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Let's take a general function $f(x)$, we can do a summation like:

$$\sum_{k=m}^n f(k)$$

And we can do an integration like:

$$\int_a^bf(k)dk$$

The basic difference between the two operation is that the first concerns only integer values of $k$ while the second is about real values of the function in an interval. So my question is: is there a similar operation that concerns rational values ? Something "between" summation and integration.

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  • $\begingroup$ You can take a (Riemann) integral over only the rational values - it's the same as taking an integral over the reals, though, for any integrable function. $\endgroup$ – Milo Brandt Jun 30 '15 at 0:37
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    $\begingroup$ Perhaps not $\endgroup$ – Count Iblis Jun 30 '15 at 3:18
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    $\begingroup$ I see integration as a generalization of summation... not as some kind of alternative. If you just set $dk = 1$, you get back summation. (Well, you get what I mean.) $\endgroup$ – user541686 Jun 30 '15 at 7:18
  • $\begingroup$ @CountIblis very interesting paper! Thanks for the link. $\endgroup$ – Renato Faraone Jun 30 '15 at 7:39
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    $\begingroup$ Since all of the answers seem to rely on Countable/Real concepts that we are more or less familiar with; how about an analysis on a set (presumably ordered) from the negative of the Continuum Hypothesis? Such things are useful in understanding the "true" mathematical structure of things like integration. For me: references for learning are appreciated. Perhaps this should be a seperate question? $\endgroup$ – rrogers Jul 1 '15 at 13:31
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Your examples can be viewd as particular cases of a general theory of integration, namely, the Lebesgue integration. In fact:

  • If $X=\{m,m+1,...,n\}$, $\mu$ is the counting measure and $f$ is positive, then $$\int_X f\ d\mu=\sum_{k=m}^nf(k).$$

  • If $X=[a,b]$, $\mu$ is the Lebesgue measure and $f$ is continuous, then $$\int_X f\ d\mu=\int_a^bf(k)\ dk,$$ where the integral in the right side is the Riemann integral.

From this point of view, the answer to your question is yes and the desired "similar operation" will also be a particular case of the Lebesgue integration. Explicitly:

  • If $X=\mathbb{Q}$, $\mu$ is some suitable measure and $f$ is measurable, then $$\int_X f\ d\mu$$ can be viewed as "something between summation and (Riemann) integration that concerns rational values".

Example: If $\mu$ is the Lebesgue measure and $f$ is Lebesgue measurable, then $$\int_\mathbb{Q} f\ d\mu=0.$$

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  • $\begingroup$ Sounds interesting! Can you show me a few other examples ? Something like the trivial integrals with one start studying calculus. :) $\endgroup$ – Renato Faraone Jun 29 '15 at 22:05
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    $\begingroup$ @RenatoFaraone From this integration perspective, to get examples satisfying your requirements, we need to integrate functions over a set that is not a interval, e.g., a subset of rational numbers. However, the ordinary theory of integration (studied in the beginning of calculus courses) deals only with functions defined on intervals (or union of such intervals). So, I don't know if it is possible to find more trivial examples. $\endgroup$ – Pedro Jun 29 '15 at 23:04
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    $\begingroup$ Anything more explicit than "some suitable" which is non-zero and distinct from counting and Lebesgue? $\endgroup$ – Keith Jun 30 '15 at 0:13
  • $\begingroup$ @Keith How about the Dirac measure concentrated at $q\in\mathbb{Q}$? In this case we have $\mu(\mathbb{Q})\neq 0$ and $\displaystyle\int_\mathbb{Q}f\;d\mu=f(q)$. $\endgroup$ – Pedro Jun 30 '15 at 14:43
  • $\begingroup$ @Pedro. Sorry - Dirac seems to me to sit "between" the zero measure and the counting measure. snulty's answer is closer to the concept. $\endgroup$ – Keith Jul 1 '15 at 1:05
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The short answer is yes, but it is not a very easy concept. The answer lies with the definition of the integral and measures. The idea of a measure is to assign a size to subsets of the reals, and use this idea of size to perform integration (among other things). The first thing to note is that there are different ways to assign a measure to subsets of the reals. Some measures behave the way that you would expect them to in that the size of the interval $(a,b)$ is $b-a$, regardless of what $a$ and $b$ are. Other measures behave differently. For example, you could say that the measure of a set is the number of integers contained therein. Under this definition of a measure, summation and integration become the same thing. We could also define a measure which heeds only rational numbers and perform integration with respect to that measure.

In summary, integration is an abstract concept. The standard integral taught in introductory calculus is just one type of integral, and the type of integral you describe is just a different type (as is summation).

For more information on this subject see here for information on measures, and here for an explanation of how integration is defined abstractly, using measures.

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    $\begingroup$ Since there are countably many rational numbers, either overall or in an interval, there is no way to get a uniform, countably-additive measure. You would need to abandon uniformity to get a measure. $\endgroup$ – Rory Daulton Jun 29 '15 at 20:59
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    $\begingroup$ I don't know of any nice measures on the rationals. @RoryDaulton is right, you cannot have both translation invariance and countable additivity for a measure on a countable set. Since measures have to be countably additive by definition, you cannot keep the translation invariance. $\endgroup$ – Alex S Jun 29 '15 at 21:05
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    $\begingroup$ @Solomonoff'sSecret : The classical construction of this complete/translation invariant/containing all intervals measure on the reals is given by the completion of the Haar measure which would give you the Lebesgue measure. The problem with this construction on $\mathbb Q$ is that $\mathbb Q$, even though Hausdorff, is not locally compact. It's also for this reason that the construction works better when we "patch the holes" in $\mathbb Q$ to recover local compactness. It's not a proof that there is no such measure on $\mathbb Q$, but it is a reason to believe why it fails to exist. $\endgroup$ – Patrick Da Silva Jun 29 '15 at 22:09
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    $\begingroup$ @RoryDaulton : $\:$ The counting measure on $\mathbb{Q}$ is "a uniform, countably-additive measure" on $\mathbb{Q}$. $\hspace{.63 in}$ $\endgroup$ – user57159 Jun 30 '15 at 0:17
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    $\begingroup$ @RickyDemer: My mistake, you are correct. I meant "there is no way to get a uniform, countably-additive probability". But of course that is not what I said. Thanks for the correction. $\endgroup$ – Rory Daulton Jun 30 '15 at 0:45
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One might say: Between sums (finitely many summands) and integrals (uncountably many "summands"), we have series (countably many summands).

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  • $\begingroup$ Well that wasn't what I'm searching for, series deals with integer values of the argument of the function. I'm searching for something between series and integrals, and I got that I must start studying measure theory I guess. $\endgroup$ – Renato Faraone Jun 29 '15 at 21:47
  • $\begingroup$ @RenatoFaraone I've written up something for you which also has the flavour of Hagen von eitzen's answer. You can sum over the rationals, but you've to be careful $\endgroup$ – snulty Jun 29 '15 at 23:32
  • $\begingroup$ This is a very good complementary answer in my opinion! $\endgroup$ – Benjamin Lindqvist Jul 2 '15 at 6:12
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Just to follow on from something inspired by john-mangual's answer:

Firstly one might know that the rationals $\mathbb{Q}$ are countable and thus can be enumerated by the positive integers $\mathbb{N}$. Let's pick an enumeration $\mathbb{Q}=\{q_1,q_2,q_3,\ldots\}$. Now let $f:\Bbb R \rightarrow \Bbb R$ such that it is defined on $\mathbb{Q}$ (or restrict to a subset if necessary), then we can sum over the rationals by its enumeration, namely $$\sum\limits_{i=1}^\infty f(q_i).$$

Now unless the above summation converges absolutely, as in unless $$\sum\limits_{i=1}^\infty |f(q_i)| <\infty,$$

then the value of this series may depend on the enumeration of $\mathbb{Q}$, and even worse due to a theorem of Riemann for what are called conditionally convergent series, if you rearrange the terms you can make the series converge to whatever you like!

However if it does converge absolutely then the sum is independent of how we choose to order the rationals, and we can sum in any order.

Maybe this will give you another insight into something else you can do, and since I don't really know any measure theory I can't help on that front!

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  • $\begingroup$ It's worth noting that a non-zero $f$ for which this sum converges absolutely cannot be continuous as every open interval contains infinitely many rationals. $\endgroup$ – Milo Brandt Jun 30 '15 at 1:08
  • $\begingroup$ @Meelo I find that fairly interesting actually, how does one go about showing that? :) $\endgroup$ – snulty Jun 30 '15 at 13:09
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Not sure that my answer is totally relevant... Ler's have a try however.

Both cases that you provide can be interpreted as distribution. The integral is a distribution with a "constant density".

$$\sum_{k=m}^n f(k)$$ is also a distribution with Dirac comb for density.

You can imagine a distribution with "masses" only at the rational numbers which will correspond to your request.

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  • $\begingroup$ I feel like what I've said might be very similar to yours and I know you've answered before me, but I'm not overly familiar with distributions, so I've put my answer in terms I understand, and maybe you can give your opinion on the similarities? $\endgroup$ – snulty Jun 29 '15 at 23:19
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Certainly for the exponential function you can define fractional derivatives or negative dervatives

$$ \frac{d}{dx^k} [e^{ax}] = a^k \; e^{ax}$$

where $k \in \mathbb{R}$. This can be extended to Fourier series so we can define such pseuo-differential operators for all of $L^2 ([0,1])$.

Wikipedia explains such a fractional deriviative behaves nicely with respect to the Laplace transform as well.

$$ \frac{d^k}{dx^k} \int_0^\infty e^{-xt} f(t) \, dt = \int_0^\infty t^ke^{-xt} f(t) \, dt$$

with some additional fine print about passing under the integral sign.

Even polynomials seem to behave nicely under this definition

$$ \frac{d^\ell}{dx^\ell} \frac{x^k}{k!} = \frac{x^{k-\ell}}{(k-\ell)!}$$

where $l! = \Gamma(l+1)$ for $l \in \mathbb{R}$ is the Gamma function.


To answer the question you asked, the rationals are countable but dense in the real line $\overline{\mathbb{Q}} = \mathbb{R}$ so there should not be much difference between summing over rationals and integrating unless your function is very chaotic. So I am arguing

$$ \sum_{x \in \mathbb{Q} \cap [0,1]} \approx \int_0^1 dx\, $$

However, there are many functions on the rational numbers were nearby inputs take radically different values (as appear in the theory of modular forms).

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  • $\begingroup$ this is answering a very different question actually, but it might be of interest $\endgroup$ – cactus314 Jun 29 '15 at 20:35
  • $\begingroup$ What exactly do you mean by your last expression? Do you mean that a function evaluated at all rationals and summed up behaves like a Riemann integral? $\endgroup$ – user190080 Jun 29 '15 at 22:29

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