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The proof of Darboux's thm

  1. Aug 4, 2007 #1

    quasar987

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    1. The problem statement, all variables and given/known data
    I'm frying my brain on the proof of Darboux's thm. It says that if a subset A of R^n is bounded and f:A-->R is bounded as well, then f is integrable on A with integral I <==> for any e>0, the exists a d>0 such that for a partition P={S_1,...,S_K} of an n-rectangle S containing A, |P|<d ==>

    [tex]\left|\sum_{i=1}^{K}f(x_i)v(S_i)-I\right|<\epsilon[/tex]

    where x_i belongs to S_i, v(S_i) is the volume of S_i, and |P|<d means that each subrectangle in P has sides lesser than d.

    The "<==" part is fairly easy.

    For the "==>" part, the author begins by proving the following lemma:

    Lemma:Let P be a partition of an n-rectangle B in R^n. Then, given e>0, there is a d>0 such that if P' is another partition of B with |P'|<d, then the sum of the volumes of the subrectangles not entirely contained in an element of P (i.e. overlapping two or more of them) is lesser than e.

    To prove the "==>" part of Darboux, the authors begins by writing |f(x)|<M, since f is bounded. Then, observe that since I equals the upper an lower integrals, we have that given e>0, there are partitions P1 and P2 such that L(f,P1) > I - e/2 and U(f,P2) < I + e/2. For P the natural refinement of P1 and P2, we have both inequalities: L(f,P) > I - e/2 and U(f,P) < I + e/2.

    Now, according to the above lemma, there exists a d>0 such that if P'={S'_1,...,S'_K} is a partition of S with |P'|<d and with {S'_1,...,S'_N} the elements of P' not entirely contained in an element of P, then

    [tex]\sum_{i=1}^{N}v(S'_i)<\epsilon/2M[/tex]

    If x_i belongs to S'_i, we have that

    [tex]\sum_{i=1}^{K}f(x_i)v(S'_i)=\sum_{i=1}^{N}f(x_i)v(S'_i)+\sum_{i=N+1}^{K}f(x_i)v(S'_i)<M(\epsilon/2M)+U(f,P)=\epsilon/2+U(f,P)<I+\epsilon[/tex]

    So far so good. But the authors then complete simply by saying, "similarily,

    [tex]\sum_{i=1}^{K}f(x_i)v(S'_i)>L(f,P)-\epsilon/2>I-\epsilon[/tex]"

    How is this inequality achieved??? After separating the series into two parts like above, we can do the same trick with the first term, but not with the second. Namely, we have

    [tex]\sum_{i=1}^{N}f(x_i)v(S'_i)>-M\sum_{i=1}^{N}v(S'_i)[/tex]

    and

    [tex]\sum_{i=1}^{N}v(S'_i)<\epsilon/2M \Rightarrow -M\sum_{i=1}^{N}v(S'_i)>-\epsilon/2[/tex]

    hence

    [tex]\sum_{i=1}^{N}f(x_i)v(S'_i)>-\epsilon/2[/tex]

    But it is not clear to me if and why it would be the case that

    [tex]\sum_{i=N+1}^{K}f(x_i)v(S'_i)>L(f,P)[/tex]
     
    Last edited: Aug 4, 2007
  2. jcsd
  3. Aug 6, 2007 #2

    quasar987

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    Are the 40 readers of my post as puzzled as I am by this, or simply were discouraged of reading by the length of the post? :smile:
     
  4. Aug 6, 2007 #3

    morphism

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    Personally I'm puzzled! :redface:
     
  5. Aug 22, 2007 #4

    quasar987

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    Last bump. I tried working on this again today and still can't make sense of

    [tex]\sum_{i=1}^{K}f(x_i)v(S_i')>L(f,P)-\epsilon/2[/tex]
     
  6. Aug 25, 2007 #5

    morphism

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    Have you tried checking the textbook's errata or contacting the author(s), or your professor? It could be an oversight.
     
  7. Aug 25, 2007 #6

    quasar987

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    I had thought of none but will do all. Thx for the advice!
     
  8. Sep 13, 2007 #7

    quasar987

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    Marsden has not replied yet, but one of my professor who is an analyst thinks it is an oversight. However, he cannot point me to another text where this thm is proven!

    Does anybody know one??
     
  9. Sep 14, 2007 #8

    morphism

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    Try Wade, An Introduction to Analysis, third edition, pp. 396. It's not exactly the same theorem, but it's similar.

    Otherwise try looking in Spivak.
     
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