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If $f^2$ is Riemann Integrable is $f$ always Riemann Integrable?

Example of a function f such that $f^2$ is Riemann-Stieltjes integrable on [a,b] but f is not.

I was considering f=$(\chi_{\mathbb{Q}}-\chi_{\mathbb{I}})$ the difference of the characteristic functions for the rationals and irrationals respectively. I know that this is not Riemann Stieltjes integrable but I'm wondering if $f^2$ is.

$f^2$=$(\chi_{\mathbb{Q}}-\chi_{\mathbb{I}})^2$=$\chi_{\mathbb{Q}}^2 +\chi_{\mathbb{I}}^2 - 2\chi_{\mathbb{I}}\chi_{\mathbb{Q}}$ = $\chi_{\mathbb{Q}} + \chi_{\mathbb{I}}$ $\equiv 1$ on [a,b].

Is a correct way of viewing this?

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marked as duplicate by Hans Lundmark, martini, mt_, Willie Wong Nov 28 '12 at 13:10

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1 Answer 1

Yes, that's perfect. It's the classical example for this.

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Thanks for your reply. I found it interesting that it doesn't depend on the $\alpha(t)$ that we integrate over. –  jimmywho Nov 28 '12 at 5:15

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