Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mathbb{R}$?

share|improve this question

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.