Is a measure space necessary for integration?

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Integrals are traditionally linked to measure spaces, such as the Lebesgue and Riemann integrals. The discussion explores the possibility of defining integration without a full measure space, questioning the necessity of measures for achieving desirable integration properties. It suggests that an alternative approach could involve assigning values to subsets based on specific properties, though these may not form a complete measure. A proposed framework includes defining an 'integral' using finite partitions of a set and calculating upper and lower bounds based on value assignments. Ultimately, the conversation highlights the complexity of integrating without a measure while considering the foundational properties required for meaningful integration.
NateTG
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Integrals are typically associated with measure spaces. For example, the Lebesgue measure for the Lebesgue integral and the Jordan measure for the Rieman integral. But it seems like it should be possible to define an analogue of integration on something weaker than a measure space. So, what is the motiviation for having integration on a measure, rather than some other method for assigning values to subsets?
 
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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.
 
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&#039;s say that \rm{sup}\{p \in \mathbb{P}, \rm{bottom}(P)\} is the &#039;integral&#039; of f on X and otherwise that f is not &#039;integrable&#039; on X.
 
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There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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