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Reimann Integration, squares and cubes of functions

  1. Nov 14, 2007 #1
    I took a short break from the rudin-crunching. I'm now doing reimann's integral. Anyhow here's a question I've having trouble with.

    Does f^2 is integrable imply that f is integrable?

    -No, take f=1 on rationals, f=-1 on irrationals on [0,1].

    Does the integrability of f^3 imply that f is integrable?

    I can't find a counterexample. I'm not asking for a proof but a counterexample if there is one and if there isn't just let me know!

    thanks for any help.
     
  2. jcsd
  3. Nov 14, 2007 #2

    mathman

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    Try f(x)=1/x and the integral from 1 to oo (infinity).
     
  4. Nov 14, 2007 #3
    sorry forgot to add that f must be a real bounded function on [a,b].
     
  5. Nov 15, 2007 #4

    mathman

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    The example I gave has f(x) real and bounded. In order not to use it you have to require both a and b to be finite.
     
  6. Nov 15, 2007 #5
    yes! a and b must be finite!
     
  7. Nov 16, 2007 #6

    mathman

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    With these restrictions it appears that f^3 integrable implies f integrable, since real numbers have only one real cube root.
     
  8. Nov 16, 2007 #7
    If [tex]f[/tex] is continous on a compact interval and [tex]g[/tex] is integrabl on that interval then [tex]f\circ g[/tex] is also integrable. That means if [tex]f^2[/tex] is integrable then [tex](f^2)^{1/2} = |f|[/tex] is integrable. And similarly [tex](f^3)^{1/3} =f [/tex] is integrable.
     
  9. Nov 17, 2007 #8
    Thanks for the help fellas.

    That's so straightforward why didn't is seeeee it.
     
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