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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to solve the following exercise but I always end up missing some vital step along the way, so any help would be much appreciated!

Let $\mathbb{Q}$ be the set of all real rational numbers, and let $I_Q = \{[a, b)_Q : a, b ∈ \mathbb{Q}\}$ where $[a, b)_Q = \{q ∈ \mathbb{Q} : a ≤ q < b\}$.

(a) Prove that $σ(I_Q) = P(\mathbb{Q})$, where $P(\mathbb{Q})$ is the collection of all subsets of $\mathbb{Q}$ and $σ(I_Q)$ is the sigma algebra generated by $I_Q$.

(b) Let $µ$ be counting measure on $P(\mathbb{Q})$, and let $ν = 2µ$. Show that $ν(A) = µ(A)$ for all $A ∈ I_Q$, but $ν \neq µ$ on $σ(I_Q) = P(\mathbb{Q})$. Why doesn’t this contradict the uniqueness of measures Theorem?

share|cite|improve this question
What is $\sigma$? – 1015 Feb 17 '13 at 23:02
Presumably the $\sigma$-algebra generated by the given collection. – Zev Chonoles Feb 17 '13 at 23:02
Precisely! Should have mentioned it. Sorry! – V. Krumov Feb 17 '13 at 23:04

For part (a), since $\mathbb{Q}$ is countable, is is enough to show that each singleton is in $\sigma(I_\mathbb{Q})$.

For part (b), look at the hypothesis that the original measure has to be $\sigma$-finite for the extension to be unique.

share|cite|improve this answer
Could you please elaborate on (b) a little more? – V. Krumov Feb 17 '13 at 23:21
@V. Krumov: clearly, if the uniqueness theorem does not hold, then one of its hypotheses must not be satisfied. In this case, the hypothesis to worry about is that the original (pre-extended) measure has to be $\sigma$ finite. – Carl Mummert Feb 17 '13 at 23:26
I really cannot see how can ν(A)=µ(A) if v=2µ? Clearly it follows for the empty set and the infinite but for all $A∈I_Q$? – V. Krumov Feb 18 '13 at 15:13
Every set in $I_\mathbb{Q}$ is infinite. – Carl Mummert Feb 18 '13 at 18:10

Your Answer


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.