Counterexample for a premeasure on a semiring over Q

[variety]

Homework Statement

Let \alpha(r)=r and let P be the family of intervals [a,b) in \mathbb{Q}. Define \mu_{\alpha}([a,b))=\alpha(b)-\alpha(a). Show by example that \mu_{\alpha} is not countably additive.

Homework Equations

\mu is countably additive if for any sequence of mutually disjoint subsets E_1, E_2, ... such that \bigcup_{i=1}^{\infty}E_i\in P, we have \mu\left(\bigcup_{i=1}^{\infty}E_i\right)=\sum_{i=1}^{\infty}\mu(E_i).

The Attempt at a Solution

This problem is driving me insane. I know I have to somehow use the incompleteness of the rationals, so I tried to use sequences which I know do not converge in \mathbb{Q}. For example, I tried defining x_{n+1}=x_n + 1/n^2 and E_n=[x_n,x_{n+1}), where x_1=0. Then \mu(E_n)=1/n^2, so \sum_{i=1}^{\infty}\mu(E_i)=\pi^2/6, which is not in \mathbb{Q}. But this does not work because \bigcup_{i=1}^{\infty}E_i=[0,\pi^2/6)\notin P.

Basically I keep running into two problems. Either this happens, or I find a sequence whose union is in P but \mu is countably additive. Any hints or suggestions would be greatly appreciated.

 

[grief]

The hint is that you need to use not only the incompleteness of the rationals, but also the fact that they have measure zero. That is, for every epsilon>0, it is possible to cover Q by countably many intervals s.t the sum of their lengths is less than epsilon. That's because Q is countable. Thus, these intervals will have union [0,1)\capQ, but the sum of their measures will be as small as you like.