In the sources I've seen, the integral is defined in non-standard calculus as the hyperreal extension of a function related to Riemann sums. E.g.,

Let $$ S(\Delta x) = \sum_{a}^{b}f(x)\Delta x$$ be a function of a Riemann sum where $\Delta x$ is the length of each partition. Then let $S^*(\Delta x)$ be the hyperreal extension of that function, which exists due to the transfer principle. Then we can talk about $S^*(\rm dx)$, where $\rm dx$ is an infinitesimal.

This is very simple and seems to work, but it doesn't really explain what's happening in the function $S^*(\Delta x)$. What does a Riemann sum with infinitesimal partitions mean? How do I sum $H$ terms, where $H$ is a hyperinteger?

Can anyone direct me to other definitions of integrals in non-standard calculus (not necessarily Riemann equivalent) that answer some of these questions? Or possibly a textbook that goes into more detail about this subject?

Edit: Let me clarify something. I understand the definition. I certainly don't think it's not well-defined.

I just feel it sort of "covers up" the really interesting bits. "Interesting bits" includes, for example, using induction on the hyperintegers.

  • 1
    $\begingroup$ What is covered up is the construction of the hyperreals. If you don't understand it, you can't claim to understand them intuitively without it being just blind faith. And remember that the hyperreals are simply the ultraproduct of the reals using an ultrafilter on the naturals. So if you think the usual reals don't make intuitive sense, then you can forget about the hyperreals making sense. I upvoted your question but I do think you chose the wrong answer, since it covers up what is covered up. $\endgroup$ – user21820 Sep 11 '17 at 8:00
  • 1
    $\begingroup$ Remember that you can't just willy-nilly integrate any function; it must be first-order definable so that the transfer principle works. See for example this post and the chain of comments, for just the surface of the iceberg that has been covered up. $\endgroup$ – user21820 Sep 11 '17 at 8:04

The integral is not the hyperfinite sum $S^{\ast}(dx)$ but rather the standard part of that hyperfinite sum. This is similar to the situation with the derivative, where one can form the quotient of infinitesimals $\frac{dy}{dx}$ where $dy$ is the $y$-increment corresponding to the $x$-increment $dx$. Here the derivative is not the quotient but rather the standard part of that quotient. Usually one denotes the infinitesimal $x$-increment by the symbol $\Delta x$ and the corresponding $y$-increment by $\Delta y$. The derivative is then defined to be $$\text{st}\left(\frac{\Delta y}{\Delta x}\right)$$ rather than the ratio itself. Here "st" is the standard part function. Then one sets $dx=\Delta x$ and defines $dy$ by $dy=f'(x)dx$ so that the derivative can be written as $dy/dx$.

I have had good experience teaching using Keisler's textbook which is available online here. A more advanced companion volume by Keisler that covers more of the theoretical material can be found here.

To answer more specifically your questions "What does a Riemann sum with infinitesimal partitions mean? How do I sum H terms, where H is a hyperinteger?": your intuition of a very small partition step and a very large number of partition points is already the right intuition. The notion of an infinite hyperinteger is merely a way of formalizing your intuition. The advantage of the hyperreal approach is that such intuitions, which remain at the level of vague musings in the epsilon, delta approach (without infinitesimals), can now be formalized rigorously. This allows one to write down proofs that are closer to the original intuition than in the epsilon, delta approach. One notion that gets clarified this way is that of a Cauchy sequence; see the related discussion I would like to know an intuitive way to understand a Cauchy sequence and the Cauchy criterion.

  • $\begingroup$ Thank you! I was familiar with Keisler's textbook, but not familiar with the additional volume. Do you know any other material on the subject of non-standard analysis/calculus, at around the same level? $\endgroup$ – GregRos Aug 11 '13 at 16:30
  • 1
    $\begingroup$ A slightly more advanced text is Goldblatt, Robert, Lectures on the hyperreals. An introduction to nonstandard analysis. Graduate Texts in Mathematics, 188. Springer-Verlag, New York, 1998. I enjoyed teaching from that, also. For a discussion of the basic intuitions, you could consult the question math.stackexchange.com/questions/405492/… $\endgroup$ – Mikhail Katz Aug 11 '13 at 16:36

Due to the transfer principle, as long as you don't compare them against each other, nonstandard analysis is completely indistinguishable from standard analysis.

In standard analysis, you have the (standard) finite summation operator: given (standard) integer limits $a \leq b$ and a (standard) real-valued function $f$ of the integers, the operator produces a real number $S$, which we often write as

$$ S = \sum_{n=a}^b f(n) $$

If we apply transfer, we get the (nonstandard) hyperfinite summation operator: given hyperinteger limits $a \leq b$ and an internal hyperreal-valued function $f$ of the hyperintegers, the operator produces a hyperreal number $S$. We might write it as

$$ S = {}^\star\sum_{n=a}^b f(n) $$

to emphasize that this is the nonstandard summation operator rather than the standard summation, but I would prefer to just write it as

$$ S = \sum_{n=a}^b f(n)$$

If you were so inclined, you don't even have to bother with the transfer principle. The nonstandard integers satisfy the (internal) principle of induction, so you could just recursively define the nonstandard summation operator in any usual way, such as

$$ {}^\star\sum_{n=a}^a f(n) = f(a) $$ $$ {}^\star\sum_{n=a}^{b+1} f(n) = f(b+1) + {}^\star\sum_{n=a}^b f(n) $$

which recursively defines the summation for all hyperintegers $a,b$ with $a \leq b$, and for all (internal) hyperreal-valued functions $f$ of the hyperintegers.

The reason you are having trouble seeing that the summation is well-defined is because you are implicitly trying to use induction over standard integers to justify recursions over nonstandard integers, but that simply doesn't work.

  • 2
    $\begingroup$ if the OP understood the notion of "internal" (as in "internal induction" and "internal hyperreal-valued functions"), he wouldn't have phrased the question the way he did. Also, understanding what you mean when you write "nonstandard analysis is completely indistinguishable from standard analysis", one needs a fairly detailed grasp of the semantic/syntactic distinction in logic, as well as the notion of first-order sentence and such. I certainly hope he benefits from your answer, but it may be helpful to provide some motivating details, as well. $\endgroup$ – Mikhail Katz Aug 11 '13 at 14:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.