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Consider the rational intervals defined as $$[a,b)_Q= \{r : a \leq r < b; r \in \mathbb{Q} \},a,b\in \mathbb{Q}, a<b. $$ Let $A$ be the class of all sets of rationals that can be produced as finite unions of rational intervals. Then $A$ is a ring not containing any sets of the form $\{a\}$. The $\sigma$-additive set function $\mu: A \to R$ is defined on the rational intervals the usual way: $\mu([a,b)_Q)=b-a$. By this $\mu$ is defined on $A$.

For every rational number $a$ the set $\{a\}$ clearly belongs to the $\sigma$-ring generated by $A$. Indeed, $\{a\} = \bigcap_{n\in\mathbb{N}} [a,a+1/n)_Q$. As a result, one has on the one hand $\mu(\{r\}) = 0$ for all rationals and, on the other hand $\mu((a,b]_Q) = \sum_{r\in[a,b)_Q}\mu(\{r\}) = \sum_{r\in[a,b)_Q} 0 = 0$ and not $b-a$ as it ought to be by the definition.

Where is the mistake?

zoli
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  • What does $rac$ mean in the definition of $[a,b)$? – Ben Millwood Dec 29 '14 at 00:45
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    You have proved that $\mu$ is not a (countably additive) measure. – Manny Reyes Dec 29 '14 at 00:51
  • But on A /mu is an additive set function and A is an algebra. – zoli Dec 29 '14 at 00:59
  • It may be finitely additive, it's just not $\sigma$-additive according to your argument – Manny Reyes Dec 29 '14 at 01:02
  • But from A an additive set function ought to be extended to the sigma algebra generated by A. – zoli Dec 29 '14 at 01:04
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    Sorry, I missed the subtlety about the sigma-algebra. It looks like you have an example of a finitely additive function on an algebra with a finitely additive extension to the $\sigma$-algebra generated thereby (restriction of Lebesgue), which is not countably additive. – Manny Reyes Dec 29 '14 at 01:23
  • I am sorry I just walked by this post. I agree that $\mu$ so defined is not $\sigma$-additive, but why is it even finitely additive? If $[a,b)\cap\mathbb{Q}$ and $[c,d)\cap\mathbb{Q}$ are disjoint rational intervals, what is the value of $\mu([a,b)\cap\mathbb{Q}\cup [c,d)\cap\mathbb{Q})$? This disjoint union is not another rational interval, right? – JacobsonRadical Apr 20 '23 at 01:30

1 Answers1

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$\mu$ is not sigma additive on $A$: Let $\{r_n : n \geq 1\}$ list all rationals in $[0, 1)$. Let $I_n$ be an interval such that $r_n \in I_n$, $I_n \subseteq [0, 1)$ and $\mu(I_n) < 1/3^n$. Let $W_n = I_n \backslash (\bigcup_{m < n} I_m)$. Then $\{W_n : n \geq 1\} \subseteq A$ is a partition of $[0, 1)$ and $\sum_{n \geq 1} \mu(W_n) \leq 1/2$.

  • I don't see why would $W_n$'s be finite sums of rational intervals. – zoli Dec 29 '14 at 23:35
  • I thought it was clear to you that $A$ is a ring. In any case, this follows from the fact that the difference of two left closed right open intervals is a union of at most two such intervals. –  Dec 30 '14 at 17:35
  • Yes, it is clear. But this argument then proves that the Lebesgue measure is not sigma-finite. (Your $I_n$'s are not necessarily "rational intervals"; the argumentation remains the same.) – zoli Dec 30 '14 at 22:09
  • No. The argument uses the fact that the rationals are countable. –  Dec 30 '14 at 22:12
  • Take the list of the $r_n$'s. Let for each $n$ $I_n=[a_n\le r_n \lt b_n)$ be a real interval such that $b_n-a_n \lt 1/3$. (Don't let $b_n$ be greater than 1.) Create your partition by the $W_n$'s. Cover [0,1) by these disjoint sets. Don't we get the same contradiction? – zoli Dec 31 '14 at 00:40
  • We have of course $1/3^n$ above (and not 1/3). Sorry! – zoli Dec 31 '14 at 09:40
  • Your $I_n$'s do not cover $[0, 1)$. –  Dec 31 '14 at 15:25