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Encouraged by this question I ask a question concerning Keisler's treatment of definite integrals using infinitesimals.

He starts with a continuous function $f$ defined on a closed interval $[a, b] $ and forms an infinite Riemann sum $\sum_{a} ^{b} f(x) \, dx$ where the infinitesimal $dx$ is obtained by partitioning the interval $[a, b] $ into $H$ parts ($H$ being a hyperinteger) and $dx=(b-a) /H$.

Next he uses the fact that $f$ has a minimum value and maximum value so that it is bounded and then proves that the infinite Riemann sum $\sum_{a} ^{b} f(x) \, dx$ is a finite hyperreal number and we can take its standard part to get a real number which is defined as the integral $\int_{a} ^{b} f(x) \, dx$.

But looking at the overall definition I don't see continuity being used in essential manner apart from what is mentioned in last paragraph. Thus the argument applies equally to any bounded function on $[a, b] $ and it's integral is defined. But as we know from standard analysis this is not the case. Not all bounded functions are Riemann integrable, but continuous functions are. Thus it appears that the overall presentation skips the important fact that continuity implies integrability. Most introductory calculus texts instead mention this fact without proof.

My question is:

Is the presentation of Riemann integral by Keisler rigorous enough or a bit hand wavy as previous paragraph indicates? And if it is hand wavy, how can the presentation be improved using infinitesimal approach to include the proof that continuity implies integrability?

Keisler's book is available for download here and the definite integrals are discussed in chapter 4.

Paramanand Singh
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There is no handwaving here.

It is true that not every bounded real function on an interval is Riemann integrable, but that is not needed here. The argument does indeed show that all infinite Riemann sums for a bounded function are well defined.

Note that Keisler starts by defining "the definite integral of $f$ from $a$ to $b$ with respect to $dx$". The "with respect to"-part matters. The point is that for an arbitrary bounded function, the definite integral with respect to an infinitesimal can depend on the infinitesimal.

Then in Theorem 1 of 4.2., Keisler shows that for $f$ continuous, the choice of the infinitesimal does not matter. This is exactly where continuity is needed.

Michael Greinecker
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  • The intent of the theorem you mention is different. It shows that the integral depends on the function and the interval and not on the infinitesimal. This holds for any Riemann integrable function and is not related to continuity. – Paramanand Singh Sep 25 '18 at 08:35
  • Also does it mean that the set of functions whose integral does not depend on choice of the infinitesimal is precisely the set of all Riemann integrable functions? – Paramanand Singh Sep 25 '18 at 08:41
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    The argument uses continuity and does not work with arbitrary bounded functions; that does of not mean one cannot use more general functions. Riemann integrability is not treated in Keisler's book, but yes, one can define the Riemann integral the same way. – Michael Greinecker Sep 25 '18 at 09:19
  • I did check the argument for proof for theorem 1. It said that the difference between two infinite Riemann sums lies between $-\epsilon(b-a) $ and $\epsilon(b-a) $ where $\epsilon $ is a positive infinitesimal. The argument is not very clear eg how does one find the difference between two sums indexed by hyperintegers and how exactly does the difference remain infinitesimal. To me the argument is essentially about uniform continuity but this is not mentioned explicitly or perhaps this is assumed as self evident. Meanwhile thanks for bearing with and clarifying my doubts! – Paramanand Singh Sep 25 '18 at 11:00
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    Well, one does these things exactly as one would do them for finite integers. That the difference is infinitesimal does relate to uniform continuity, but on a compact interval there is no difference and even proving that is straightforward. – Michael Greinecker Sep 25 '18 at 11:20
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    Generally, a real function is uniformly continuous if it maps infinitesimally close hyperreal numbers to infinitesimally close hyperreal numbers. It is continuous if whenever a hyperreal number is infinitesimally close to a real number, the corresponding values are infinitesimally close. Now on the interval $[a,b]$, infinitesimally close hyperreal numbers have the same standard part and have therefore values infinitesimally close to the value of their common standard part by ordinary continuity. But then their values must be infinitesimally close. – Michael Greinecker Sep 25 '18 at 11:20