Can one define the Lebesgue integral with nonstandard analysis?

Question

Reading Keisler's books about non-standard analysis, I have begun to suspect the possibility that there may not be an adequate characterization of the Lebesgue integral and of measure theory in general using non-standard analysis.

Question: Is this an actual weakness of non-standard analysis? Or is Keisler not doing the subject justice by failing to mention that the Lebesgue integral and its generalizations can also be well understood in the framework of nonstandard analysis?

Explaining in detail how the construction of the Lebesgue integral carries over to non-standard analysis would be too broad and too much to ask or expect of any answer, so I will without hesitation accept answers only giving references to sources which discuss this issue. This is why I have tagged the question (reference-request).

Background: I have been reading Keisler's books about non-standard analysis, the introductory Calculus: An Infinitesimal Approach and the more advanced and rigorous Foundations of Infinitesimal Calculus. The argument for why differentiation is more intuitive with non-standard analysis is fairly clear to me, however, I am not convinced yet that the same is true of integration.

Specifically, both of Keisler's books only mention how to perform Riemann integration using non-standard analysis, but say nothing about Lebesgue integration (i.e. general integration with respect to an arbitrary measure, not just the Lebesgue measure on $\mathbb{R}$ and its subsets). In fact, the Lebesgue integral is not mentioned at all. To me at least, this seems like an unusually large omission, at least from Foundations of Infinitesimal Calculus, which seems to be fairly rigorous.

Answer 1

Thinking about how to answer the title question... I think there is actually a fairly straightforward path.

By the regularity theorem, if $X$ is a standard Lebesgue measurable set and we pick a positive infinitesimal $\epsilon$, there exists a closed set $C$ and an open set $U$ such that:

• $C \subseteq X \subseteq U$
• $\mathop{\mathrm{std}}(\mu(U)) = \mu(X)$
• $\mathop{\mathrm{std}}(\mu(U \setminus C)) = 0$

This now makes it obvious how to define the Lebesgue measure.

Every open subset of the reals is a disjoint union of open intervals; let $\lambda$ be the function computing the total length. Then,

A standard set $X$ is said to be Lebesgue measurable if and only if there exists a closed set $C$ and an open set $U$ such that

• $ C \subseteq X \subseteq U$
• $\lambda(U \setminus C) \approx 0$

The Lebesgue measure of such a set is defined to be $\mu(X) = \mathop{\mathrm{std}}(\lambda(U))$

(note that the standard part of a positive infinite hyperreal is the extended real $+\infty$)

This can be viewed as an expression of Littlewood's first principle:

Every measurable set is nearly a finite sum of intervals

I imagine you can even restrict the $U$ above to be a (hyper)finite sum of open intervals.

I'm too tired to continue, but once we have the Lebesgue measure, I can't imagine there being any difficulty in defining the integral by approximating a standard function by a nonstandardard simple function.