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F is integrable if and only if its positive and negative parts are

  1. Feb 1, 2012 #1

    Fredrik

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    1. The problem statement, all variables and given/known data

    Problem 2.6.3. in "Foundations of modern analysis", by Avner Friedman. Let f be a measurable function. Prove that f is integrable if and only if f+ and f- are integrable, or if and only if |f| is integrable.

    2. Relevant equations

    Friedman defines "integrable" like this: An a.e. real-valued measurable function ##f:X\to\overline{\mathbb R}## is said to be integrable if there's a sequence ##\langle f_n\rangle## of integrable simple functions that's Cauchy in the mean and such that ##f_n\to f## a.e.

    "Cauchy in the mean" means that it's a Cauchy sequence with respect to the L1 norm (which hasn't been defined at this point). In other words, ##\langle f_n\rangle## is Cauchy in the mean if for all ##\varepsilon>0## there's an ##N\in\mathbb Z^+## such that for all ##n,m\in\mathbb Z^+##,
    $$\newcommand{\dmu}{\ \mathrm{d}\mu}
    n,m\geq N\ \Rightarrow\ \int|f_n-f_m|\dmu<\varepsilon.$$
    3. The attempt at a solution

    I haven't looked at the "Cauchyness" yet. I'm still just focusing on the part about convergence a.e. If ##f^+, f^-## are integrable, it's easy to show that f is, because there are sequences ##\langle (f^\pm)_n\rangle## that converge a.e. to ##f^\pm##, and we just need to find another sequence ##\langle f_n\rangle## that converges a.e. to f. All we have to do is to define ##f_n=(f^+)_n-(f^-)_n##, and the result ##f_n\to f## a.e. follows from
    $$|f_n-f|=|(f^+)_n-(f^-)_n-f^+-f^-|\leq|(f^+)_n-f^+|+|(f^-)_n-f^-|.$$ The proof that |f| is integrable is very similar. It's the converses of these results that I'm struggling with. In particular, how do I prove that if f is integrable, then its positive and negative parts are integrable too?

    It actually looks impossible to me. At least if I start with the "obvious" idea. Let ##\langle f_n\rangle## be a sequence of integrable simple functions that's Cauchy in the mean and such that ##f_n\to f## a.e. Then break each ##f_n## into positive and negative parts, and define the sequences ##\langle (f^\pm)_n\rangle## by ##(f^\pm)_n=(f_n)^\pm##. (I'll just write ##f^\pm_n## from now on).

    Consider an x such that f(x)=0. We have ##f^\pm(x)=0##, but I don't see a reason why we can't have something like ##f^\pm_n\to\pm 1##.

    On the other hand, if I assume that both f and |f| are integrable (instead of just f), I can prove that f+ and f- are integrable with a simple triangle inequality argument. :confused:
     
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  3. Feb 1, 2012 #2

    micromass

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    So your problem is: If [itex]f_n\rightarrow f[/itex] a.e. with [itex]f_n[/itex] simple, then also [itex]f_n^+\rightarrow f^+[/itex] a.e.

    This is what you want to prove right??

    But [itex]f^+=f\vee 0[/itex]. So what you actually need to prove is that if [itex]x_n\rightarrow x[/itex], then [itex]x_n\vee 0\rightarrow x\vee 0[/itex]. That doesn't sound too complicated to prove.

    Am I missing something?
     
  4. Feb 1, 2012 #3

    Fredrik

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    Yes, that's what I want to prove, and no I don't think you missed anything. I kept getting confused by inequalities like
    $$|f^+_n(x)-f^+(x)|\leq |f_n(x)-f(x)|+|f^-_n(x)-f^-(x)|.$$ I see now that when x is such that f(x)>0, I can make the second term =0, not just <ε, by choosing ε<f(x) and N such that n≥N implies ##|f_n(x)-f(x)|<\varepsilon##. So it looks like what I missed is that I have to choose ε small enough. Thanks for helping me figure that out.
     
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