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Measurability and Lebesgue integral of complex functions

  1. Oct 18, 2014 #1
    Hi, friends! I think that it is correct to say that, given a measure space ##X##:
    • if ##f:X\to\mathbb{R}##, ##\tilde{f}:X\to\mathbb{C}## and ##\forall x\in X\quad f(x)=\tilde{f}(x)##, then ##f## is measurable if and only if ##\tilde{f}## is;
    • ##f:X\to\mathbb{C}## is Lebesgue integrable if and only if both ##\text{Re}f:X\to\mathbb{R}## and ##\text{Im}f:X\to\mathbb{R}## are; in that case ##\int_X f(x)d\mu=\int_X \text{Re}f(x)d\mu+i\int_X \text{Im}f(x)d\mu##.
    Am I right? ##\infty## thanks!
     
  2. jcsd
  3. Oct 18, 2014 #2

    mathman

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    Second statement looks OK.

    First statement is confusing - I can't understand the relationship and/or difference between these two functions.
     
  4. Oct 18, 2014 #3
    I mean: the Borel algebra of ##\mathbb{R}## and ##\mathbb{C}## are different, but I have convinced myself of that equivalence because I think that the Borel sets of ##\mathbb{R}## are Borelian as subsets of ##\mathbb{C}## and any Borel set belonging to ##\mathbb{C}## contained in ##\mathbb{R}\subset\mathbb{C}## is a Borel set considered as a subset of ##\mathbb{R}##; therefore I think that our function, which can be considered as a complex or real function defined on the measure space ##X## with measure ##\mu##, is measurable with respect to the Borel algebra of ##\mathbb{R}## (in the sense that the counterimage of any Borel set is ##\mu##-measurable), i.e. as a real function, if and only if it is with respect to the Borel algebra of ##\mathbb{C}##, i.e. as a complex function.
    Thank you so much again!
     
  5. Oct 19, 2014 #4

    mathman

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    Since the domain (X) is the same real line in either case, the issue you are raising is not relevant. X is the measure space. The range of the function does not come into play.
     
  6. Oct 20, 2014 #5
    Thank you so much!
     
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