# Iterated integrals as pre-sheaves

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 $\mathbb{R}^d$?

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.

• 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? – md2perpe Oct 1 '18 at 15:41

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 $$E\times X$$.

• 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? – md2perpe Oct 6 '18 at 8:22
• That depends on the measure you put on N and the specific subspace. The measure might already be concentrated on a proper subspace. – Michael Greinecker Oct 6 '18 at 8:31
• Can you reproduce ordinary iterated integration using this? – md2perpe Oct 6 '18 at 8:51
• Yes, that is possible. – Michael Greinecker Oct 6 '18 at 8:56
• 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 $f \mapsto \int \left( \int f(x_1, x_2, x_3) \, dx_3 \right) \, dx_2$? – md2perpe Oct 6 '18 at 9:22