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If f is integrable over E iff |f| is integrable over E.

  1. Jan 17, 2010 #1
    1. The problem statement, all variables and given/known data
    Royden Chapter 4, Problem 10a

    Show that if f is integrable over E then so is |f| and [tex]\left|\int_E f \right| \leq \int |f|[/tex].
    Does the integrability of |f| => the integrability of f?



    2. Relevant equations

    [tex]f^+ = max\{f, 0\}[/tex]
    [tex]f^- = max\{-f, 0\}[/tex]

    [tex]|f| = f^+ + f^-[/tex]

    A function is integrable if [tex]\int f < \infty[/tex]



    3. The attempt at a solution

    I have [tex]\left|\int_E f \right| \leq \int |f|[/tex]

    [tex]\left|\int_E f\right| = \left| \int_E f^+ - \int_E f^- \right| \leq \left| \int_E f^+ \right| + \left| \int_E f^- \right| = \int_E |f|[/tex].

    Next I want to show that if f is integrable over E then so is |f|.

    [tex]\int_E f < \infty [/tex]

    [tex]\int_E f^+ - \int_E f^- < \infty [/tex]

    f+ and f- are finite because the expression a-b is only finite if both a and b are finite.

    [tex]\int_E f^+ < \infty [/tex]
    [tex]\int_E f^- < \infty [/tex]

    Hence:

    [tex]\int_E |f| = \int_E f^+ + \int_E f^- < \infty [/tex].


    Does the integrability of |f| => the integrability of f?


    I would say yes for similar reasons. I don't feel very confident about my answers here. Is the reasoning correct?

    Also part b says that the improper Riemann integral (a limit) may exists for a function when the Legesgue integral fails to exists. and gives (sin x)/x as an example. Is the problem with (sin x)/x that f+ and f- are not finite??
     
  2. jcsd
  3. Jan 17, 2010 #2
    To see that |f| is integrable => f is integrable, think about the inequality you proved.
     
  4. Jan 17, 2010 #3
    The proof of the inequality

    [tex]
    \left|\int_E f \right| \leq \int |f|
    [/tex]

    is correct if you assume that f is integrable.

    And what if both are infinite? Does this contradict the hypothesis that f is integrable?

    By the way, a function is Lebesgue integrable iff is absolutely Lebesgue integrable (that is f is integrable iff |f| also is).

    The reason regarding the existence of improper Riemann integrals is pretty much what you said, but note that the how problem ties with your quoted statement above.
     
  5. Jan 17, 2010 #4
    And that was given, so it's just very simple.

    Yes.
     
  6. Jan 17, 2010 #5
    So you have it.
     
  7. Jan 17, 2010 #6
    But I don't get it. Graphically that is. The integral on R of higher dimensions is analogue to summation of R. Now if f = sinx /x the summation (if x belongs to N) is finite but |sinx/x| is not finite. Doesn't the same apply for the integral? Let x belong to R of higher dimensions. Wouldn't the same happen? Because I have never seen what you said. I have seen and tried to prove what you are asking but never what you said.
     
  8. Jan 17, 2010 #7
    Sorry, you have to more clear. I don't understand your reply.
     
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