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From: matt@oddjob.UChicago.UUCP (Matt Crawford)
Newsgroups: net.math
Subject: Re: f(x) = (if x = p/q then 1/q else 0) integrable ??
Message-ID: <591@oddjob.UChicago.UUCP>
Date: Mon, 4-Feb-85 18:32:57 EST
Article-I.D.: oddjob.591
Posted: Mon Feb  4 18:32:57 1985
Date-Received: Wed, 6-Feb-85 00:16:31 EST
References: <350@decwrl.UUCP> <ucbvax.4467> <589@oddjob.UChicago.UUCP> <watmath.11212>
Reply-To: matt@oddjob.UUCP (Matt Crawford)
Organization: U. Chicago: Astronomy & Astrophysics
Lines: 33
Keywords: oops

In article <watmath.11212> wjafyfe@watmath.UUCP (Andy Fyfe) quotes me:
>In article <589@oddjob.UChicago.UUCP> matt@oddjob.UUCP (Matt Crawford) writes:
>>All you people who say "yes, it is integrable ..." may be deceiving
>>some readers.  The *Riemann* integral exists only if the limit of
>>the finite sums exists as the `mesh size' (largest difference between
>>any two mesh points) goes to zero.  I can always choose a mesh of
>>all rational or all irrational points and get different answers.
>>Therefore, the function is not *Riemann* integrable.  And in the
>>words of Richard Feynman:  "Lebesgue, Schlemesgue!"

and then goes on to give a correct explanation of why the function
*is* Riemann integrable.  I stand by the first portion of my state-
ment above (because all previous respondents had just said that it
was integrable because it was a constant (zero) almost everywhere).
As for the second part, well I guess I fucked up there.  Sure enough,
if you estimate the integral by Simpson's rule with 2^n points the
estimate is (n+2)/2^(n+1) --> 0 as n-->infinity.
>
>Now if f(x) = 1 on the rationals, it wouldn't be Reimann integrable.
>This is because any interval making up the mesh necessarily contains
>both an irrational and a rational, so the upper sum is always 1, while
>the lower sum is always 0.
>
>What do you mean by a mesh of all rational, or irrational points?
>
We aren't using the same definition, but they are equivalent.  I was
applying the definition from my 1st year calculus course which in
essence said that the limit of the sum of f(x_i)*(x_{i+1}-x_i) must
go to the same constant as the max of (x_{i+1}-x_i) goes to zero,
no matter how you choose the x_i's.  ("_" denotes subscript.)
_____________________________________________________
Matt		University	crawford@anl-mcs.arpa
Crawford	of Chicago	ihnp4!oddjob!matt