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Homework Help: Measure theory

  1. Aug 15, 2007 #1
    1. The problem statement, all variables and given/known data
    Consider a measure f mapping from a family of sets A to [0,infinity]

    Let the measure be finitely additive and countable subadditive.

    Prove that f is countably additive on A.


    3. The attempt at a solution
    To show equality from an inequality we do ie.

    a<=b>=a so a=b

    I tried this strategy with measures and sets but couldn't see it through. I might need to construct another set but don't see what to construct.
     
  2. jcsd
  3. Aug 15, 2007 #2

    matt grime

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    Isn't this just the fact that a countable sum is the limit of its finite subsums (by definition)?

    Or, in the thing you're doing with a<=b<=a (which is what you should have written, rather than writing b>=a twice), a is the infinite sum of the measures, and b is the measure of the infinite union of the sets. What you need to do is show b=>a-e for any e, which is just saying something about sums being limits of partial subsums.
     
    Last edited: Aug 15, 2007
  4. Aug 17, 2007 #3
    I prefer to show reverse inequality.
     
  5. Aug 17, 2007 #4

    matt grime

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    This would show the reverse inequality, if by that you mean it would show that a<=b<=a: one is clear b<=a, and the other would show that b>a-e for any e. thus b=>a.
     
  6. Aug 17, 2007 #5
    subtractions are avoided when dealing with measures because the element of infinity is in the set which cannot be subtracted. So equivalently show b+e>a for any e? To show b>=a
     
  7. Aug 17, 2007 #6

    matt grime

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    Uh, what on earth does that mean? Just treat the case of a collection of sets having infinite measure separately if you must, but it is not important - the sequence either converges or diverges.

    Look, I have a sequence of measurable (i.e. of finite measure) sets

    S_1, S_2, S_3,...

    And I know that m(S_1 u S_2 u S_3 .....) => m(S_1 u S_2 ... u S_n)

    for any n. Now do you understand why this is just a statement about the sum of a series? I am *not* subtracting any measures, the -e is purely from what it means for an infinite sum to be the limit of its partial subsums.


    OK, let's put it this way: if any of the S_i has infinite measure, so does the union and there is nothing to prove. So suppose each S_i has finite measure.

    Let X_n be the union of S_1,..,S_n, and let X be the union of all the S_i

    Then m(X_n)=m(S_1)+..+m(S_n) by hypothesis, and

    What do we know? m(X) is less than or equal to the limit of m(X_n) by assumption, and m(x) is greater than or equal to m(X_n) for all n - it's just a question about sequences:

    suppose that a_n tends to a and that b satisfies a<=b and b=>a_n for all n, then b=a (note this even includes the case when a is infinity.
     
    Last edited: Aug 17, 2007
  8. Aug 17, 2007 #7
    I am unsure about the statement 'b=>a_n for all n'. Do you mean b>a_n for any finite n?

    I am a bit rusty with proof of infinite sequences and series.
     
  9. Aug 18, 2007 #8
    I think I see where you are getting at. You have laid out most of the solution in post 6 and recquires one more step or so to complete it. All I have to prove is that m(x) is greater than or equal to m(X_n) for all n.
     
  10. Aug 18, 2007 #9
    In fact to prove 'm(X) is greater than or equal to m(X_n) for all n.' All one needs is the monatonicity of measure since X_n is a subset of X for all n so m(X_n)<=m(X) for all n. => lim(n->infinity) m(X_n)<=m(X)

    but m(X)<= lim(n->infinity) m(X_n) as given so m(X)=lim(n->infinity) m(X_n) = infinite series as each partial sum is a finite series. Take n-> infinity and you get an infinite series.

    The assumption in this is that all sets S_i in X are disjoint and rightly so they should for equality. It seems like this is a nice problem to test your knowledge of infinite series and sequences.
     
  11. Aug 18, 2007 #10

    matt grime

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    I meant what I wrote. And how can there be anything other than a 'finite n' since n is an element of the natural numbers?

    That is a bit worrying if you're doing measure theory.
     
  12. Aug 19, 2007 #11
    This statement is correct?

    'since X_n is a subset of X for all n so m(X_n)<=m(X) for all n implies lim(n->infinity) m(X_n)<=m(X)'
     
  13. Aug 20, 2007 #12
    Matt, just an observation. I never recall you make a post telling me I am right. Is not replying your way of saying that there is nothing wrong with the post?
     
  14. Aug 20, 2007 #13

    matt grime

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    What do you want me to say?

    "since X_n is a subset of X for all n so m(X_n)<=m(X) for all n"

    is obviously true by the definition of measure.

    m(X_n) is an increasing sequence of real numbers bounded above be m(X), so of course the limit of m(X_n) is less than m(X), again by the basic definitions of analysis.

    Does that ease your mind?
     
  15. Aug 21, 2007 #14
    If its true then it would be good to just say so. No further explanation is needed. I should be grateful that you're here in the first place.
     
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