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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    $\begingroup$ 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. $\endgroup$ Aug 9, 2011 at 12:50
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    $\begingroup$ 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". $\endgroup$ Aug 9, 2011 at 15:59
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    $\begingroup$ @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. $\endgroup$ Aug 9, 2011 at 18:47
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    $\begingroup$ 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. $\endgroup$ Aug 9, 2011 at 21:22
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    $\begingroup$ @PietroMajer: I think you misunderstood the question. It is not about generalities like: suppose I need to teach a graduate analysis course which does integration and distributions, can I teach it without mentioning Riemann integrals at all? The question is about specific results (like the first example) in distribution theory where all of a sudden one needs to pull out of a hat some Riemann sums to do the trick. By the time one reaches this point, one must have mastered Lebesgue integration so this resorting to older knowledge (Riemann theory) may seem strange. $\endgroup$ Feb 14, 2020 at 19:36

6 Answers 6


I will only consider the temperate situation (involving the spaces $\mathscr{S}$, $\mathscr{S}'$ and $\mathscr{O}_{\rm M}$) and I will only discuss the first example as well as a "dual" version with a derivative instead of an integral to be commuted with a distribution pairing. I will show that Riemann integration (and its dual, i.e., the classical definition of derivative as the limit of a ratio) not only is unnecessary but perhaps is a crime against Laurent Schwartz, or rather against the spirit of his theory of distributions.

Let us first go over some notational conventions. When talking about a distribution $F\in\mathscr{S}'(\mathbb{R}^n)$, I will instead say $F(x)\in\mathscr{S}'_x(\mathbb{R}^n)$ in order to view it as generalized function $F(x)$ of a variable $x$ that must be named. This name will also appear as a subscript in the corresponding function space. In the same vein, let me now take a Schwartz function $\phi(x)\in\mathscr{S}_x(\mathbb{R}^n)$ meaning let me take an element $\phi\in\mathscr{S}(\mathbb{R}^n)$ but with the extra emphasis on giving the variable a name, here $x$. I will not denote the distribution pairing by $\langle F,\phi\rangle$ which is rigorous but lacks expressive power, nor by the integral $$ \int_{\mathbb{R}^n}F(x)\phi(x)\ d^nx $$ which is not rigorous, but has the expressive power needed for "multilinear algebra" in infinite dimension. Instead, I will keep the best of both worlds and write $$ \langle F(x),\phi(x)\rangle_{x}\ . $$

These conventions settled, I can state the temperate variant of Fubini's Theorem for Distributions. Let $F(x)\in\mathscr{S}'_x(\mathbb{R}^m)$, let $G(y)\in\mathscr{S}'_y(\mathbb{R}^n)$, and let $\eta(x,y)\in\mathscr{S}_{x,y}(\mathbb{R}^{m+n})$. Then

a) $\langle G(y),\eta(x,y)\rangle_{y}\in\mathscr{S}_x(\mathbb{R}^m)$;

b) $\langle F(x),\eta(x,y)\rangle_{x}\in\mathscr{S}_y(\mathbb{R}^n)$;

c) we have the equalities $$ \langle F(x),\langle G(y), \eta(x,y)\rangle_{y}\rangle_{x} =\langle G(y),\langle F(x), \eta(x,y)\rangle_{x}\rangle_{y} =\langle (F\otimes G)(x,y),\eta(x,y)\rangle_{x,y} $$ where $(F\otimes G)(x,y)$ heuristically is the generalized function $F(x)G(y)$ of the composite variable $(x,y)$ in $\mathbb{R}^{m+n}$.

One also has a stronger version of the theorem which includes:

d) The element of $\mathscr{S}_{x}(\mathbb{R}^{m})$ constructed in a) is a hypocontinuous bilinear function of $G$ in $\mathscr{S}'_{y}(\mathbb{R}^{n})$ and $\eta$ in $\mathscr{S}_{x,y}(\mathbb{R}^{m+n})$. Likewise, the element of $\mathscr{S}_{y}(\mathbb{R}^{n})$ constructed in b) is a hypocontinuous bilinear function of $F$ in $\mathscr{S}'_{x}(\mathbb{R}^{m})$ and $\eta$ in $\mathscr{S}_{x,y}(\mathbb{R}^{m+n})$.

Now the OP's first example asks us to show that for $F$ a temperate distribution in $\mathbb{R}^n$ and for $\psi,\phi$ Schwartz functions on $\mathbb{R}^n$, we have the identity $$ \int_{\mathbb{R}^n}\langle F(y),\psi(x-y)\rangle_{y}\ \phi(x)\ d^n x =\langle F(y),(\phi\ast\psi_0)(y)\rangle_{y} $$

Low tech proof: Use finite Riemann sums to approximate the integral which reduces the question to a trivial algebraic fact, i.e., linearity.

High tech proof: In my previous notations, the identity means $$ \langle 1(x),\langle F(y),\phi(x)\psi(x-y)\rangle_{y}\rangle_x =\langle F(y),\langle 1(x),\phi(x)\psi(x-y)\rangle_{x}\rangle_y $$ where the strange notation $1(x)$ is for the constant function of $x$ equal to 1 seen as a temperate distribution in the usual way (by Lebesgue integrating against it). Since $\eta(x,y)=\phi(x)\psi(x-y)\in\mathscr{S}_{x,y}(\mathbb{R}^{2n})$, this trivially follows from the algebraic fact/formula expressed by the above distributional Fubini theorem, part c).

Although not in the original question, let me also consider another "Stokes-Theorem/Fundamental-Theorem-of-Calculus"-dual result that everyone learns in a PDE course with distributions, regarding convolutions. In the same setting as above, we have that $(F\ast\psi)(x)$ is a smooth function of temperate growth which I will write as the statement $(F\ast\psi)(x)\in\mathscr{O}_{{\rm M},x}(\mathbb{R}^n)$.

Low tech proof: One uses induction, starting with a similar commutation identity, i.e., derivation under the distibutional pairing sign, in order for the derivative to hit the test function inside the pairing. Usually, one proves this by tedious estimates on derivatives, etc.

High tech proof: Take an arbitrary $\phi(x)\in \mathscr{S}_{x}(\mathbb{R}^{n})$ and consider $$ \phi(x)(F\ast\psi)(x)=\phi(x) \langle F(y),\psi(x-y)\rangle_{y}=\langle F(y),\phi(x)\psi(x-y)\rangle_{y}\ . $$ Again because $\eta(x,y)=\phi(x)\psi(x-y)\in\mathscr{S}_{x,y}(\mathbb{R}^{2n})$, the distributional Fubini, part a), with $F$ instead of $G$ gives that $$ \phi(x)(F\ast\psi)(x)\in \mathscr{S}_{x}(\mathbb{R}^n)\ . $$ This is true for all $\phi$, so by the multiplier space characterization of $\mathscr{O}_{\rm M}$, we have that the convolution is a smooth temperate function.

Remark 1: One can show the intermediate identity $\partial_z^\alpha (F\ast \psi)(z)=(F\ast\partial^{\alpha}\psi)(z)$ used in the low tech proof, with the same method as above by replacing $1(x)$ by a derivative of a delta function in $x$ located at $z$.

Remark 2: This is pure algebra.

Addendum: The map $\mathscr{S}'\times\mathscr{S}\rightarrow\mathscr{O}_{\rm M}$, $(F,\psi)\mapsto (F\ast \psi)$ is hypocontinuous. This follows from the newly added part d) above of Fubini's Theorem for distributions and from the topology of $\mathscr{O}_{\rm M}$ being defined by the seminorms $||\phi \cdot||$ where $\phi$ ranges over $\mathscr{S}$ and $||\cdot||$ ranges over continuous seminorms of $\mathscr{S}$. Since convergent sequences form bounded sets, this shows the construction is jointly sequentially continuous.

Also note that proving Fubini's Theorem for distributions (and the Kernel Theorem etc.) is very easy, if one does what algebraists say one should never do: take a (Schauder) basis, e.g., that of Hermite functions. Some hints about this are given in my answer


with some more pointers given in


Finally, to see more involved applications of this kind of reasoning with distributions, you can also have a look at my recent article "A Second-Quantized Kolmogorov–Chentsov Theorem via the Operator Product Expansion" about pointwise multiplication for random Schwartz distributions.

  • $\begingroup$ Good (though I'd still try to suppress the argument "$x$" in your pairing...). :) $\endgroup$ Jan 27, 2020 at 17:17
  • $\begingroup$ is it in order to be more "coordinate free"? $\endgroup$ Jan 27, 2020 at 18:08
  • $\begingroup$ Well, perhaps something like that. In a similar sense to my feeling that $\int_X f(x)\,dx$ can just as usefully be written with fewer marks and no dummy variables, as $\int_X f$. And thinking that "$f$" is a function, not "$f(x)$", although "$x\to f(x)$" is a verbose way of writing it. This does become mildly relevant in talking about $y\to f(x,y)$ as a function-valued function of $x$, and so on. :) $\endgroup$ Jan 27, 2020 at 18:40
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    $\begingroup$ @paulgarrett: I agree (provided I understood you correctly). There is no need for naming the variable if one just has one $\mathbb{R}^n$ to deal with and we pair distributions and test functions on that space. However, suppose one considers in Einstein notation some contraction of tensors $A_{ijk}B_{il}C_{jklm}D_{mnopq}$ or rather the infinite dimensional analogue. Schwartz called that Volterra composition in his ICM contribution and his two papers on vector-valued distributions. It seems hard to me to avoid naming indices/variables in this multilinear algebra game. $\endgroup$ Jan 27, 2020 at 19:41
  • $\begingroup$ Yes, depending on the notational corner one has painted oneself into... Still, often a multilinear map can be correctly described as the multilinear map described on monomial tensors by $v\otimes \lambda\to \lambda(v)$ (for example, for trace...). :) $\endgroup$ Jan 27, 2020 at 19:53

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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    $\begingroup$ 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. $\endgroup$
    – Terry Tao
    Aug 10, 2011 at 3:13
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    $\begingroup$ I also thought at first that resorting to Riemann sums was THE way to prove this kind of Fubini for two integrals, one which is honest and the other being in the distribution sense. This is misguided however. In fact this is a particular case of Fubini's Theorem for distributions applied to a tensor product distribution $1\otimes u$ where "1" is the constant function equal to one seen as a distribution in the variable $x$, while $u$ is a generalized function of the variable $y$. $\endgroup$ Jan 23, 2020 at 15:33

An instance where the Riemann integral cannot be replaced by the Lebesgue 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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    $\begingroup$ 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. $\endgroup$ Feb 5, 2013 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.

  • $\begingroup$ 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). $\endgroup$ Feb 2, 2014 at 16:18
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    $\begingroup$ The best way to prove the identity is to say that it is a particular case of Fubini's Theorem for distributions which itself is strongly related to the Kernel Theorem. $\endgroup$ Jan 23, 2020 at 15:21

A very belated addition to this interesting thread: as in many situations, as we can see in hindsight, "(most useful) integrals" are uniquely determined by their restrictions to "good" functionals on compactly-supported continuous functions... via Riesz-Markov-Kakutani theorem on functionals on such spaces of functions.

As a not-often-mentioned corollary, whenever it makes sense, a Riemann-sum type of integral (giving the functional on compactly-supported continuous...) uniquely determines everything in any reasonable direction.

After all, while we surely do appreciate Lebesgue's great insight of how to construct integrals of a much wider class of functions, for many purposes we do not care about the construction (after we've maybe proven some basic set-up things).

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    $\begingroup$ Reminds me of how I unlearned my first impression of the distributional derivative. I first learned the notion this way: 1) write the paring as a formal integral, 2) do some heuristic integration by part, 3) realize that now you have something that makes sense rigorously, 4) use the latter as the rigorous definition. Now I prefer to view it as the unique extension of the classical derivative from say $\mathscr{S}$ to $\mathscr{S}'$. $\endgroup$ Feb 1, 2020 at 0:01
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    $\begingroup$ @AbdelmalekAbdesselam, yes, nowadays when teaching I try to emphasize that although it was a superb insight of Schwartz to in one stroke define distributions as dual spaces, that mainly just give instant rigor to the idea of defining things by extension-by-continuity. $\endgroup$ Feb 1, 2020 at 14:38
  • $\begingroup$ Indeed, these days I have to worry about continuity a lot because I do probability theory with random Schwartz distributions. Various constructions need to be Borel measurable, which often is simply due to continuity. $\endgroup$ Feb 1, 2020 at 20:43

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.


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