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Refer to this question, what would be such an $F$ if $E=\mathbb Q$ ? I mean, $\mathbb Q$ has measure null and thus the measure is finite. Therefore, for all $\varepsilon>0$, there exist $[a_1,b_1],...,[a_N,b_N]$ s.t. $$m\left(E\Delta \bigcup_{i=1}^N [a_i,b_i]\right)<\varepsilon.$$ But I can't find them if $E=\mathbb Q$. What would be such cubes $[a_i,b_i]$ if $E=\mathbb Q$ ?

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$$ \mu(A\Delta \Bbb Q) = \mu(A\setminus \Bbb Q) + \mu(\Bbb Q\setminus A) = \mu(A) + 0. $$

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  • $\begingroup$ why $\mu(A\backslash Q)=\mu(A)$ ? If $A$ is an interval that contain $\mathbb Q$, then $A=\mathbb R$... so $\mu(A\backslash \mathbb Q)=\mu(A)$ look strange to me... And if $A$ is a finit interval, how can $A$ approximate $\mathbb Q$ ? $\endgroup$ – idm Nov 3 '15 at 16:07
  • $\begingroup$ @idm $A\setminus \Bbb Q$ are all irrational numbers in $A$, to take set differences we do not have to have set inclusion $\endgroup$ – Ulysses Nov 4 '15 at 7:30

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