matt grime said:
And what properties would this 'assignment of values' have? I suspect it might necessarily be a measure for it to have the nice properties of integration that you want.
I was originally thinking of something like:
If A and B are valued sets then
If A \subset B then A's complement in B is also valued.
If A \cap B = \null then V(A)+V(B)=V(A \union B).
And, V(A)\geq 0.
This is not necessarily a measure since it is not necessarily an algebra.
But it's not that difficult to hack up an abstract notion of 'integral' that doesn't even require that:
Let's say I have some set X, with M \subset P(X) and some
function V:M \rightarrow R, where R is a linearly ordered complete ring.
Then let \mathbb{P} be the set of all finite partitions of X that are subsets of M and that do not contain the empty set.
Now, let P \in \mathbb{P} be some partition of X that is a subset of M. Then (for lack of a better term) let the top of P[/tex] be <br />
\rm{top}(P)=\sum_{p \in P} V( p) \times \rm{max}(f(p))<br />
where \rm{max}(f(p)) is the supremum of the image of p in f.<br />
And, let the bottom of P be<br />
\rm{bottom}(P)=\sum_{p \in P} V(p) \times \rm{min}(f(p))<br />
where \rm{min}(f(p)) is the infimum of the image of p in f.<br />
Then if \rm{sup}\{p \in \mathbb{P}, \rm{bottom}(P)\}=\rm{inf}\{p \in \mathbb{P}, \rm{top}(P)\}, let's say that \rm{sup}\{p \in \mathbb{P}, \rm{bottom}(P)\} is the 'integral' of f on X and otherwise that f is not 'integrable' on X.
