From: jbaez@riesz.mit.edu (John C. Baez)
Subject: Re: Lebesgue integral (was: Couple of questions
Organization: MIT Department of Mathematics, Cambridge, MA
Date: Tue, 15 Sep 92 06:07:16 GMT

rickert@mp.cs.niu.edu (Neil Rickert) writes:
>In article <1992Sep15.020445.26398@unixg.ubc.ca> ramsay@unixg.ubc.ca (Keith Ramsay) writes:

>>I've heard that much (nearly all, perhaps) of what is done with L^2,
>>regarded as Lebesgue square-integrable functions, modulo functions
>>supported on sets of measure zero, is just as nicely done (if not
>>better) by regarding L^2 as the formal completion (relative to the L^2
>>norm) of a convenient dense subset (where the Lebesgue measure theory
>>is not needed). I'm not enough of an analyst to confirm this.

I wouldn't want to say you can do everything better, since someone
will probably come up with something you can't do better, but it is a
very nice approach and one I often use myself.  Just take, say,
Schwartz functions, which we can integrate a la
Riemann, and complete 'em in the L2 norm.  Now say you want to show
something.  Well, what do you want to show?  Say you want to do 
Fourier transforms and prove the Plancherel theorem (that the Fourier
transform is an isometry on L2.)   The easiest way
is to note:

1)  Schwartz functions are in L2
2)  Schwartz functions are dense in L2
3)  The Fourier transform of a Schwartz function is Schwartz
4)  The Fourier transform is an isometry on Schwartz functions

and you conclude by easy abstract nonsense that there is a unique way
of extending the Fourier transform from Schwartz functions to an
isometry on all of L2, thus simultaneously defining the Fourier
transform and proving Plancherel.  

The nice thing is that 1-4 are all really easy and only use Riemann
integration.  Check out Reed and Simon vol. 2 - Fourier Analysis,
Self-adjointness.  They do all this in essentially 3 pages, pages 2-4!  
In those same two pages they prove the Fourier inversion theorem.  

In my book with Segal and Zhou the real power of this approach shows
through when we define L2 of an infinite-dimensional Hilbert space
with Gaussian "measure".  The point is, we all know how to integrate
polynomials in finitely many variables against a Gaussian, so using
that formula we can define their integral in the infinite-dimensional
case and define the L2 by completion.  That's really the best way of
making sense of these "measures" in QFT - realize that they are just
a trace on an algebra of "nice" functions, in this case polynomials.

>This leaves the physicist with a function that he can't be sure he can
>get a Fourier series for, and a solution to a differential equation
>which has a Fourier series but is not a function since it is in some
>mysterious formal completion space.

I don't think so.  I just sketched how you can do Fourier transforms
(Fourier series are easier).  As for "mysterious formal completions",
1) it's not so mysterious, 2) elements of L2 as defined in the
old-fashioned way are "mysterious equivalence classes" of functions
mod null functions - they sure aren't functions - they don't have
values at points - so we're not losing much.

Of course a true analyst should know lots of different approaches.


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