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Counting measure

  1. Dec 1, 2008 #1
    1. The problem statement, all variables and given/known data
    In the measure space {X,S,u} where u is the counting measure
    S= all subsets of X
    fn(x)=[tex]\chi[/tex]{1,2,,,..n}(x) where [tex]\chi[/tex] is the characteristic (indicator) function.

    Does fn(x) converge
    b.almost uniformly
    c.in measure

    2. Relevant equations

    3. The attempt at a solution
    My guess would be
    a.pointwise.yes, since it goes eventually to 1 , but it's hard to demonstrate this
    b. almost uniformly : yes?
    c. in measure: yes? follows directly from b if the answer to b is yes
  2. jcsd
  3. Dec 1, 2008 #2
    a) Yes. Take [tex]x \in X[/tex]. It shouldn't too difficult to show that eventually [tex]|f_n ( x ) - 1 | = 0[/tex].
    b) See (c)
    c) No. It's easy to see that the only possible limit is [tex]f\equiv1[/tex]. Now, for any [tex]1>\varepsilon>0, n \in \mathbb{N}[/tex], [tex]\{ | f_n - 1 | > \varepsilon \} = \{ n + 1, n + 2, \ldots \}[/tex]. What is the measure of this set? What does this say about almost uniform convergence?
  4. Dec 1, 2008 #3
    I think the only problem is that we have to find the cardinality of (n+1,n+2,n+3...) as n goes to infinity. As n goes to infinity, there is no number larger than n and in fact no n+1,n+2.....can exist?
  5. Dec 1, 2008 #4
    That's not a very precise way to think about it. Think about it...without a doubt [tex]n\to\infty[/tex], but at any "stage" of this limit, [tex]n<\infty[/tex] so the set I wrote about is well-defined, and can in fact be mapped bijectively to the natural numbers. What does that imply about its cardnality?
  6. Dec 1, 2008 #5
    I completely understand the solution now.

    One last question
    How to prove that the metric space (L,d)
    where L=all measurable functions
    d(f,g)=[tex]\int{\frac{|f-g|}{1+|f-g|}[/tex] is complete?

    I really have no idea how to show this since I have to show that the limit of ARBITRARY cauchy sequence is another measurable function. I think it is very difficult.
    Last edited: Dec 1, 2008
  7. Dec 1, 2008 #6


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    Is that the complete question? What domain are your measurable functions defined on?
  8. Dec 1, 2008 #7
    Sorry, my bad (X,S,u) is a FINITE measure space and L is the set of FINITE measurable functions. no information other than that.
    no mention of lebesgue measure or borel set
  9. Dec 1, 2008 #8


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    Science Advisor
    Homework Helper

    That helps. Otherwise you wouldn't have a metric. Now your metric is equivalent to the L_1 metric, isn't it? Sorry, it's been a long time since my Real Analysis classes and I don't have this stuff at the tip of my tongue anymore.
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