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In nonstandard analysis, we can define an integral by first dividing [0,1] into infinite intervals of infinitesimal length and then taking the appropriate hyperfinite sum. More formally, let $N \in {}^*\mathbb{N}$ be any hyperinteger and let $G=\{{n \over N} : n \in \{1, ..., N\}\}$. For a function $f:[0.1]\to\mathbb{R}$ and $A \subseteq [0,1]$ we define the grid integral by $$st(\sum_{g \in G \cap \ {}^*A} {}^*f(g) \cdot {1 \over N})$$ It is known that for Riemann integrable functions and intervals this coincides with the riemann integral. (see Nonstandard Methods in Ramsey Theory and Combinatorial Number Theory page 36) Does the same hold true for lebesgue measurable (or integrable) functions and measurable sets?

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  • $\begingroup$ Is your $G$ of null-measure? I'm not familiar with hyper integers $\endgroup$ Commented Jun 30, 2023 at 22:46
  • $\begingroup$ @JulesBesson Standard measure theory does not apply here. $G$ is not even a subset of $\mathbb{R}$. It is a hyperfinite set of hyperreals. It does not make sense to say $G$ is Lebesgue measurable set or a null set in the normal sense. $\endgroup$ Commented Jun 30, 2023 at 22:49
  • $\begingroup$ Ok I'll leave it to the experts then :) I think I have a grasp of what hyper integers are, via ordinals but this does not seem to be complete. $G$ would not happen to be countable by any chance? $\endgroup$ Commented Jun 30, 2023 at 22:50
  • $\begingroup$ No it's uncountable. $\endgroup$ Commented Jun 30, 2023 at 23:02

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This does not hold for Lebesgue measurable functions.

Let $f$ denote the indicator function of the irrationals. Then $f$ has Lebesgue integral $1$ on the interval $A = [0,1]$. However, since $f$ is zero on all ratios of integers, by Transfer its extension $~^\star\!f$ is zero on all ratios of hyperintegers. All terms under the sum are zero, so the grid integral equals $0$ instead of $1$.

And a minor correction: the quantity above generally does not coincide with the Riemann integral either. Di Nasso, Goldbring and Lupini only claim that the grid integral coincides with the usual Riemann integral in the case when $A$ itself is an interval.

You can find a nonstandard treatment of the Lebesgue integral in Chapter V.16. of Goldblatt's Lectures.

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  • $\begingroup$ It may be worth making it clear that the approach Goldblatt presents indeed passes via hyperfinite partitions, but not in as naive a matter as appears in the question. $\endgroup$ Commented Jul 1, 2023 at 19:41

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