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Integrator as an average?

  1. Mar 6, 2014 #1
    Hello all,

    I'm going through Foundations of Mathematical Analysis by Johnsonbaugh and Pfaffenberger, and I read a curious line that I was hoping someone here could clear me up on (perhaps I'm thinking too much into it).

    The begininning of the section on the Lebesgue integral introduces a sequence of Riemann integrable functions which converge (somehow) to a function, ƒ,which is not Riemann integrable. The authors go on to write....

    "Why is ƒ not Riemann integrable? The fault lies not with the average used, namely α(x)=x, but rather with the sets averaged over, i.e., intervals."

    where α(x) is the integrator. Generally, when I think of α(x), I think of it as a weighting function. So, what do the authors here mean that α(x)=x is the "average used"?
     
  2. jcsd
  3. Mar 7, 2014 #2

    Stephen Tashi

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    Science Advisor

    I don't have copy of that book. You should state the example.

    One guess is that the quotation is an informal reference to the fact a net quantity (like total dollars) can be computed from the total number of things involved ( like total people) times the average of that quantity per thing (like "mean cost per person"). An average can be viewed as a constant "weight" that makes a product yield the desired result.
     
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