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The product integral is the multiplicative version of standard integrals. Indefinite products are the discrete counterpart to this integral; they multiply iterations on a function $f(x)$ by each other.

Is there a simple relation between the two? If so, what is this relation?

Motivation

This may give an alternative (although fairly roundabout) to calculating summations. It can be seen in the indefinite product link above that the indefinite product is related to summations by a fairly simple relation. If there exists another fairly simple relation between product integrals and indefinite products, this would provide an alternative to summations, via product integrals.

Secondly, I'm interested in exploring indefinite products, and this may provide an easy alternative.

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    It's been awhile, but I was interested in that, as well as "Type I", found in Wikipedia here: https://en.wikipedia.org/wiki/Product_integral#Type_I. I'll try to give this some more thought. I did find some ways of relating products found in the book "Non-Newtonian Calculus" by Grossman & Katz.2018-03-19

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This is a more concise version of an answer I posted to this related question. In any case, I claim that the "Lebesgue theory of product integrals" addresses what you are interested in, namely a connection to indefinite products, for at least two reasons:

  1. In the case that the integrand is a (suitable) simple function, the product integral reduces to an indefinite product. The rest of the theory then generalizes from this case.
  2. In the special case when the underlying measure is counting measure, the product integral reduces to an indefinite product.

Admittedly the usefulness of this answer will probably depend on the extent to which you are already familiar with the theory of Lebesgue integration. If you are not familiar at all, in what follows, think of the ground space $X$ as $\mathbb{R}$, $\mathscr{F}$ as the collection of all subsets of $\mathbb{R}$ which can be formed by taking unions and intersections of sequences of intervals, and $\mu$ a function which assigns to each interval the value corresponding to the interval's length. Also think of a simple function as a step function or a possibly fancier version of a step function.

For a simple function with non-negative coefficients $a_k$ for the indicator functions of the measurable sets $A_k$, one can (and should) define the product integral as:

$ \prod f^{d\mu} := \prod_{k=1}^n a_k^{\mu(A_k)} \,, \quad \text{where }f = \sum_{k=1}^n a_k \mathbf{1}_{A_k}(x) \,. $

(Choosing the convention $0^0 = 1$ is helpful, if not necessary, here.) Taking logarithms, we see that: $\log \left( \prod f^{d\mu} \right) = \sum_{k=1}^n \log(a_k) \mu(A_k) = \int \log f d\mu \implies \prod f^{d\mu} = \exp\left( \int \log f d\mu \right) \,.$

The relevant expression equals $\int \log f d\mu$ because $f$ being a simple function implies that $\log f$ is also a simple function, and the Lebesgue integral of a simple function is just a sum.

Since $\exp$ and $\log$ are continuous functions and strictly increasing functions, continuous functions commute with limits of sequences, and essentially all of the basic results involving the Lebesgue integral revolve around limits of increasing sequences, more or less anything we might ever want to show regarding the basic properties of product integrals follows nearly directly from the theory of Lebesgue integrals. For example, the product integral of any measurable function, when it exists (and is non-zero), can be approximated arbitrarily closely by the product integrals of approximating simple functions (and all of those are just finite products).

When $\mu$ is counting measure or any measure absolutely continuous with respect to counting measure, then any product integral is just a (possibly infinite) product, the same way that any regular integral is just a (possibly infinite) sum. In particular the relationship of an indefinite product to an indefinite sum is just a special case of the general relationship of a product integral to a regular integral, $\prod f^{d\mu} = \exp(\int \log f d\mu)$.