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Measurability with respect to completion

  1. Nov 20, 2013 #1
    How can one show that a positive function with a Lebesgue integral is measurable with respect to the complete sigma algebra?
     
  2. jcsd
  3. Nov 22, 2013 #2
    Just as always, you must show that [tex] \{x : f(x)< \alpha \} [/tex] is a set in the sigma algebra for any [tex] \alpha \in \mathbb{R}. [/tex]
     
  4. Nov 22, 2013 #3
    Ok, but how does the Lebesgue integral aspect factor in to the argument?
     
  5. Nov 23, 2013 #4
    You need to show the function is measurable with respect to the lebesgue measure. Thus given [tex]\alpha \in \mathbb{R} [/tex] you must show that
    [tex]\{x|f(x)< \alpha \} [/tex]
    is a lebesgue measurable set.
     
  6. Nov 23, 2013 #5
    Also, a couple of comments are in order regarding your initial post.

    1) It doesn't make sense to ask whether a function is measurable with respect to a sigma algebra. What you should be asking is how to show that the function [tex] f [/tex] is measurable with respect to the measure space (which according to your post I can only assume is)
    [tex] (\mathbb{R}, \mathcal{M} ,m) , [/tex]
    that is, the real line together with the sigma algebra of all lebesgue measurable sets, and the lebesgue measure. Doing this is simply a matter of definition which I have given in the previous post.

    2) It doesn't make sense to call a sigma algebra complete (unless this means something specific which I am unaware of). I can only assume the word complete here is referring to the fact that the space which I had previously mentioned is a complete measure space, that is for any subset [tex] E \in \mathcal{M} [/tex] and [tex] A \subseteq E [/tex] we have [tex] m(E)=0 \Rightarrow A \in \mathcal{M}. [/tex]

    Thus your post should have read:

    How does one show a positive function is measurable with respect to the measure space [tex] (\mathbb{R}, \mathcal{M} ,m) [/tex]
     
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