[FOM] Category and Measure

[joeshipman@aol.com]

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-----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.
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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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