The Extended Integral

  • Thread starter Buri
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Question 1: Can someone please explain to me on page 2 how is it that the inequality half-way through the page follows from the comparison property?

I don't see how. To me it just seems like an obvious fact since for each i, m(f)v(R_i) is less than or equal to the integral of f over R_i just by definition of the ordinary integral. I don't see how the comparison property comes into play here.

Question 2: The text says on page 3, "This corollary - corollary 15.5 - tells us that any theorem we prove about extended integrals has implications for ordinary integrals. The change of variables theorem, which we prove in the next chapter, is an important example."

I don't see how. First of all, the extended integral is only defined for open sets. In this corollary the set S is completely arbitrary so it doesn't necessarily have to be open. Second, the function is bounded which in the case of an extended integral doesn't necessarily have to be so - it could be unbounded in which case the ordinary integral doesn't make sense. So how could this corollary show that ANY theorem we prove about extended integrals have implications for ordinary integrals?

Any help would be appreciated. Thanks!
 

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  • #2
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Anyone?
 
  • #3
chiro
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Hey Buri.

I read your attachments but I couldn't see a precise definition for the extended integral. Have you got a definition that is specific with regards to the appropriate set as a comparison to an ordinary integral (ie extended integral in terms of Integral (over D) f).

Any topological information (with regards to the sets) would be nice as well.
 
  • #4
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Here's the definition that Munkres gives:

Let A be an open set in R^n; let f : A -> R be a continuous function. If f is non-negative on A, we define the (extended) integral of f over A, to be the supremum of the numbers {integral of f over D, as D ranges over all compact rectifiable subsets of A}, provided this supremum exists. In this case, we say that f is integrable over A (in the extended sense). More generally, if f is an arbitrary continuous function on A, set f+ (x) = max {f(x),0} and f- (x) = max{-f(x),0}. we say that f is integral over A (in the extended sense) if both f+ and f- are.
 
  • #5
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I also have this theorem as an alternate formulation of the definition:

Let A be open in R^n; let f : A -> R be continuous. Chose a sequence C_n of compact rectifiable subset of A whose union is A such that C_n subset Int(C_n+1) for each N. Then f is integrable over A if and only if the sequence of integral of |f| over C_n is bounded. In this case, integral of f over A is equal to the limit (as n goes to infinity) of integral of f over C_n.
 
  • #6
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I'll post the pages of the chapter tomorrow when I'm at school (my internet connection at home is way to slow to do it..)
 
  • #7
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Here are the pages I said I would post. Sorry I took longer.
 

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  • #8
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And the following 3 pages that lead up to the 3 pages I posted at the start of the thread..
 

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  • #9
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So I'd appreciate any help I could get. Thanks!
 
  • #10
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C'mon guys, can anyone help?
 
  • #11
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Question 1: Can someone please explain to me on page 2 how is it that the inequality half-way through the page follows from the comparison property?

I don't see how. To me it just seems like an obvious fact since for each i, m(f)v(R_i) is less than or equal to the integral of f over R_i just by definition of the ordinary integral. I don't see how the comparison property comes into play here.
Yes, this inequality is a simple consequence of the definition. But I think Munkres had the following in his mind:

[tex]m_{R_i}(f)v(R_i)=m_{R_i}(f)\int_{R_i}{1}=\int_{R_i}{m_{R_i}(f)}\leq \int_{R_i}{f}[/tex]

But your method is as good (not to say that it's even better!)

Question 2: The text says on page 3, "This corollary - corollary 15.5 - tells us that any theorem we prove about extended integrals has implications for ordinary integrals. The change of variables theorem, which we prove in the next chapter, is an important example."

I don't see how. First of all, the extended integral is only defined for open sets. In this corollary the set S is completely arbitrary so it doesn't necessarily have to be open.
Well, the corollary does not tell us that every integral is an extended integral, that would not be true! However, the corollary tells us that every regular integral can be "transformed" to an extended integral. So every regular integral can be calculated by just considering the associated extended integral.

Let's give an example: say that you have proven that for extended integrals the following hold
[tex]\int_G{af}=a\int_G{f}[/tex]
where a is just a real number. Of course, we know that this also holds for regular integrals, but let us assume that we don't know that. Then the property of extended integrals immediately implies the property for regular integrals as follows:
[tex]\int_S{af}=~(extended)~\int_{int S}{af}=~(extended)~a\int_{int S}(f)=a\int_S(f)[/tex]
This shows how (almost) every property of extended integrals implies a property of regular integrals. But we did NOT say that extended integrals are regular integrals!!

Second, the function is bounded which in the case of an extended integral doesn't necessarily have to be so - it could be unbounded in which case the ordinary integral doesn't make sense. So how could this corollary show that ANY theorem we prove about extended integrals have implications for ordinary integrals?
Well, I think it's clear why we ask the function to be bounded: the regular integral is only defined for bounded functions. In fact, the whole idea of extended integrals is to make it suitable for unbounded functions!!

But you are probably correct that not EVERY theorem of extended integrals have implications for ordinary integrals (I can find very pathological examples). But a lot of theorems do!! So if I were Munkres I would of said: "This corollary tells us that a lot of theorems that we prove for extended integrals, have implications for ordinary integrals." One example, will of course be the change of variables theorem...
 
  • #12
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Finally this all makes sense to me :) Thanks a lot micromass. You can't imagine how much I appreciate your help. Thanks again! :)
 

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