Reimann Integration, squares and cubes of functions

[SiddharthM]

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.

 

[mathman]

Try f(x)=1/x and the integral from 1 to oo (infinity).

 

[SiddharthM]

sorry forgot to add that f must be a real bounded function on [a,b].

 

[mathman]

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.

 

[SiddharthM]

yes! a and b must be finite!

 

[mathman]

With these restrictions it appears that f^3 integrable implies f integrable, since real numbers have only one real cube root.

 

[Kummer]

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.

 

[SiddharthM]

Thanks for the help fellas.

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