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Re: f is constant?
Posted:
Feb 28, 2005 7:27 PM
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In article <d00b3u$6rt$2@mailhub227.itcs.purdue.edu>,
Dave Seaman <dseaman@no.such.host> wrote:
>The fundamental theorem does not fail. In fact, the Lebesgue version of
>the FTOC is stronger than the Riemann version.
There are several Lebesgue versions. One is: if f is differentiable
everywhere on [a,b] and f' is in L^1[a,b], then
f(b) - f(a) = int_a^b f'(x) dx
Another is
If f is absolutely continuous on [a,b] then f' exists almost
everywhere on [a,b] and
f(b) - f(a) = int_a^b f'(x) dx
Neither of those versions applies where f is the Cantor function, since
the Cantor function is not differentiable everywhere and is not absolutely
continuous on [0,1].
>It's also a theorem that a function is Riemann integrable iff it is
...bounded and...
>continuous almost everywhere, which would seem to indicate that the
>Cantor function is Riemann integrable.
The Cantor function is continuous, so of course it's Riemann integrable.
But that's irrelevant, since we're talking about integrating f', not f.
The derivative of the Cantor function is not everywhere defined, and
therefore not Riemann integrable.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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