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Suppose $(a_k)$ is some enumeration of rationals, and consider intervals $J_k=(a_k-\frac{1}{2^{k}}, a_k+\frac{1}{2^k})$. It is trivial to see that

$$ \lambda(\bigcup_{k=1}^{\infty}J_k)\leq 2 $$

where $\lambda$ is the Lebesgue measure on $\mathbb{R}$.

Also it is not very hard to see that for any $\varepsilon$, we can find some enumeration of rationals $(a_k)$ such that

$$ \lambda(\bigcup_{k=1}^{\infty}J_k)\geq 2-\varepsilon $$

My question is in the opposite direction. What is the infimum of the above measure, taken over all possible enumerations of the rationals. Clearly it must be at least $1$, but it is actually equal to $1$?

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For a given enumeration $\{a_n:n\in\mathbb{N}\}=\mathbb{Q}$ we define $J_n=(a_n-2^{-n},a_n+2^{-n})$ and $$ m(a)=\lambda\left(\bigcup_{n\in\mathbb{N}} J_n\right) $$ For a given $\varepsilon>0$ there exist $N\in\mathbb{N}$ such that $\sum_{n\geq N} 2\cdot 2^{-n}<\varepsilon$. Let $a_1=0$. For a given $n<N$ take any $q\in\mathbb{Q}$ such that $(q-2^{-n-1},q+2^{-n-1})\subset J_n$ and set $a_{n+1}=q$. From construction it follows that $$ \bigcup_{1\leq n\leq N}J_n=J_1 $$ This gives us the estimate $$ m(a)\leq\lambda\left( \bigcup_{1\leq n\leq N}J_n\right)+\lambda\left( \bigcup_{n\geq N}J_n\right)\leq\lambda(J_1)+\sum_{n\geq N}\lambda(J_n)=1+\sum_{n\geq N} 2\cdot 2^{-n}<1+\varepsilon $$ Thus for any $\varepsilon>0$ there exist an enumeration $a$ such that $m(a)<1+\varepsilon$. On the other hand, for any enumeration $a$ we have $m(a)\geq \lambda(J_1)=1$. Therefore $$ \inf_a\; m(a)=1 $$

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