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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 $0<a<b<1$. 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 $a<b$.

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? – Patrick Da Silva Nov 24 '12 at 5:42
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. – user44532 Nov 24 '12 at 14:37
The algebra would just consist of finite unions of intervals. – Lukas Geyer Nov 24 '12 at 15:32
@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. – user44532 Nov 24 '12 at 15:58
Algebras are closed under taking set differences, so for $0<a<b<c<1$ 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 $0<a<b<1$. 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. – Lukas Geyer Nov 24 '12 at 21:58

1 Answer 1

I think you should really just try to consider the measure that gives you length first. You'll get something like $\mu(X \cap (a,b)) = b-a$ and it'll be finitely additive, i.e. the measure of a pair of disjoint intervals is going to be the sum of lengths.

EDIT : The thing below doesn't help you answer your question because the trace of the Lebesgue measure is the zero measure (since $\mathbb Q$ has zero measure). I was a little tired when I added this comment. But there is absolutely no problem in defining a measure over $\mathbb Q$ which gives "length" as a weight to an interval of rational numbers. It just won't coincide with the Lebesgue measure but that's absolutely not a problem.

In a more theoretical setting, since $\mathbb Q \subseteq \mathbb R$, if you assume the construction of the Lebesgue measure, you can obtain a measure on $\mathbb Q$, called the trace of the measure on $\mathbb R$ (at least the word trace is used in French... I'm not sure about English, maybe it's something like "induced measure", I don't know), and since the Lebesgue measure's $\sigma$-algebra is generated by intervals, the trace of that $\sigma$-algebra is precisely the $\sigma$-algebra you wanna work with.

When you have $(X,\mathfrak T, \mu)$ a measure space and $X' \in \mathfrak T$, the trace of the measure space $(X, \mathfrak T, \mu)$ on $X'$ is defined as $(X', \mathfrak T', \mu')$ such that $$ \mathfrak T' = \{ Y \cap X' \, | \, Y \in \mathfrak T \} $$ and $$ \mu' = \left. \mu \right|_{\mathfrak T'}. $$ Hope that helps,

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But if you restrict Lebesgue measure to the intervals of rational numbers, you get the zero measure, since the whole space is countable. – Lukas Geyer Nov 24 '12 at 21:51
@Lukas : Yeah.. so that's probably not the way to go huh! But still, length would be what you want. Maybe just not the measure induced from $\mathbb R$. – Patrick Da Silva Nov 25 '12 at 3:01
@PatrickDaSilva: Yes indeed. I've worked it out now. Thanks. – user44532 Nov 28 '12 at 22:30
@CYC : Glad to hear it! – Patrick Da Silva Nov 29 '12 at 5:03

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