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From: wjafyfe@watmath.UUCP (Andy Fyfe)
Newsgroups: net.math
Subject: Re: f(x) = (if x = p/q then 1/q else 0) integrable ??
Message-ID: <11212@watmath.UUCP>
Date: Wed, 30-Jan-85 22:54:50 EST
Article-I.D.: watmath.11212
Posted: Wed Jan 30 22:54:50 1985
Date-Received: Thu, 31-Jan-85 02:47:28 EST
References: <350@decwrl.UUCP> <ucbvax.4467> <589@oddjob.UChicago.UUCP>
Reply-To: wjafyfe@watmath.UUCP (Andy Fyfe)
Organization: U of Waterloo, Ontario
Lines: 34
Summary: 

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!"
>_____________________________________________________
>Matt		University	crawford@anl-mcs.arpa
>Crawford	of Chicago	ihnp4!oddjob!matt

       { 1/q, x=p/q
f(x) = {		, f(0) = 1 (though it doesn't matter)
       { 0, otherwise

Now f(x) is in fact Riemann integrable over [0,1].  The reason f(x)
if Riemann integrable over [0,1] is the following:  Choose some n>0.
Then only a finite number of x make f(x) > 1/n.  I can make a mesh around
those points arbitrarily narrow, so that their upper sum is made less
than epsilon.  And the total of the remainder of the upper sums is less
than (1/n)*(1-0), so the upper sum is less than (epsilon + 1/n).  The lower
sum is always zero, and thus, in the limit (as n goes to infinity), we get
the Riemann integral to be 0.

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?

--andy fyfe		...!{decvax, allegra, ihnp4, et. al}!watmath!wjafyfe
			wjafyfe@waterloo.csnet