Iterated integrals as pre-sheaves

Question

Let $n \in \{ 1, 2, 3, \ldots \}$ be fixed and set $N = \{ 1, \ldots, n \}$. Let $X_1, \ldots, X_n$ be measure spaces and for $I = \{ i_1, \ldots, i_m \} \subseteq N$ set $X^I = X_{i_1} \times \cdots \times X_{i_m}$ and let $\int f \, dx^I$ denote the integral $\int \cdots \int f(x_1, \ldots, x_n) \, dx_{i_1} \, \cdots \, dx_{i_m}.$

Then, if $J \subseteq I$ we have a linear map $\operatorname{res}_{I \to J} : L^1(X^I) \to L^1(X^J)$ defined by $f \mapsto \int f \, dx^{I \setminus J}.$ This map works as restriction morphisms for a presheaf.

Question: Can this presheaf be generalized in some natural way to cases where $N$ is not a discrete space but continuous like $\mathbb{R}^d$?

Answer 1

I'm not sure why one would want to mess with $N$. It seems to me that there is a natural sheaf of $L^p$ functions on a (Borel) measure space $X$, in the way that you have defined. Indeed, this is a pre-sheaf and in fact it is flabby. In the special case that $X$ is a product of measure spaces, then there is a natural restriction map to the inclusion of any of its summands.

Answer 2

I don't really know anything about presheaves, but the following might be what you are looking for:

Since that is a case in which thing work particularly smoothly, I will do things in terms of probability measures. Let $(N,\mathcal{N},\nu)$ be a probability space and $(X,\mathcal{X})$ a measurable space. Let $\kappa:N\times\mathcal{X}\to[0,1]$ be a transition probability; that is, $\kappa(n,\cdot)$ is a probability measure for fixed $n$ and $\kappa(\cdot,E)$ is measurable for fixed $E$. Now there is an induced measure $\tau$ on $\mathcal{N}\otimes\mathcal{X}$ given by $$\tau(A)=\int\int 1_A(n,x)~\mathrm d\kappa(n,\cdot)~\mathrm d\nu.$$ Now you can just take the $L_1(\tau)$. For $E\in\mathcal{N}$, you can let $L^1(E)$ be the space of elements of $L_1(\tau)$ that vanish outside $E\times X$. If $E\subseteq F$, you get a restriction map $r_{EF}:L_1(F)\to L_1(E)$ that takes an element of $L_1(F)$ and changes its value to zero outside $E\times X$.