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Help with Apostol's "Calculus, vol. 1", Section 1.18

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  1. Jun 21, 2015 #1
    • Member warned about not using the homework template
    In section 1.18 ("The area of an ordinate set expressed as an integral"), Apostol proves two theorems. the first, theorem 1.10, deals with the area of a function's ordinate set; the second, theorem 1.11, deals with the area of the graph of the function of theorem 1.10. (I have attached two excrepts from Apostol's book, one per theorem.)

    I am having problems understansing Apostol's logic in theorem 1.11 where he states:

    "The argument used to prove Theorem 1.10 also shows that Q’ is measurable and that a(Q’) = a(Q)."

    I don see how he could argue this, being that ##Q'=\{(x,y)|a \le x \le b, 0 \le y < f(x) \}## which implies that ##S \subseteq Q'## is not true for all step regions S (since S may contain a point of the graph of ##f(x)## which Q', by definition, can't).

    Thanks in advanced for any help.
     

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  3. Jun 22, 2015 #2

    RUber

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    I think the method is the same, but the definition of S would have to change if a point in S was equal to f(x).
    So for the measure of Q', you would have sets S' and T', but the result would be the same--that the area of a line (or curve) is zero.
     
  4. Jun 22, 2015 #3

    Zondrina

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    You can make the points ##(x, y)## in the set ##Q'## sufficiently close to the graph of ##y = f(x)##. So the areas are indeed equal.
     
  5. Jun 24, 2015 #4
    RUber, I see what you're saying, however the new step regions (S' and T') would produce functions bearing the following relationship with the function##f(x)##: ##s'(x) < f(x) \le t'(x)##. Which doesn't help since the definition of the integral requires "##\le##" for both step function inequalities.

    Zondrina, The idea is making the of make the points equal since by definition of Q': ##Q \neq Q'##. So no matter how close I get too the points of ##f(x)##, I would not be able to put all points of both regions in a one-to-one correspondence in order to then argue by congruence of regions.
     
    Last edited: Jun 24, 2015
  6. Jun 24, 2015 #5

    RUber

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    You can define a sequence s_n(x) such that the limit as n goes to infinity of s_n(x) = f(x). In this way, you will still satisfy the requirement that the intervals will converge on the true integral.
     
  7. Jun 24, 2015 #6
    I get what you're saying. The only problem is that Apostol still hasn't mentioned sequences. The only relevant thing I can think of that has been covered in the book till now is the Least Upper Bound Property of Numbers. Which would simply state that ##f(x)## is the supremum for ##Q'##
     
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