[FOM] Category and Measure
joeshipman@aol.com
joeshipman at aol.com
Tue Sep 18 16:17:49 EDT 2007
****
-----Original Message-----
From: Timothy Y. Chow <tchow at alum.mit.edu>
Joe Shipman wrote:
> But, can someone explain what's so useful about meager sets when
> working with a measure space like the real numbers?
The pointwise limit of a sequence of continuous functions is continuous
except on a set of first category. So for example if f is everywhere
differentiable, then its derivative f' is continuous except on a set of
first category. I don't think there is a way to say the "same thing"
using measure.
****
That is a very interesting duality, since if f is Riemann integrable
(not Lebesgue integrable because that concept presupposes measure) then
it is continuous almost everywhere in the sense of measure not category.
Putting these together, this seems to say that there is a function f
that is everywhere differentiable, whose derivative f' is NOT a Riemann
integrable function (because if f' were always Riemann integrable, then
you COULD replace category by measure in your statement above).
Such an example would make the best possible case for using the
Lebesgue integral instead of the Riemann integral (because it already
shows that the Riemann integral fails to be the inverse of
differentiability for a *single* function, while the familiar arguments
for using Lebesgue rather than Riemann involve its superior properties
for *sequences* of functions).
Can anyone point to an explicit example of an everywhere differentiable
function such that the meager set of points where the derivative fails
to be continuous is not measure 0 and therefore not Riemann integrable?
(Preferably a function whose derivative is bounded, since we want the
Riemann integrability to fail purely because of the measure-category
issue.)
-- JS
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