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.

suitably_understood, $\frac{d}{dx}\int^x_{x_o} f(t)\,dt=f(x)$, for nice functions $f$, etc.? It's not really about aconstructionof an integral, but thispropertyrelating differentiation and "integration". – paul garrett Aug 9 '11 at 15:59anything at allyou'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