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Counterexample for a premeasure on a semiring over Q

  1. Nov 23, 2009 #1
    1. The problem statement, all variables and given/known data
    Let [tex]\alpha(r)=r[/tex] and let [tex]P[/tex] be the family of intervals [tex][a,b)[/tex] in [tex]\mathbb{Q}[/tex]. Define [tex]\mu_{\alpha}([a,b))=\alpha(b)-\alpha(a)[/tex]. Show by example that [tex]\mu_{\alpha}[/tex] is not countably additive.

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

    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 [tex]\mathbb{Q}[/tex]. For example, I tried defining [tex]x_{n+1}=x_n + 1/n^2[/tex] and [tex]E_n=[x_n,x_{n+1})[/tex], where [tex]x_1=0[/tex]. Then [tex]\mu(E_n)=1/n^2[/tex], so [tex]\sum_{i=1}^{\infty}\mu(E_i)=\pi^2/6[/tex], which is not in [tex]\mathbb{Q}[/tex]. But this does not work because [tex]\bigcup_{i=1}^{\infty}E_i=[0,\pi^2/6)\notin P[/tex].

    Basically I keep running into two problems. Either this happens, or I find a sequence whose union is in [tex]P[/tex] but [tex]\mu[/tex] is countably additive. Any hints or suggestions would be greatly appreciated.
  2. jcsd
  3. Nov 23, 2009 #2
    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)[tex]\cap[/tex]Q, but the sum of their measures will be as small as you like.
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