sum of the lengths, is less than $ \varepsilon > 0 $

Sorry for ask this question, but I want to see an example

  • $\begingroup$ For (i), note that $\lim_{n\rightarrow \infty} m(A_n) = 0$. For (ii), note that $A$ contains $\mathbb{Q}$, so cannot be covered by finitely many intervals. $\endgroup$ Oct 19, 2011 at 17:20
  • $\begingroup$ Is there something wrong in the definition of $A_n$? For every $n\in\mathbb{N}$ and $x\in\mathbb{R}$ there is a rational number in $(x-1/2^{n+1},x+1/2^{n+1})$. Hence $A_n=\mathbb{R}$ for every $n$? $\endgroup$
    – LostInMath
    Oct 19, 2011 at 17:21
  • $\begingroup$ It's really small but it says $n+i$ not $n+1$ in the definition. $\endgroup$
    – Mr. Fox
    May 24, 2020 at 22:15

1 Answer 1


Hint (for ii): Show that if $E$ is Jordan null, then the (topological) closure of $E$ is Jordan null. From this you can show that each Jordan null set is nowhere dense. Finally, show that $A$ cannot be written as a countable union of nowhere dense sets. (In fact, $A$ is easily large enough for this, since the complement of $A$ is a first category set.)

Additional Hints: Assume $E$ is Jordan null. Let $\varepsilon > 0$ be given. Cover $E$ with open intervals $I_{1}, \; I_{2}, \; ..., \; I_{n}$, where the sum of the lengths of these open intervals is less than $\frac{\varepsilon }{2}$. Take the closure of both sides of $E\subseteq \bigcup\nolimits_{n=1}^{N}I_{n}$, using the fact that the closure operator is monotone with respect to set inclusion and distributes over a finite union. This gives

$$\overline{E}\;\;\subseteq \;\;\overline{\bigcup\nolimits_{n=1}^{N}I_{n}} \;\; = \;\; \bigcup\nolimits_{n=1}^{N}\overline{I_{n}} \;\;= \;\; \left( \bigcup\nolimits_{n=1}^{N}I_{n}\right)\; \cup \;\mathcal{E},$$

where $\mathcal{E}$ is the finite set of endpoints of the intervals $I_{1}, \; I_{2}, \; ..., \; I_{n}$. Now note that $\mathcal{E}$ can be covered by a finite collection of open intervals whose lengths add to less than $\frac{\varepsilon }{2}$.

  • $\begingroup$ Thanks Thomas, with what you said the problem it´s done. But now, sorry for be so stupid, but I'm not, not at all comfortable with this, and still I have a bit scary. And I want to know how can I prove that if E is a jordan null set, it´s closure also will be. <.< $\endgroup$
    – Susuk
    Oct 19, 2011 at 17:42

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