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Is a measure space necessary for integration?

  1. Dec 15, 2005 #1

    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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  3. Dec 15, 2005 #2

    matt grime

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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.
     
  4. Dec 15, 2005 #3

    NateTG

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    I was originally thinking of something like:
    If [itex]A[/itex] and [itex]B[/itex] are valued sets then
    If [itex]A \subset B[/itex] then [itex]A[/itex]'s complement in [itex]B[/itex] is also valued.
    If [itex]A \cap B = \null[/itex] then [itex]V(A)+V(B)=V(A \union B)[/itex].
    And, [itex]V(A)\geq 0[/itex].
    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 [itex]X[/itex], with [itex]M \subset P(X)[/itex] and some
    function [itex]V:M \rightarrow R[/itex], where [tex]R[/tex] is a linearly ordered complete ring.
    Then let [itex]\mathbb{P}[/itex] be the set of all finite partitions of [itex]X[/itex] that are subsets of [itex]M[/itex] and that do not contain the empty set.
    Now, let [itex]P \in \mathbb{P}[/itex] be some partition of [itex]X[/itex] that is a subset of [itex]M[/itex]. Then (for lack of a better term) let the top of [itex]P[/tex] be
    [tex]\rm{top}(P)=\sum_{p \in P} V( p) \times \rm{max}(f(p))[/tex]
    where [itex] \rm{max}(f(p))[/itex] is the supremum of the image of [itex]p[/itex] in [itex]f[/itex].
    And, let the bottom of [itex]P[/itex] be
    [tex]\rm{bottom}(P)=\sum_{p \in P} V(p) \times \rm{min}(f(p))[/tex]
    where [itex] \rm{min}(f(p))[/itex] is the infimum of the image of [itex]p[/itex] in [itex]f[/itex].
    Then if [tex]\rm{sup}\{p \in \mathbb{P}, \rm{bottom}(P)\}=\rm{inf}\{p \in \mathbb{P}, \rm{top}(P)\}[/tex], let's say that [itex]\rm{sup}\{p \in \mathbb{P}, \rm{bottom}(P)\}[/itex] is the 'integral' of [itex]f[/itex] on [itex]X[/itex] and otherwise that [itex]f[/itex] is not 'integrable' on [itex]X[/itex].
     
    Last edited: Dec 15, 2005
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