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In his notes:, Pete Clark outlines an axiomatic approach to the Riemann Integral. He doesn't use the language of sheafs, but it seems implicit in his definition before Theorem 1. He goes on to show that the fundamental theorem of calculus follows, and then that such an integral exists.

I'm a little shaky on sheaf-theoretic language (I haven't been taught this material formally), so I just want to run the following reformulation past any experts out there to make sure its correct.

Given a sheaf $F$ on $\mathbb{R}$ containing the sheaf of bounded real-valued continuous functions on X, can the Riemannian integral be defined axiomatically as:

$I: F \rightarrow \mathbb{R}$ such that

i. $f,g \in F$ and $f\le g \implies If\le Ig$

ii. $f,g \in F$ and $supp(f) \cap supp(g)$ is a point or empty, then $I(f+g)=If+Ig$

iii. $I(C)=c(b-a)$ where $C$ is the constant function on the interval $(a,b)$ with value $c$.

This then begs the question as to whether the Lesbegue integral can be also defined in a similar fashion, say by replacing $\mathbb{R}$ by some measurable space, the sheaf of continuous real-valued functions by the sheaf of simple functions, and iii. by the usual lesbegue integral of an indicator function where now $I$ is positive homegenous.

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Why is this tagged category-theory when there's a perfectly good sheaf-theory tag? – Zhen Lin Mar 26 '12 at 2:31
@zhen:good question. I've changed it. – Mozibur Ullah Mar 26 '12 at 2:36
A belated response: though I am conversant with sheaves and favor a relatively sheafy approach to various types of geometry (in particular, I start to long for sheaves whenever I see any talk of "atlases"), I really did not have sheaves in mind here. As Zhen Lin's answer shows, integrals are in fact not very sheafy. – Pete L. Clark Jul 16 '13 at 8:31

The blunt answer to the question in your title is no: the integrable functions on the real line do not form a sheaf. For example, consider the constant function $1$ on the real line $\mathbb{R}$. Under any reasonable definition, $1$ is not integrable, since $\int_\mathbb{R} 1$ is not finite. But there is an open cover $\{ U_\alpha \}$ of $\mathbb{R}$ such that the restriction of $1$ to each $U_\alpha$ is integrable over $U_\alpha$.

Of course, it is easy enough to fix the problem: we either demand that the base space be compact, or otherwise weaken the ‘integrable’ requirement to ‘locally integrable’.

The question in the body of your post is rather different. First of all, what is a map from a sheaf to a set? Such a thing makes no sense as-is. You could talk about a sheaf morphism $I : \mathscr{F} \to \underline{\mathbb{R}}$, where $\underline{\mathbb{R}}$ is the constant $\mathbb{R}$ sheaf, but this is no good: if $U$ and $V$ are disjoint open sets, then $\underline{\mathbb{R}} (U \cup V) \cong \mathbb{R} \times \mathbb{R}$, so that means $I$ takes "functions" to locally constant functions, which is certainly not what an integral does.

If you must insist on using sheaf language, then what you are looking for must be a presheaf morphism $I : \mathscr{F} \to \mathbb{R}$. But even this is not useful: by the definition of presheaf morphism, $I$ must commute with the restriction maps; but that means that if $f \in \mathscr{F} (U)$ and $V \subseteq U$, then $I(f) = I(f |_V)$, which is surely not what you want for an integral!

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I've amended the title in light of your first comment, but I don't see why a map from a sheaf to a set is so problematic, can you please explain. – Mozibur Ullah Mar 26 '12 at 3:19
@Mozibur: If you're not going to make it into a sheaf or presheaf morphism, then why bother with the language of sheaves? – Zhen Lin Mar 26 '12 at 3:29

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