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The question is

Let $\{I_1,I_2,...,I_n\}$ be a finite family of intervals in $\Bbb{R}$ such that $\Bbb{Q} \bigcap [0,1]\subset \bigcup_{j=1}^NI_j$, show $\sum_{j=1}^N|I_j|\ge1$ where $|I_j|$ is the Lebesgue measure of $I_j$. Is this true when the faimily of intervals is infinite?

I have been thinking about this problem for an hour and could not get a clear clue. My guess is that if $\sum_{j=1}^N|I_j|<1$, then $\bigcup_{j=1}^NI_j\bigcap[0,1]$ must not be a cover of $[0,1]$ and hence some rationals are left out, but how do I show this formally? Can anyone provide a hint or reference? Thank you!

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1 Answer 1

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In the expression $\Bbb{Q} \bigcap [0,1]\subset \bigcup_{j=1}^NI_j$ take closure on both the sides and find the measure of both the sides. Note that taking closure and taking measure are monotone.

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  • $\begingroup$ Thank you. That's a brilliant proof. Now $\overline {\Bbb{Q} \cap [0,1]} = [0,1] \subset \overline {\bigcup\nolimits_{j = 1}^N {{I_j}} } = \bigcup\nolimits_{j = 1}^N {\overline {{I_j}} } $, thus $1 \le \sum\limits_{j = 1}^N {\left| {\overline {{I_j}} } \right|} = \sum\limits_{j = 1}^N {\left| {{I_j}} \right|} $ $\endgroup$
    – Tony
    Jul 9, 2015 at 4:45
  • $\begingroup$ I still have a question. In the infinite case $\bigcup\nolimits_{j = 1}^\infty {\overline {{I_j}} } \ne \overline {\bigcup\nolimits_{j = 1}^\infty {{I_j}} } $ but $\bigcup\nolimits_{j = 1}^\infty {\overline {{I_j}} } \subseteq \overline {\bigcup\nolimits_{j = 1}^\infty {{I_j}} } $. Is it because of this that $1 \le \sum\limits_{j = 1}^\infty {\left| {{I_j}} \right|} $ does not hold in the infinite case? $\endgroup$
    – Tony
    Jul 9, 2015 at 4:49
  • $\begingroup$ We have countable sub additivity no? please ask your this doubt little bit more precisely. I cant get. $\endgroup$
    – GA316
    Jul 9, 2015 at 5:07
  • $\begingroup$ Thank you for the reply. In the finite case after taking closure on both sides, we have $[0,1] \in \overline {\bigcup\nolimits_{j = 1}^N {{I_j}} }=\bigcup\nolimits_{j = 1}^N {\overline {{I_j}} }$, then we can take the measure and use the subadditivity. However, in the infinite case, after taking the closure on both sides, we only have $[0,1] \in \overline {\bigcup\nolimits_{j = 1}^\infty {{I_j}} } \ne \bigcup\nolimits_{j = 1}^\infty {\overline {{I_j}} }$, then we cannot use the countable sub-additivity here? $\endgroup$
    – Tony
    Jul 9, 2015 at 5:24
  • $\begingroup$ are you thinking about proving the same result for infinite family of intervals? for infinite family this result is not true. $\endgroup$
    – GA316
    Jul 9, 2015 at 6:48

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