# Can distribution theory be developed Riemann-free?

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.

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I guess it is true that, even if you begin with the Lebesgue integral, you can then prove (using uniform continuity) that when a function is continuous on $[a,b]$ then Riemann sums converge to the Lebesgue integral. And you might want to do this because it is sometimes useful, as in your examples. But of course this is far short of developing the Riemann integral from scratch. And you can think of this convergence result as just one more theorem about the Lebesgue integral. – Gerald Edgar Aug 9 '11 at 12:50
Isn't the crucial point that one has a/the fundamental theorem of calculus, that is, that suitably_understood, $\frac{d}{dx}\int^x_{x_o} f(t)\,dt=f(x)$, for nice functions $f$, etc.? It's not really about a construction of an integral, but this property relating differentiation and "integration". – paul garrett Aug 9 '11 at 15:59
@Gerald Edgar: It's true that one doesn't need much Riemann integration theory to make the sort of argument I'm asking about work, but then again once you've proved that Riemann sums converge to anything at all you've done most of the work needed to build the theory. Aside from that, I find it strange that apparently the rest of analysis can be developed without using Riemann sums to calculate Lebesgue integrals, but here it is not clear how to proceed without them. – Paul Siegel Aug 9 '11 at 18:36
@paul garrett: Actually, I don't think either of the two results discussed in my question make any use of the fundamental theorem of calculus. The proofs come down to showing that two integrals are equal, and Riemann sums are used to rip apart the first integral and reassemble it into the second integral. Due to the topology on the space of test functions it is crucial to the argument that Riemann sums converge uniformly; I'm starting to suspect that this property is close to the heart of the matter. – Paul Siegel Aug 9 '11 at 18:47
I am aware of, but don't know the details about, something called "gauged integrals", which have absolutely nothing to do with what's usually called "gauge theory". Rather, they are close in spirit to Lebesgue integration but allow for things like conditional convergence and indefinite integrals. I don't know if this is what you're looking for. – Theo Johnson-Freyd Aug 9 '11 at 21:22

An instance where the Riemann integral cannot be replaced by the Lebesque integral is the following theorem: A sequence $x_i$ in the interval $[0,1]$ is equidistributed if and only if for each Riemann integrable function $f$ we have $$\lim_{n\to\infty}\frac1n\sum_{i=1}^n f(x_i) = \int_0^1f(x)dx.$$

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Of course, if you want a characterization involving all Riemann integrable functions you have to use the Riemann integral. But continuous functions (or, by Weyl's criterion, exponentials) would be enough. – Jochen Wengenroth Feb 5 '13 at 8:49

The equations of the examples follow by interchanging integrals and duality brackets as follows: $$\langle u_y,\int \psi(y,x)dx\rangle=\int\langle u_y,\psi(y,x)dx\rangle.$$ Since $u$ is a continuous linear functional, this Fubini-type formula follows if the integral converges in the space of test functions; more accurately, if the Riemann sums converge in the topology of the space of test functions in the $y$ variable. To actually prove the convergence is somewhat tedious but straightforward; in a course on distribution theory one has to do this in sufficient detail. In case $u$ is a (locally) integrable function, this argument proves, for the particular case at hand, Fubini's formula via distribution theory.

I do not see an issue of Riemann vs. Lebesgue integration here. For fixed $y$ the integral is just a Riemann and/or Lebesgue integral of a test function.

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On the other hand this kind of equation if exactly what the Pettis-integral is all about. So the underlying theorem here is: Riemann integrable vector valued functions are Pettis integrable (with identical integral values of course). – Johannes Hahn Feb 2 '14 at 16:18

Regarding the first example: there is essentially no way to get around the "Riemann integration". Often when you use it (for example, to characterize monotonic functions) the distributions in question are measures. In this case, you can use Fubini's theorem to interchange the order of integration. Otherwise the statement you're trying to prove is:

$\int \int u(x) \phi(y - x) dx \psi(y) dy = \int u(x) (\int \phi(y) \psi(y + x) dy) dx$

where the $dx$ integral is to be viewed in the sense of distributions. A priori from the definition of a distribution, this formula is only clear when $\phi$ is a delta function (in which case the convolution is just a translation) or a linear combination thereof. So to prove the general case you will have to use the continuity in the definition of a distribution to pass from the limit by approximating $\phi(x) dx$ with point masses. This is essentially an exercise which is often done in Riemann integration, although you have to keep track of the error and make sure the convergence is in $C^k$.

In Rudin's book, this discrete approximation is actually how he constructs the Lebesgue measure from scratch, and it's interesting that it actually even works in the measure case because you are bypassing the use of Fubini's theorem. From that point of view it isn't completely a "Riemann integral" approach.

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Indeed, the Riemann integrability of continuous functions can be re-interpreted as an assertion that certain finite linear combinations of Dirac masses converge in the vague topology (and hence in the sense of distributions) to the uniform distribution, which makes it clear why this fact could be useful in the theory of distributions, as it links continuous distributions with their discrete counterparts. – Terry Tao Aug 10 '11 at 3:13