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A question on proving countable additivity

  1. Jul 9, 2010 #1
    This question comes from the proof of Lemma 9.3 of Bartle's "The Elements of Integration and Lebesgue Measure" in page 97-98. This proof is shown as the image below.

    Form (9.1) mentioned in the lemma is: [tex](a,b], (-\infty,b], (a,+\infty), (-\infty,+\infty)[/tex].

    My question is: although [tex]I_j[/tex] constructed in P98 is a bit fatter than [tex](a_j,b_j][/tex], I doubt the assertion that the left endpoint a, and in turn the compact interval [a,b], is also covered by [tex]\{I_j\}[/tex], as the proof in the text claimed (I drew a red underline). Is my doubt correct (this means the text is incorrect), or point a can be proved to be covered by [tex]\{I_j\}[/tex] (how)? Thanks!
    PS: the establishment of the converse inequality does not need the coverage of the whole [a,b]. A small shrink, say [tex][a+\epsilon,b][/tex], is sufficient to get the inequality.
  2. jcsd
  3. Jul 9, 2010 #2


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    Definitely seems to be an error in the text. If [tex]a_i=a+\frac{1}{n}[/tex] and [tex]\epsilon=\frac{1}{2}[/tex] and [tex]\epsilon_i=\frac{1}{2^{i+1}}[/tex] the [tex]I_j[/tex] never include [tex]a[/tex]

    It seems true that you can pick to [tex]\epsilon_i[/tex]'s so that you get a covering... we know that the [tex]a_i[/tex] have to get arbitrarily close to [tex]a[/tex], so you can pick one really close to [tex]a[/tex] to add one of your larger values of [tex]\epsilon_i[/tex] to
  4. Jul 9, 2010 #3
    Thank you Office_Shredder!
  5. Jul 10, 2010 #4
    I have not gone through the entire arguement, but the condition 9.2 and the ordering of the a's and b's assumed would require that [tex] a=a_1[/tex] and [tex]b_i=a_{i+1}[/tex].
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