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I'm wondering if the integral of the product of two real-valued functions on a given space is also necessarily also an inner product for every function that can be defined on that space. E.g., are the Riemann integral, Lebesgue integral, Darboux integral, Riemann-Stieltjes integral, Daniell integral, Haar integral, Henstock-Kurzwell integral, Ito integral, and Young integral (see this article) all inner products on their respective spaces?

**I'm thinking about inner products as more general versions of integrals of products, but maybe I'm thinking about this the wrong way. Is it better to think about integrals of products as more general versions of inner products?

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Just check the axioms. Regardless of how you define definite integrals, the question is whether the following results hold for $\Bbb R\mapsto\Bbb R$ functions:$$\int fgdx=\int gfdx,\,\int(\alpha f+\beta g)hdx=\alpha\int fhdx+\beta\int gh dx,\,f\ne0\implies\int f^2 dx>0.$$The first is trivial; the second holds if the integrals involved are finite; the third is valid on appropriate spaces of functions. It doesn't really matter which kind of definite integral you have in mind.

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  • $\begingroup$ Thank you! And I guess that by the same token that given a space $P$ of real-valued functions defined on some set $S$, with inner product $\langle *, * \rangle: ~ P \times P \to \mathbb{R}$, one can define the integral of a function $f \in P$ as $\langle \mathbb{1}, f \rangle$, where $\mathbb{1}: ~ S \to \mathbb{R}$ is given by the rule $\mathbb{1}(s)=1$ for all $s \in S$? $\endgroup$
    – Wallace
    May 6, 2019 at 9:58
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    $\begingroup$ @Wallace If you make such a definition, then you would have to verify that it satisfies the properties of most types of integrals. If so, then why not? $\endgroup$
    – Allawonder
    May 6, 2019 at 11:08
  • $\begingroup$ @Allawonder, thanks for the response. Are there any properties of integrals you have in mind specifically? Any you think this probably wouldn't satisfy? $\endgroup$
    – Wallace
    May 6, 2019 at 14:20
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    $\begingroup$ @Wallace I can't speak for Allawonder, but results such as FTC, IBP etc. would be expected for suitable functions. $\endgroup$
    – J.G.
    May 6, 2019 at 14:31
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    $\begingroup$ @Wallace I don't think it wouldn't probably satisfy any particular property. What I meant is that however you choose to define the integral of a function, it should behave somehow like other integrals, otherwise what's the point calling it an integral? $\endgroup$
    – Allawonder
    May 6, 2019 at 19:39

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