Reimann Integration, squares and cubes of functions

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Discussion Overview

The discussion revolves around the integrability of functions, specifically exploring whether the integrability of the squares and cubes of a function implies the integrability of the function itself. The scope includes theoretical aspects of Riemann integration and counterexamples.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions if the integrability of f² implies the integrability of f, providing a counterexample with f defined as 1 on rationals and -1 on irrationals over [0,1].
  • Another participant suggests f(x) = 1/x as a potential counterexample for the integrability of f³, but later clarifies that f must be a real bounded function on [a,b].
  • A participant emphasizes that both a and b must be finite for the discussion to hold.
  • Another participant proposes that under the restrictions of boundedness and finiteness, the integrability of f³ implies the integrability of f, citing the uniqueness of real cube roots.
  • One participant states that if f is continuous on a compact interval and g is integrable on that interval, then f∘g is also integrable, leading to the conclusion that if f² is integrable, then |f| is integrable, and similarly for f³.

Areas of Agreement / Disagreement

Participants generally agree on the need for boundedness and finite limits for the discussion, but there remains uncertainty regarding the implications of integrability between f, f², and f³, with multiple viewpoints presented.

Contextual Notes

Limitations include the dependence on the definitions of integrability and the specific conditions of boundedness and finite intervals, which are not universally applicable.

SiddharthM
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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.
 
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Try f(x)=1/x and the integral from 1 to oo (infinity).
 
sorry forgot to add that f must be a real bounded function on [a,b].
 
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.
 
yes! a and b must be finite!
 
With these restrictions it appears that f^3 integrable implies f integrable, since real numbers have only one real cube root.
 
If [tex]f[/tex] is continuous 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.
 
Thanks for the help fellas.

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

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