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We know how to sum or average a finite number of terms: sums.

We know how to sum a countable infinite number ${\beth_0}$ of terms: series.

We know how to sum ${\beth_1}$ terms: integrals.

How to sum ${\beth_2}$ terms: ???

One "concrete" example please.

Let ${\mathbb{R}^\mathbb{R}}$ be the set of all functions $f:\mathbb{R} \to \mathbb{R}$.

Let ${x_0} \in \mathbb{R}$. Let ${\mathbb{R}^\mathbb{R}}\left( {{x_0}} \right)$ be the subset of all functions in ${\mathbb{R}^\mathbb{R}}$ having ${{x_0}}$ in their domains of definition. There are still ${\beth_2}$ of them.

Is the "functional mean image" of ${x_0}$ under all functions in ${\mathbb{R}^\mathbb{R}}\left( {{x_0}} \right)$ that we can formally write as

$\int\limits_{{\mathbb{R}^\mathbb{R}}\left( {{x_0}} \right)} {{\text{D}}f\,f\left( {{x_0}} \right)\,\,} $

(well) defined? If not, why?

Same question in the set of all bijections from $\mathbb{R}$ to $\mathbb{R}$.

This hypothetical "functional mean image", to be compared to the usual ${\beth_1}$ integral

$\int\limits_\mathbb{R} {{\text{d}}xf\left( x \right)} $

  • may be (well) defined in some branch of mathematics I (or you) do not know;
  • may be an unidentified mathematical object;
  • may not exist.

Anything welcome. My apologies if it is trivial but I sincerely do not know. Thanks.

Fabrice Pautot
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    Relevant related question on MO: http://mathoverflow.net/questions/1388 – mrf Apr 22 '16 at 08:56
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    Possibly functional integrals or (in QM) path integrals. – Raymond Manzoni Apr 22 '16 at 09:00
  • @mrf. Thanks for the pointer. I check... – Fabrice Pautot Apr 22 '16 at 09:07
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    @ Raymond: wonderful answer. My question precisely comes from some seemingly nasty functional integrals arising from a collision between dynamical system theory and Bayesian probability theory. Please check MO questions: http://mathoverflow.net/questions/232043/whats-the-probability-distribution-of-a-deterministic-signal-functional-integ , http://mathoverflow.net/questions/236527/is-there-a-bayesian-theory-of-deterministic-signal-prequel-and-motivation-for-m and http://mathoverflow.net/questions/236619/existence-of-probability-distributions-measures-spaces-and-mathematical-expectat. – Fabrice Pautot Apr 22 '16 at 09:12
  • But I'd like to get more precise answers please, e.g. your "functional mean image" is well known as dflgjdfljdfg, etc. – Fabrice Pautot Apr 22 '16 at 18:10
  • Well from your questions at MO it seems that you know better functional integrals than I do! :-)
    I know a little (Feynman) path integrals and for them the functions have to be restrained : instead of using variation calculus to find the path $f$ returning the least action $S$ between two points in space-time we add all the contributions $e^{iS/\hbar}$ between these two points over the possible paths verifying $f(x_1)=y_1,f(x_2)=y_2$.     – Raymond Manzoni Apr 23 '16 at 09:48
  • What bothers me here is that you don't impose any condition on your $f:\mathbb{R}\to \mathbb{R}$ functions ; shouldn't then the weight of $f(x_0)=y_0$ and $,f(x_0)=y_0+d;$ be the same for any real $d$? In my first Wikipedia link one finds "Most functional integrals are actually infinite but the quotient of two functional integrals can be finite" but I don't see how this may help here... – Raymond Manzoni Apr 23 '16 at 09:48
  • Perhaps you might have a look at this post and maybe also some other discussions about [meta-tag:cross-posting]. – Martin Sleziak Apr 23 '16 at 11:25
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    If somebody told me about summing $\mathfrak c$-many elements, I probably would not think of integrals first. There is also this type of sum (see also related questions) which make sense for any cardinality of the index set. – Martin Sleziak Apr 25 '16 at 06:46
  • We only have finite sums, the rest is just an illusion. – Count Iblis Jun 15 '16 at 01:33
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    With measure theory you can integrate in every measurable space, despite its cardinality. As an example you can consider $(\mathbb{N},\mathcal{P}(\mathbb{N}),\mu)$ where $ \mu$ is the only measure such that $\mu({n})=1, \forall n \in \mathbb{N}.$ In this case integration formally reduces to series in a very natural way. This means that integral is not the way to sum over cardinality $2^\omega$ but rather a way to generalize sums to every cardinal. If you are looking for a concrete measure or $\sigma$-algebra structure on the set $\mathbb{R}^\mathbb{R}$, well that's another question. – artful_dodger Jul 03 '17 at 05:36

1 Answers1

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Path integrals in quantum mechanics take averages of all the ways a particle could move from one point to another.

William Grannis
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