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This has probably been asked before, but I cannot find it. Define $\mu$ on $(\mathbb{R}, B(\mathbb{R})$ by letting $\mu (A)$ be the number of rational numbers in $A$. Show that $\mu$ is a $\sigma$-finite under which each open subintervall of $\mathbb{R}$ has infinite meassure.

How can I construct this sequence of $(A_i)$ covering $X$? Can I construct sets with only one rational number in it, still covering X? Is there such a thing as "closest" rational number, for another rational number... Part2 seems to be true since open intervalls are homeomorphic with $\mathbb{R}$?

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up vote 5 down vote accepted

Let $\{r_n\}$ an enumeration of $\Bbb Q$, and let $A_n:=\{r_n\}\cup\Bbb Q^c$. These ones are sets of measure $1$ and $\bigcup\limits_nA_n=\Bbb R$.

An open interval contains infinitely many rationals.

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