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Fubini's theorem "my simpler proof"

  1. Nov 1, 2014 #1
    This summer I was studying multivariable calculus and it did bother me how multiple integrals were for some reason reducible to iterated integrals. I thought of a proof, which in my opinion is fairly simple. I think it contains the same information that fubini's theorem states (at least to my knowledge). I wanted someone to tell me if all the steps are valid.
    Let ##R=[a,b]\times[c,d]##

    For simplicity let ##lim_{i,j} \to \infty \sum_{i}^{} \sum_{j}^{} f(x,y) \Delta x \Delta y = \sum_{i}^{} \sum_{j}^{} f(x,y) \Delta x \Delta y##

    Fubini's theorem says:

    ##\int\int_R f(x,y)dA = \int_{c}^{d} \int_{a}^{b} f(x,y) dx dy##

    This can be shown by the associative property of addition in the following manner:

    ##\sum_{i}^{} \sum_{j}^{} f(x,y) \Delta x \Delta y = \sum_{j}^{} \Big(\sum_{i}^{} f(x,y) \Delta x \Big) \Delta y ##


    By Riemann definition of integration

    ##\Big(\sum_{i}^{} f(x,y) \Delta x \Big)=\int_{a}^{b} f(x,y)dx=A(y) ##

    Then

    ##\sum_{j}^{} \Big(\sum_{i}^{} f(x,y) \Delta x \Big) \Delta y =\sum_{j}^{} A(y) \Delta y = \int_{c}^{d} A(y) dy##

    Therefore

    ##\sum_{j}^{} \sum_{i}^{} f(x,y) \Delta x \Delta y = \int_{c}^{d} \Big(\int_{a}^{b} f(x,y)dx \Big) dy##

    Which by definition of iterated integrals ##= \int_{c}^{d} \int_{a}^{b} f(x,y)dx dy##

    This procedure can also be used to show that the order of integration can be flipped around.

    I wanted to know if this is valid, and if it in any way simplifies stuff?

    I haven't seen and don't have the skills to understand Fubini's actual proof. But I just read in Stewarts calculus book that the proof is "too complicated". Is my reasoning correct? Can I follow this line of thought in the meanwhile?

    Thanks.
     
    Last edited: Nov 1, 2014
  2. jcsd
  3. Nov 2, 2014 #2

    Stephen Tashi

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    That isn't the definition of Riemann integration. The definition involves taking a limit.
     
  4. Nov 2, 2014 #3
    The primary problems with your proof are that (1) the 2-D version of the Riemann integral is not defined by an iterated sum and (2) the interchange of the limits involved - a process that you have hidden behind convenient notation - cannot be so easily dismissed.
     
  5. Nov 2, 2014 #4
    Stephen: I mentioned at the beginning of my proof that I was going to omit the limit symbol for simplicity (as a typer).

    gopher_p: what do you mean by an interchange of the limits involved? The only fiddling around I've done with limits can be considered valid rules of limits. Also, I'm pretty sure the 2-D version of the Riemann integral is an iterated sum.


    edit: I think I see what you mean by the interchange of limits. I still think it's valid though, hehe.
    Please correct me if I'm wrong. Thanks.
     
    Last edited: Nov 2, 2014
  6. Nov 2, 2014 #5
    The conclusion of Fubini's Theorem is true, and you can interchange the limits. You cannot, in the process of proving Fubini's theorem, hand-wave that away. It's not necessarily that you're saying something that is incorrect, it's that you're not saying all of the important things that PROVE that what you're saying is correct.

    To quote Seinfeld, "You yada-yada'd over the best part."
     
  7. Nov 2, 2014 #6
    gopher:

    Lol. I see what you're saying now. Just out of curiosity, I'd like to know a little bit about what those "important things" are. Is fubini's theorem really a hard one to prove as my textbook says?

    Thanks.
     
  8. Nov 2, 2014 #7
    I wouldn't say that Fubini's theorem is "hard" to prove. It's just that (1) the statement of the "standard" version of the theorem and its proof are typically given in the context of the Lebesgue integral and measure theory, which is well beyond the scope of most introductory calculus texts and (2) any definition of the Riemann integral which would lend itself to an "easy" proof of a version of Fubini for Riemann integrable functions uses ideas from elementary analysis - infimums, supremums, partitions, etc. - that are also slightly more advanced (not by much, though) than what is typically covered in introductory calc.

    To put in perspective, the proof of the sum rule for single-variable integrals is pretty easy given that you know a few elementary results regarding finite sums and limits. It's harder, though, if you don't have those results. Note that one of the key components there - the sum rule for limits - is easy and intuitive, but its proof is not trivial. And the sum rule for limits would be very hard to prove rigorously without the right definition of the limit.

    At the end of the day, most of the important results of (standard) analysis are about limits. When you omit or suppress the limit-related details, you really aren't doing analysis anymore.

    Now if you're just looking for an explanation for why Fubini's theorem should work - i.e. one that a physicist or engineer would be satisfied with (and reasonably so) - then I think you're on the right track with the explanation that you've given. You'd still want to explain why you can turn the general Riemann sum for your integral into an iterated sum, but I don't think that'd be too hard. Your question, though, was whether or not your explanation constituted the key elements of a proof. And what I am saying is that any proof of the theorem will be focused significantly on the limits involved. After all, the theorem really is all about how doing the "all at once" limit of the integral is equivalent to doing the "one-at-a-time" limits of the iterated integrals and how the order of the ""one-at-a-time" limits in the latter integrals does not matter.
     
  9. Nov 2, 2014 #8
    Gopher:

    Thanks a lot. You're answer has been really helpful. I was wondering though, my textbook mentions that Cauchy had provided a proof for continuous functions while Fubini provided the general proof. Do you know anything about this? I was searching for Cauchy's proof and found nothing.
     
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