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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?

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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
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1 Answer

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.

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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
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Every set in $I_\mathbb{Q}$ is infinite. –  Carl Mummert Feb 18 '13 at 18:10
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