Re: I don't understand this
From: David C. Ullrich (ullrich_at_math.okstate.edu)
Date: 07/15/04
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Date: Thu, 15 Jul 2004 09:46:27 -0500
On Thu, 15 Jul 2004 14:02:17 +0000 (UTC), alex_s@yahoo.com (Alex)
wrote:
>Suppose f is a bounded real function on [a,b], and f^2 is an integrable
>on [a,b]. I know that it doesn't follow that f is integrable on [a,b],
>since we can show this by counterexample. But I don't see why we can't
>prove that it follows using the theorem 6.11 from Rudin's book, the
>theorem which we use to prove that from the condition that f^3 is an
>integrable f is also integrable.
>
>The theorem: Suppose f is integrable on [a,b], m<=f<=M, g is continuous
>on [m,M] and h=g(f(x)) on [a,b]. Then h is an integrable on [a,b].
Presumably to show that f^3 integrable implies f integrable we
use the theorem with h(x) = cuberoot(x), because then f = h(f^3).
But there _is_ no h such that f = h(f^2) in general, because...
************************
David C. Ullrich
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