# Why are Darboux integrals called Riemann integrals?

As far as I have seen, the majority of modern introductory real analysis texts introduce Darboux integrals, not Riemann integrals. Indeed, many do not even mention Riemann integrals as they are actually defined (with Riemann sums as opposed to Darboux sums). However, they call the Darboux integrals Riemann integrals. Does anyone know the history behind this? I can understand why they use Darboux - I find it much more natural and the convergence is simpler in some sense (and of course the two are equivalent). But why do they call them Riemann integrals? Is this another mathematical misappropriation of credit or was Riemann perhaps more involved with Darboux integrals (which themselves may be misnamed)?

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See this. They are equivalent approaches, and they are very similar, but not identical. Could you provide a source where that is done? –  Pedro Tamaroff Sep 25 '12 at 18:14
I'm sorry, I'm not sure what you're asking, provide a source where what is done? –  Curtis Sep 25 '12 at 18:20
"As far as I have seen, the majority of modern introductory real analysis texts introduce Darboux integrals, not Riemann integrals. Indeed, many do not even mention Riemann integrals as they are actually defined (with Riemann sums as opposed to Darboux sums). However, they call the Darboux integrals Riemann integrals. " –  Pedro Tamaroff Sep 25 '12 at 18:45
In Rudin's book, which many would argue is the standard. See pp. 120, 121 (the first two pages of Chapter 6: Integration.) –  Curtis Sep 25 '12 at 21:12
Indeed, Darboux's name is not even in Rudin's index. Another example is the appendix in Stein and Sakarchi's Fourier Analyis. Both of these call all the sums and upper integrals after Riemann, and call the integrable functions Riemann integrable. I believe there are other textbooks, but I don't have direct access to them now. –  Curtis Sep 25 '12 at 21:22