TITLE: Can distribution theory be developed Riemann-free?
QUESTION [25 upvotes]: I imagine most people who frequent MO have been indoctrinated into the point of view that the Riemann integral can be safely discarded once one has taken the time to develop the Lebesgue integral.  After all the two integrals agree more or less whenever they are both defined, and the Lebesgue theory is well known to be more robust and flexible in a lot of important ways.
However, I have recently encountered an apparent counter-example to the extreme view (which perhaps nobody actually holds) that the Riemann integral is entirely dispensable as a technical tool.  The context is the theory of distributions.  It is not uncommon that when one wants to generalize an operation from test functions to distributions that there are two natural choices: the operation can  either be defined "directly" or by specifying how it pairs with test functions.  Here are two basic examples:

The first example involves the convolution of a distribution $F$ with a test function $\psi$.  The direct definition is given by $F \ast \psi(x) = \langle F, \psi_x \rangle$ where $\psi_x(y) = \psi(x-y)$.  The definition by pairing stipulates that for any test function $\phi$, $\langle F \ast \psi, \phi \rangle = \langle F, \phi \ast \psi_0 \rangle$.
The second example involves the Fourier transform of a (tempered) distribution $F$.  The direct definition is given by $\hat{F}(\xi) = \langle F, e_\xi \rangle$ where $e_\xi(x) = e^{2 \pi i \xi x}$.  The definition by pairing just sets $\langle \hat{F}, \psi \rangle = \langle F, \hat{\psi} \rangle$ for any appropriate test function $\psi$.

In both of these examples, and others like them, all of the authors that I have consulted (including Folland and Taylor) prove that the direct definition agrees with the definition by pairing by carrying out a calculation with Riemann sums.
So I am left wondering if there decent proofs of these results for ordinary Lebesgue-abiding citizens.  This question is a little problematic since the Lebesgue integral and the Riemann integral agree on the relevant space of functions, but if there isn't a good affirmative answer then it seems to me that there should be a convincing explanation why measure theoretic tools aren't strong enough to make the argument work.

REPLY [2 votes]: In the fifties and sixties an elementary treatment of distributions which goes all the way up to the Fourier transform for tempered distributions and convolutions was developed by several mathematicians, in particular, J. Sebastião e Silva.  Elementary means using only the tools of a univariate freshman analysis course (in the one dimensional case), and of a second term course for the multivariate case.  Neither the Lebesgue nor the Riemann integral is required  and there is no use of  functional analysis (in particular, duality theory for non-normed locally convex spaces).  The only facts about integration used are those concerning the integral of continuous functions on a compact interval and its simple properties.  This can be developed (as in Dieudonné's texts) by first integrating step functions with intervals as sets of constancy and then extending to uniform limits thereof  (of course, here you need the fact that continuous functions on compact intervals are uniformly continuous but it is hard to see how one can get any useful theory of analysis without this).
The starting point is to construct, for a compact interval $J$, a superspace $C^{-\infty}(J)$ of $C(J)$ and a linear operator $D$ thereon with the following properties:
1)  if $x$ is continuously differentiable, then $Dx$ is the classical derivative;
2) for each distribution $x$, there exist an integer $n$ and a continuous function $X$ so that $x=D^n X$;
3) if $x$ is a distribution such that $D^n x= 0$, then $x$ is a polynomial of degree less then $n$.
Of course, it follows from the Schwartz theory that such objects exist but the point here is that one can construct a model using only the tools mentioned above.  Furthermore, this system is categorical in the sense that there is only one model.  This is important philosophically since it means that one can deduce from it anything that can be deduced from the Schwartz approach.
In order to construct such a model, one only requires the tools mentioned above (see below).  It is then fairly standard to extend to the case of multivariate distributions and distributions on open subsets (rather than compact intervals).    This has been carried out in detail by Sebastião e Silva, who went on to develop a theory of definite integrals, integrals of distributions depending on a parameter and hence the Fourier transform, and convolutions for 
distributions, all at this elementary level.
Let me close with a brief description of his construction of $C^{-\infty}([0,1])$.
Let $I$ denote the operation which associates to each continuous function that primitive which vanishes at zero.  Then one considers, for each $n$, the space of $n+1$-tuples of continuous functions $(x_0,....,x_n)$ and takes its quotient with respect to those families for which $I^n x_n + I^{n-1} x_{n-1} +... +x_0=0$.  The required space of distributions is then the union of those over $n$.  The differentation operator is the right shift.
This is, of course, a very cursory description but a precise, elementary and easily readable treatment with many examples can be found in the texts of Sebastião e Silva and these  are readily available online (at jss100.campus.ciencias.ulisboa.pt ,  go to "Publicações", then to "Textos didacticos" vol. III, Theory of Distributions).
Apologies: I can't get the special portuguese symbols to work but the texts are in english by the way.