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A set of measure 0

  1. Apr 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Let [itex](X_n)[/itex] be a sequence of measurable subsets of [itex]\mathbb R[/itex] such that

    [tex]\sum_{i=1}^\infty m(X_i) < \infty[/tex]

    Define

    [tex]X = \bigcap_{i=1}^\infty \left( \bigcup_{j=i}^\infty X_j \right)[/tex]

    Prove that m(X) = 0.

    2. Relevant equations
    Theorem. Let [itex](E_n)[/itex] be a sequence of measurable sets such that [itex]E_{n+1} \subseteq E_n[/itex] and [itex]m(E_1) < \infty[/itex]. Then

    [tex]m\left(\bigcap_{i=1}^\infty E_i \right) = \lim_{i \to \infty} m(E_i)[/tex]


    3. The attempt at a solution
    Define [itex]E_i = \bigcup\limits_{j=i}^\infty X_j[/itex]. Then by the aforementioned theorem,

    [tex]m(X) = \lim_{i \to \infty} m(E_i)[/tex]

    My only problem is showing that the limit is in fact 0. I haven't used that [itex]\sum m(X_i) < \infty[/itex]. Any tips?
     
  2. jcsd
  3. Apr 1, 2009 #2
    you can say that m(X) <= m(E_{i}) for each i, and
    m(E_{i}) = lim m(X_{j}) = 0 since the sum was finite.
     
  4. Apr 1, 2009 #3
    I don't understand why [itex]m(E_i) = \lim m(X_j)[/itex]. We have that

    [tex]E_i = \bigcup_{j=i}^\infty X_j[/tex]

    so

    [tex]m(E_i) \le \sum_{j=i}^\infty m(X_j)[/tex]

    I do agree that [itex]\lim m(X_j) = 0[/itex].
     
  5. Apr 1, 2009 #4
    are these intervals strictly nested or can there be a smallest interval?

    [edit] i need clarifying: what exactly is INT(UNION(X_i)) with two indexes i and j?
     
    Last edited: Apr 1, 2009
  6. Apr 1, 2009 #5
    sorry thats only true if the E_{i} where increasing, but

    [tex]lim_{i\rightarrow\infty}\left(\sum^{\infty}_{j=i}m(X_{j})\right) =lim_{i\rightarrow\infty}m(X_{i}) [/tex].

    recall that if an infinite series converges you can make the remainder sum arbitrarily small.
     
  7. Apr 1, 2009 #6
    You're right. That didn't occur to me. Thanks for the tip.
     
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