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Two conditions of existence for Lebesgue integral

  1. Oct 16, 2014 #1
    Dear friends, I read in Kolmogorov-Fomin's that the following property of measurable real or complex valued functions ##\varphi,f## defined on measure space ##X##, proven in the text for ##\mu(X)<\infty## only, is also valid if ##X=\bigcup_n X_n## is not of finite measure, but it is the union of a countable sequence of measurable sets of finite measure ##X_n## (which we can suppose such that ##X_1\subset X_2\subset ...##): if ##\varphi## is Lebesgue integrable on ##X## and ##\forall x\in X\quad|f(x)|\leq\varphi(x)## then ##f## is Lebesgue integrable on ##X##.

    Given the http://librarum.org/book/10022/159 [Broken] of Lebesgue integral ##\int_X g(x)d\mu:=\lim_n \int_{X_n}g(x)d\mu## for such a measure space, I know, from the property above for ##X_n## such that ##\mu(X_n)<\infty##, that if ##\int_{X_n}\varphi(x)d\mu## exists then ##\int_{X_n}f(x)d\mu## also exists, but how can we know that if ##\lim_n\int_{X_n}\varphi(x)d\mu## exists then ##\lim_n\int_{X_n}f(x)d\mu## exists?
    ##\infty## thanks!
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Oct 17, 2014 #2

    mathman

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    I haven't worked out the details, but I suspect you should work with the integral of φ over Xm-Xn for m>n. This -> 0 as m and n become infinite. Since |f| ≤ φ, the integral of f over the same domain will also -> 0 under the same condition.
     
  4. Oct 17, 2014 #3
    Thank you so much!!!
     
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