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Let $X=\mathbb{Q}\cap[0,1]$. Let $\mathcal{A}$ be the algebra generated by the collection of all sets of the form $(a,b)\cap X$ where $0. I would like to find out whether there is a finitely additive measure $\mu:\mathcal{A}\rightarrow[0,\infty]$ such that $\mu(X\cap(a,b])=b-a$ for all $a,b\in(0,1)$ with $a.

Remark: Previously I was considering the $\sigma$-algebra generated by this collection of sets and whether there is a measure having those same properties but I soon found out that there isn't (basically because $X$ is countable).

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    What's the issue with just taking length?2012-11-24
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    I'm unsure about what to do because I was thinking that if I am to check the conditions for being a finitely additive measure I will need to know how the sets in the algebra look like.2012-11-24
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    The algebra would just consist of finite unions of intervals.2012-11-24
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    @LukasGeyer: Would that be intervals of every type, i.e. open, closed, half-open? I know that if the generating set consists of half-open intervals, then the algebra generated will consist of all finite unions of half-open intervals, but I'm not sure what happens when the generating set consists of open intervals.2012-11-24
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    Algebras are closed under taking set differences, so for $0 you get that $(a,b] = (a,c) \setminus (b,c) \in \mathcal{A}$, and similarly you get $[a,b), \, [a,b] \in \mathcal{A}$ for all $0. There is a little issue with not allowing $a=0$ and $b=1$. As a consequence, intervals with endpoints $0$ or $1$ are not in $X$, but unions of intervals of this form like $[0,a] \cup [b,1]$ will be. In any case, length measure works or all of them, if you only require finite additivity.2012-11-24

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