Measure on algebra generated by subsets $\mathbb{Q}\cap(a,b)$ 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).
 A: 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,
