# Counterexample for a premeasure on a semiring over Q

1. Nov 23, 2009

### variety

1. The problem statement, all variables and given/known data
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.

2. Relevant 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 $$\mu\left(\bigcup_{i=1}^{\infty}E_i\right)=\sum_{i=1}^{\infty}\mu(E_i)$$.

3. 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.

2. Nov 23, 2009

### 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)$$\cap$$Q, but the sum of their measures will be as small as you like.