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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 $\mathbb{R}^d$?

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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.

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  • $\begingroup$ Could you elaborate using an example where $N=\mathbb{R}$ and $X_t=\mathbb{R}$ with Lebesgue measure for $t \in N$? What would $\int f \, dx^{[a,b]}$ be defined? $\endgroup$ – md2perpe Oct 1 '18 at 15:41
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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 $E\times X$.

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  • $\begingroup$ What happens if $F$ is a finite-dimensional space and $E$ is a proper subspace, it has fewer dimensions? Will $r_{EF}$ be the zero map? $\endgroup$ – md2perpe Oct 6 '18 at 8:22
  • $\begingroup$ That depends on the measure you put on N and the specific subspace. The measure might already be concentrated on a proper subspace. $\endgroup$ – Michael Greinecker Oct 6 '18 at 8:31
  • $\begingroup$ Can you reproduce ordinary iterated integration using this? $\endgroup$ – md2perpe Oct 6 '18 at 8:51
  • $\begingroup$ Yes, that is possible. $\endgroup$ – Michael Greinecker Oct 6 '18 at 8:56
  • $\begingroup$ Could you please show how to chose $(N, \mathcal{N}, \nu)$ and $(X, \mathcal{X})$ to reproduce iterated integration from $L^1(\mathbb{R}^3)$ to $L^1(\mathbb{R}^2)$ and then to $L^1(\mathbb{R})$ by integrals like $f \mapsto \int f(x_1, x_2, x_3) \, dx_3$ and $f \mapsto \int \left( \int f(x_1, x_2, x_3) \, dx_3 \right) \, dx_2$? $\endgroup$ – md2perpe Oct 6 '18 at 9:22

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