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Is an Integral a measure

  1. Mar 18, 2010 #1
    I just beginning to study measure theory. so far from what i understand so far , can we say in general, an integral is a measure, (ie it is nothing but a set function. a mapping [tex] F : \mathcal{F} \rightarrow \mathbb{R} [/tex] where [tex] \mathcal{F} [/tex] is a family of sets.

    does it make sense to say in general an integral of a function F, is [tex] \int_{A} F d\mu [/tex] is the measure of the image of the [tex] F [/tex] over some set [tex] A [/tex] using the measure [tex] \mu [/tex]. with the condition the image of [tex] F [/tex] over [tex] A [/tex] must be measurable using the measure [tex] \mu [/tex]

    so for example the two that i know are lebesgue-integral, lebesgue-stieltjes integral, are basically are general integrals using different measures.

    if we could say the above then where would the riemann integral fit in to this.
    sorry if this is a bit vague, i'm trying to get my head around this stuff
     
    Last edited: Mar 18, 2010
  2. jcsd
  3. Mar 18, 2010 #2
    No, I would say your description is wrong. Why not wait a couple more weeks in the course, then try to fomulate it again?
     
  4. Mar 18, 2010 #3
    I'm not actually doing a course, i work full-time, its something i'm trying to learn through self-study. i go back re-learn what i thought i knew and re-formulate it in the next week or so
    thanks
     
  5. Apr 25, 2010 #4
    okay here is my second attempt:
    A measure is a set function [tex] \mathcal{F} \to \Re [/tex]. where [tex] \mathcal{F} [/tex] is a sigma-algebra. the invariants it must satisfy are it countable additive and the measure of a null-set is zero.
    Now integral [tex] \int_{B} f d\mu [/tex] is nothing but a special case of a measure, where it calculates the measure of the set described by [tex]f [/tex] over the set [tex] B [/tex]. The arbitrary measure [tex] \mu [/tex] is use to calculate the measure of this set.
    Since we viewing integral is as measure it must satisfy the invariants of a measure such as being countable additive. The Riemann-integral only satisfies this condition only a small class of functions, this why the Lebesgue integral is introduced, to over come some of these shor-coming
     
  6. Apr 26, 2010 #5
    I'd say you're on the right track.

    Even though the integral of a real-valued (measurable) function over a subset D of the domain on which the function is defined is not in general a measure, it is certainy possible (and easy) to define functions whose integral will be a measure.

    E.g. a strictly positive real-valued function will have an integral that is a measure, but the intgral of a sine-wave will not fulfill the axioms of a measure.

    You can, in fact, view any real-valued positive function as a quotient between two different measures on the same sigma-algebra: see the "Radon–Nikodym theorem", it was of great help for me understanding measures in relation to measurable functions.
     
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